what is graphical representation of motion

Motion Graphs: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: March 31, 2022

When trying to explain how things move, physicists don’t just use equations – they also use graphs! Motion graphs allow scientists to learn a lot about an object’s motion with just a quick glance. This article will cover the basics for interpreting motion graphs including different types of graphs, how to read them, and how they relate to each other. Interpreting motion graphs, such as position vs time graphs and velocity vs time graphs, requires knowledge of how to find slope. If you need a review or find yourself having trouble, this article should be able to help.

What We Review

Types of Motion Graphs

There are three types of motion graphs that you will come across in the average high school physics course – position vs time graphs, velocity vs time graphs, and acceleration vs time graphs. An example of each one can be seen below.

The position vs time graph (on the left) shows how far away something is relative to an observer.

The velocity vs time graph (in the middle) shows you how quickly something is moving, again relative to an observer.

Finally, the acceleration vs time graph (on the right) shows how quickly something is speeding up or slowing down, relative to an observer.

Because all of these are visual representations of a movement, it is important to know your frame of reference. We learned in our introduction to kinematics that two people can observe the same event but describe it differently depending upon where they stand. If this or anything about the position, velocity, and/or acceleration is still a bit confusing, revisit our kinematics post and our acceleration post before moving on.

Describing Motion with Position vs Time Graph s

We typically start with position-time graphs when learning how to interpret motion graphs – generally because they’re the easiest to try to picture. Let’s look at the position vs time graph from above. We see that our vertical axis is Position (in meters) and that our horizontal axis is Time (in seconds). This means we know how far away an object has moved from our observer at any given time. This particular graph shows an object moving steadily away from our observer.

Position vs Time Graph for Multi-Stage Motion

Let’s consider the graph and images below. We are still considering a position vs time graph, but this time we are looking at motion that changes. The car begins by moving 5 meters away from the observer in the first 5 seconds. After that, the car remains stopped 5 meters away from the observer for another 5 seconds. Finally, the car turns around and moves for 5 seconds back to its original position, 0 meters from the observer.

There are two key points that we can take from the example above. The first is that our position vs time graph shows how far away we are at any given time and nothing else. It cannot tell us distance or displacement – we would have to do a little mathematics to find those out. The second is that the change in position is not always positive. Here, we’ve defined moving to the right as positive. So, in the beginning, when the car was moving to the right, its position increased. In the end, when it moved back to the left, it was moving in the negative direction.

Position vs Time Graph for Passing an Observer

This implies that the position could, potentially, go below the x-axis. Let’s look at an example combining these two points in practice.

This time, our car started to the right, and drove straight past our observer to the left. At t=0\text{ s} , the car was 10\text{ m} to the right of our observer, so its position was x=10\text{ m} . As it passed the observer, its position was x=0\text{ m} at t=5\text{ s} . The car then ended its journey 10\text{ m} to our observers left at t=10\text{ s} so that its final position was x=-10\text{ m} .

Finding Distance and Displacement from a Position vs Time Graph

Example 1: constant position vs time graph.

We’ll continue working from the graph above as we have already pulled the important values from it. Because we have a simple, straight line we only need the values from the very beginning and very end of the car’s journey, which we already pulled out above:

• t_{1}=0\text{ s}
• x_{1}=10\text{ m}
• t_{2}=10\text{ s}
• x_{2}=-10\text{ m}

Finding Distance From A Position-Time Graph

As we learned from our introduction to kinematics lesson, we know that the equation for distance is:

The problem here is that we didn’t pull out any d values from our position vs time graph, only x values. We can still use those, though. In general, if you take the absolute value of an x value, it can be thought of as a d value and plugged into our distance equation. So the d values we’ll be using are:

• d_{1}=\lvert x_{1} \rvert =\lvert 10\text{ m} \rvert = 10\text{ m}
• d_{2}=\lvert x_{2} \rvert =\lvert -10\text{ m} \rvert = 10\text{ m}

Now we can plug these values into our equation and solve for our distance.

Finding Displacement From A Position vs Time Graph

Our equation for displacement is:

In this case, we will be using x_{2} as x_{f} and x_{1} as x_{i} as they represent the end and beginning of our movements, respectively. When we plug in our values, we find:

In this case, the negative sign makes sense as our line is moving down the graph and the car moved from right to left, which we had previously defined as positive to negative.

Example 2: Changing Position vs Time Graph

Now that we know the basics of finding distance and displacement from a position vs time graph, let’s get a bit more in-depth. We’ll return to the graph about the car that moved forward, stopped, and then turned around and returned to its original position. The graph has been copied below for convenience.

How to Find Distance From A Position vs Time Graph

Finding distance from these graphs can get a bit complex as you’ll need to find several different values. If you’ll notice, the slope of our graph changes regularly – the line seems to turn. Each segment with a unique slope requires our attention. So, we’ll need to look at t=0\text{ s} through t=5\text{ s} ,  t=5\text{ s} through  t=10\text{ s} , and  t=10\text{ s} through  t=15\text{ s} .

We’ll want to look at the position value on the left and right of each side of those segments and find the absolute value of the delta between those values. These will serve as the d values that we will plug into our distance equation.

• d_{1}=\lvert 5\text{ m}-0\text{ m} \rvert =5\text{ m}
• d_{2}=\lvert 5\text{ m}-5\text{ m} \rvert =0\text{ m}
• d_{3}=\lvert 0\text{ m}-5\text{ m} \rvert =5\text{ m}

We can now plug all of these values into our equation and solve for distance.

How to Find Displacement From A Position vs Time Graph

Finding displacement from a graph that changes how it’s moving is a bit easier than finding the distance. Because displacement only concerns the distance between the starting and ending positions of an object’s motion, we only need to find the position at the rightmost point on the graph ( t=15\text{ s} ) and the leftmost point on the graph ( t=0\text{ s} ). The positions at these times will serve as our x_{f} and x_{i} values respectively. 

• x_{f}=0\text{ m}
• x_{i}=0\text{ m}

Now that we have these values, we can plug them into our displacement formula and solve:

Finding Velocity from a Position vs Time Graph

Now that we know how to find distance and displacement from a position vs time graph, we can start finding another value – velocity. If you think about it, these distances and displacements that we’re finding are occurring over some amount of time (as given by the graph) and all we really need to find velocity is displacement and time. So let’s start with a simple graph – the one of an object moving steadily away.

The displacement for the movement depicted by this graph would be \Delta x=25\text{ m}-0\text{ m}=25\text{ m} and because our time here moves from t=0\text{ s} to t=5\text{ s} , we have a change in time of \Delta t=5\text{ s} . This is enough information for us to solve for the velocity using the equation we learned before:

One very important thing you may notice if you’re savvy with slopes is that the slope of this graph is also equal to 5 . (If you are not particularly savvy with slopes, I would recommend reviewing how to solve for slope as we’ll be relying on that knowledge for most of what remains of this post.) This similarity is no mere coincidence. The velocity of any movement will always be equal to the slope of the position-time graph at that time.

Proving Velocity is the Slope of Position vs Time Graphs

The slope of any given straight line can be found with the equation

Here, m is the slope, y_{2} and y_{1} are two different position values, and x_{2} and x_{1} are the time values corresponding to the two position values.

Let’s begin by selecting two points off of our graph above (being sure to include the units when we do). Let’s take (2\text{ s},10\text{ m}) and (4\text{ s},20\text{ m}) and plug the values in:

This setup of subtracting the rightmost value from the leftmost value should look a bit familiar. Another way to think of it is as taking a final value and subtracting an initial value, much like a delta. In fact, this is equivalent to a change in position over a change in time – the definition of velocity.

Let’s make sure our slope works out to be a velocity value before we jump to any conclusions, though. If the slope of this graph is also the velocity of the same motion, two things need to be true. First, we need a numerical value of 5 . Second, we need our units to be in m/s. Let’s solve the equation above and see what we get.

We have a value that matches the velocity we solved for in both numerical value and physical units. We could have selected any two points on our graph and received the same result. The part of this that truly proves the slope of any position-time graph is the velocity is that the units of our slope work out to be m/s – the units for velocity.

Example 1: Finding Velocity from a Position vs Time Graph

Let’s try finding our velocity again, but using the slope formula. We’ll reuse the graph below that we saw earlier in this article.

We’ll want to begin by selecting the points we want to use. This graph is a straight line which means its slope never changes so it won’t particularly matter what two points we choose. Since we have a point where our x value is zero and a separate point where our y value is zero, we may as well use those to make the mathematics easier. So, the values we’ll be using here are:

• y_{2}=0\text{ m}
• y_{1}=10\text{ m}
• x_{2}=5\text{ s}
• x_{1}=0\text{ s}

Now, all we need to do is set our velocity equal to our slope, plug in our values, and solve for our velocity:

Here, we get a negative velocity of v=-2\text{ m/s} . If we look at our graph, we see it has a negative slope, so we should have expected this negative velocity from the start. If you ever get a positive when you expected a negative or vice versa, check to make sure you plugged your values into your formula in the correct order. That simple mistake has thrown many scientists off course.

Example 2: Finding Velocity with Changing Motion

Being able to find the velocity of a simple, straight position vs time graph is all well and good, but there will be times when you’ll have to split a graph apart. Let’s revisit the graph below as an example of this.

We already said before that we could split this graph up into a few different chunks based on when the slope changes. We know what happens when we have a positive slope and what happens when we have a negative slope, though, so let’s look at just the middle section where it’s flat. Here, the values we can pull from the line segment are:

• y_{2}=5\text{ m}
• y_{1}=5\text{ m}
• x_{1}=10\text{ s}

If we plug these values into our slope formula, we can find that

Since the segment was a flat line with a slope of 0 , the velocity also had to be 0\text{ m/s} . If we recall, this graph depicted a car that moved in the positive direction, stopped and remained motionless, and then moved back in the negative direction. The middle segment of this graph, the one that we looked at, corresponds to when the car was stopped so again. Therefore, it makes sense that we would see a velocity of 0\text{ m/s} .

Describing Motion with Velocity vs Time Graph s

Velocity-time graphs are relatively similar to position-time graphs, and just as important in the study of motion graphs. We still have our time in seconds along the x-axis, but now we have our velocity in meters per second along the y-axis. Let’s consider the velocity-time graph below. 

To find the velocity of an object at any given time here, we simply need to read the value from the graph. There’s no mathematics to do or formulas to use. So, for example, at t=2\text{ s} the velocity is v=4\text{ m/s} because that is the value we read off of the graph. Similarly, the velocity is t=4\text{ s} is v=8\text{ m/s} . The fact that those two values differ and that the slope here is positive tells us that the motion in this graph is an object moving away from an observer and getting faster – like a car leaving a stoplight. We can also show more complicated motions and dip below the x-axis.

Velocity vs Time Graph with a Change in Direction

Let’s imagine a scenario for the graph to the right. We see that the graph starts with the object’s top velocity of v=10\text{ m/s} and then seems to get lower. The object reaches a velocity of v=0\text{ m/s} at t=2.5\text{ s} . While it may make sense to say that the object is now at the same point as the observer, we can’t actually infer that. All we can tell from here is that the object is momentarily at rest relative to the observer. The velocity then continues decreasing to v=-10\text{ m/s} , implying that the object is now moving in the negative direction. A real-life scenario for this may be that you observe someone pulling into a long driveway, stopping briefly at the end, and then backing down it.

Velocity vs Time Graph for Multi-Stage Motion

Now, let’s return to our car from before that moved in the positive direction, stopped, and then came back. Since the slope in each segment of the position graph was constant, we assumed that the car’s movements had a constant velocity and that it had zero velocity when it stopped. The velocity-time graph for this motion would look a bit like this:

You can see that the velocity remains a constant v=1\text{ m/s} while the car moves to the right, changes to v=0\text{ m/s} while the car stops, and then becomes v=-1\text{ m/s} while the car moves back to the left. 

Finding Displacement from a Velocity vs Time Graph

Much like how we could find a velocity from a position-time graph, we can find displacement from a velocity-time graph. This process will be a bit different. Instead of finding the slope of the velocity graph, we will be finding the area under the velocity graph. This may sound counterintuitive, but we can prove that it works by checking our units. Let’s say that an object moves at 5\text{ m/s} for 10\text{ s} . The velocity-time graph for this motion would look like this:

Proving Displacement is the Area under Velocity vs Time Graphs

To prove that the area under this velocity-time graph is the object’s displacement, let’s start with figuring out the displacement. The equation for displacement is \Delta x=vt . In this case, we know v=5\text{ m/s} and t=10\text{ s} . Therefore, \Delta x=(5\text{ m/s})\cdot (10\text{ s})=50\text{ m} .

Now that we know we’re looking for a displacement of 50\text{ m} , let’s try finding the area under the curve. Specifically, this is the area between the line of our graph and the x-axis. We’ll start by drawing a shape – in this case, a rectangle. We’ll also include values for its base and height.

It’s worth noting here that the units along each axis were also included for the base and height of the rectangle. The equation for the area under the curve is the one you would use to find the area of a rectangle, A=bh . So, let’s pull down our values and solve our equation:

• b=10\text{ s}
• h=5\text{ m/s}

As a result, we obtained the same numerical value of 50 , but more to the point we obtained the correct units. The area under the curve of a velocity graph will always be a displacement. Let’s look at a couple of more examples. If you’re uncertain about your ability to remember the equations for the area of a rectangle or triangle, it may be worth writing them in your notes or referencing a formula sheet such as this one .

Example 1: Finding Displacement for Multiple Velocities

The graph above was pretty simple, so let’s look at some more complex motion graphs. We can return to the velocity-time graph for our car that moved to the right, paused, then moved back to the left.

We already know that our displacement for this motion is 0\text{ m} . Let’s start by sectioning off our graph here into shapes we cam find the area of. Again, we’re looking for the area between the line of the graph and the x-axis.

It seems strange to have a negative value for the height of a shape as you’ve likely been told that area should always be a positive value. We’ll see why having a negative height when the graph is below the x-axis is both allowed and important. Now that we have all of our rectangles we can start finding their area.

Let’s begin with the rectangle farthest to the left.

• b=5\text{ s}
• h=1\text{ m/s}

Now we can solve for the area of our middle rectangle. This may seem like a trick question as it is, essentially, just a flat line, but we’ll still want to include it.

• h=0\text{ m/s}

Finally, let’s find the area of the rectangle on the right. This has a negative value for its height so it should also have a negative area, strange as that may seem.

• h=-1\text{ m/s}

Now that we know the area of all three rectangles, we’ll want to add those areas together to find the total area under the velocity-time curve and therefore also our total displacement.

Now, we see the expected value of 0\text{ m} that we’d found before. It’s important to note that this was only possible because one of our rectangles had a negative area.

Example 2: Finding Displacement with Changing Velocity

We know we can use rectangles to find the area under a velocity-time graph, but not all graphs are horizontal lines. Sometimes, graphs are diagonal which requires us to find the area of a different shape – a triangle. Consider the velocity-time graph.

We can create a rectangle on the right where the velocity is constant, but the area where it’s increasing will not look like a rectangle at all. Instead, this is where we’ll have to create a section that is a triangle.

We now have two separate shapes. Much like when we had three separate rectangles, we’ll find the area of each shape individually and then add those two areas together to find the overall displacement for this motion. Let’s start with the triangle.

• h=10\text{ m/s}

Now, we can find the area of the rectangle portion.

Finally, we’ll add these two values to find our total displacement:

Finding Acceleration from a Velocity vs Time Graph

At this point, it may not shock you to learn that the slope of a velocity-time graph can tell us just as much as the area under its curve. Instead of displacement, though, the slope of a velocity graph will tell us an object’s acceleration. Let’s consider the graph.

The velocity of the object being shown in this graph is steadily increasing by 2\text{ m/s} every 1\text{ s} . With that information, we can prove that a=\Delta v/\Delta t=(2\text{ m/s})/(1\text{ s})=2\text{ m/s}^2 . Now that we know what our acceleration should be, let’s try to find it by finding the slope of the velocity time graph.

Proving Acceleration is the Slope of Velocity vs Time Graphs

If you remember from earlier, the slope of any given straight line can be found with the equation:

Let’s begin by selecting two points off of our graph above (being sure to include the units when we do). Let’s take (2\text{ s},4\text{ m/s}) and (4\text{ s},8\text{ m/s}) and plug the values in.

Again, this should look like a set of delta values – change in velocity over change in time, specifically. This is the definition of acceleration. While you may already be able to see how this will turn into proof that acceleration is the slope of a velocity graph, let’s keep going. What we’ll be looking for when we solve this equation this time will be a numerical value of 2 and units of m/s 2 .

Notice that we have a value that matches the acceleration we solved for before. We could have selected any two points on our graph and received the same result. The part of this that truly proves the slope of any velocity-time graph is the acceleration is that the units of our slope work out to be m/s 2 – the units for acceleration.

Example 1: Finding Negative Acceleration

Let’s consider the velocity vs time graph.

We can see that the graph has a constant, negative slope so we can choose any two points we want and we should get the correct acceleration, which should also be a negative value. Whenever possible, it’s worth choosing values with zeros, so let’s select the points (0\text{ s},5\text{ m/s}) and (5\text{ s},0\text{ m/s}) . Now that we have our points, let’s pull out the values we need and plug them into our slope formula to solve for the acceleration of this object.

• y_{2}=0\text{ m/s}
• y_{1}=5\text{ m/s}

Example 2: Finding Multiple Accelerations

Let’s consider a more complex example with the velocity-time graph below.

This graph has a change in its slope. This means we have two separate sections that we can look at: before t=5\text{ s} and after t=5\text{ s} .

Let’s consider the after t=5\text{ s} portion of the graph. Assuming that our acceleration will be negative because our velocity values are always negative is a common mistake among budding physicists. We’ll see here that this isn’t always true. Let’s choose the points (5\text{ s},-10\text{ m/s}) and (10\text{ s},-5\text{ m/s}) and plug these values into our slope formula.

• y_{2}=-5\text{ m/s}
• y_{1}=-10\text{ m/s}
• x_{2}=10\text{ s}
• x_{1}=5\text{ s}

If we plug these values into our slope formula, we can find that:

We see that even though our velocity values are negative, our slope is still positive so our acceleration must still be positive. Be careful when looking at motion graphs and making early assumptions. Things are sometimes more complicated than they appear.

Describing Motion with Acceleration vs Time Graph s

Last but not least, we can describe an object’s motion with an acceleration vs time graph. These will likely be graphs with zero slope while you are starting your study of motion graphs. You may find them becoming more complicated if you pursue a career in physics, but for now, we can keep things simple. The below graph is a standard example of an acceleration graph you may see.

This graph actually shows acceleration due to gravity on Earth’s surface at a constant value of 9.81\text{ m/s}^2 . The other acceleration-time graph you’re likely to see in a high school physics class may look more like this:

This would indicate that the object’s velocity is not changing, or perhaps that it isn’t moving at all. 

It is worth noting that the area under the curve of an acceleration vs time graph is equal to an object’s velocity, much like how the area under a velocity vs time graph is the displacement. Most high school physics classes won’t spend much time on this idea, but as you progress through your physics career this idea may come up. If you’d like to prove it to yourself, you could follow the same proof we used when proving the relationship between a velocity-time graph and an object’s displacement.

Pairing Motion Graphs

The last skill we’ll cover for motion graphs is determining which pair of graphs represent the same motion. We can make more than one graph to describe any given motion. For example, we had both a position-time graph and a velocity-time graph for our car moving to the right, pausing, and then coming back to the left. We can extend this idea to include acceleration graphs too. Let’s consider the three graphs you were presented with at the beginning of this article.

We can see the position vs time graph on the left has a constant, positive slope. Since we know that velocity is the slope of a position vs time graph, our velocity must therefore be a constant, positive value. Indeed, we see that the velocity vs time graph here is constantly at 5\text{ m/s} . If we continue following this logic, we can assume that our acceleration should be zero as the slope of our velocity vs time graph is zero. And, again, we see that this holds true as the acceleration vs time graph is constant at 0\text{ m/s}^2 .

While it may be easy to see that these three motion graphs are connected after looking at them for a few moments, you’ll want to be able to compare more complex graphs throughout your physics career. There are a few steps you can take to achieve this goal.

The Steps to Pairing Motion Graphs

Step 1: observe the shape and make a prediction.

Odds are, you’ll see three different shapes when looking at position graphs, three shapes when looking at velocity graphs, and only two shapes when looking at acceleration graphs. Without even looking at the numbers on these graphs, the shapes can tell you a lot. Here are the shapes of each graph you may see throughout your high school physics career:

All of these examples are of positive values but know that they may be flipped. This would simply mean that the slope values are negative instead of positive. Just from glancing at the graphs above, even though they don’t have any number, you could match them up just by looking at the slopes.

Corresponding Shapes on Motion Graphs

Here’s a chart of how the different shapes match up.

You may notice that the arrows here go both ways and for good reason. While you’ll need to do some math to know exactly which parabola-shaped position-time graph matches a diagonal-shaped velocity-time graph, those two shapes will always go together, regardless of which you start with.

Step 2: Decide if it is Positive or Negative

There is one more step you can take in matching up motion graphs before you start doing any actual math, which is looking for positive or negative slopes. Let’s look at this position-time graph.

We can see that the slope here is negative. The curve is above the x-axis so the values are positive, but the slope itself is negative. We know that this shape of the position-time graph will go with a flat velocity-time graph, but we need to pick the right one. We may need some mathematics for this, but let’s first try to narrow down our options. Let’s say you had to pick between the three velocity-time graphs below.

We’ve decided that our graph should be a straight, horizontal line. All three of these graphs match that description. We also said that our slope is negative, and only two of the velocity-time graphs have negative values. So, without doing any mathematics, we know that the positive-time graph above will have to be paired with the middle velocity-time graph or the right velocity-time graph. To find out for sure, we’ll have to add some numbers and do some math.

Step 3: Calculate the Slope and Compare

Let’s take that same position-time graph and the two velocity-time graphs we couldn’t decide between. As you’ll see, they now have some numbers so that we can do some math to actually match the correct graphs.

We’ll need to begin by finding the slope of the position-time graph. To keep things simple, let’s use (0\text{ s},5\text{ m}) and (5\text{ s},0\text{ m}) . Now let’s pull out some values and solve for slope.

So, the velocity-time graph that matches our position-time graph here should have a value of -1\text{ m/s} . If we look, the velocity-time graph on the left has its line moving through -1\text{ m/s} so that must be the correct velocity-time graph.

We used these three steps of looking at the shape, looking at positive or negative, and then calculating the slope to go from a position-time graph to a velocity-time graph, but they can help us do much more than that. They can help us match up all three kinds of graphs or any pair of motion graphs – even a position-time graph and an acceleration-time graph.

Example 1: Which Pair of Graphs Represent the Same Motion?

Let’s consider the position-time graph below and try to match it to the correct velocity vs time graph.

Observe the Shape

Right off the bat, we know we have a parabola-shaped position-time graph. From that, we can narrow down our velocity-time graph options. A parabola position-time graph always goes to a diagonal velocity-time graph so we can cross off the middle graph immediately. Now we’re left with two different choices.

You may think you have the right answer already (and indeed you might), but let’s think through this carefully. We already matched up our shapes so now it’s time to compare our positives and negatives.

Decide if it is Positive or Negative

The parabola in the position-time graph points upward so it has a positive slope. That means our velocity-time graph needs to be positive. If you’re thinking too carefully about slope, you may be drawn to the graph on the left. While it’s true that the graph on the left has a positive slope, it actually contains negative values. The values are what’s important here, not the slope. Instead, the graph on the right is the correct choice here. We knew we needed a diagonal velocity-time graph with positive values and we only had one option. You may often encounter examples like this, but be careful to always check your instincts before answering too quickly.

It is also worth noting here that a helpful trick for recognizing whether a velocity-time graph could give a positive slope to a position-time graph is if the curve of the velocity-time graph is over the x-axis. The same is true in reverse; a velocity-time graph with a curve below the x-axis will match with a position-time graph with a negative slope.

Example 2: Match the Velocity-Time Graph

The same principles that we just used above can also help us transition from a velocity-time graph to an acceleration-time graph. Let’s consider the set of motion graphs below.

From the start, we can see that we have a diagonal-shaped velocity-time graph so we can eliminate our middle acceleration-time graph. Although we will want a flat graph, the middle one is on the x-axis, which would imply that our velocity-time graph has zero slope. If that were the case, it would be flat instead of diagonal.

Now that we are again down to two graphs, let’s look at the positives and negatives. The velocity-time graph has a negative slope so we’ll want an acceleration-time graph with a curve below the x-axis. This leaves us with only one option – the graph on the right. 

Again, we didn’t need to get to the mathematics. You could check the values if you wanted to, but often looking just at the shapes of your graphs will be enough. Just make sure you always think through both the shape of your motion graphs and if your positives and negatives line up.

Physicists use motion graphs to visualize data all the time. While the different types and shapes may be confusing at first, getting comfortable with them will help you make connections between the kinematics terms you’ve learned so far. It can also help you simplify problems by being able to visualize what the problem is asking you in a different way. If you take the time to get comfortable reading each time of motion graph, deriving different values from them, and matching them up, you’ll be well on your way to visualizing data the way research scientists do every day.

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2.8 Graphical Analysis of One-Dimensional Motion

Learning objectives.

By the end of this section, you will be able to:

• Describe a straight-line graph in terms of its slope and y -intercept.
• Determine average velocity or instantaneous velocity from a graph of position vs. time.
• Determine average or instantaneous acceleration from a graph of velocity vs. time.
• Derive a graph of velocity vs. time from a graph of position vs. time.
• Derive a graph of acceleration vs. time from a graph of velocity vs. time.

A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of position, velocity, and acceleration versus time to illustrate one-dimensional kinematics.

Slopes and General Relationships

First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable . If we call the horizontal axis the x x -axis and the vertical axis the y y -axis, as in Figure 2.44 , a straight-line graph has the general form

Here m m is the slope , defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter b b is used for the y -intercept , which is the point at which the line crosses the vertical axis.

Graph of Position vs. Time ( a = 0, so v is constant)

Time is usually an independent variable that other quantities, such as position, depend upon. A graph of position versus time would, thus, have x x on the vertical axis and t t on the horizontal axis. Figure 2.45 is just such a straight-line graph. It shows a graph of position versus time for a jet-powered car on a very flat dry lake bed in Nevada.

Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity v - v - and the intercept is position at time zero—that is, x 0 x 0 . Substituting these symbols into y = mx + b y = mx + b gives

Thus a graph of position versus time gives a general relationship among displacement(change in position), velocity, and time, as well as giving detailed numerical information about a specific situation.

The Slope of x vs. t

The slope of the graph of position x x vs. time t t is velocity v v .

Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension .

From the figure we can see that the car has a position of 525 m at 0.50 s and 2000 m at 6.40 s. Its position at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.

Example 2.17

Determining average velocity from a graph of position versus time: jet car.

Find the average velocity of the car whose position is graphed in Figure 2.45 .

The slope of a graph of x x vs. t t is average velocity, since slope equals rise over run. In this case, rise = change in position and run = change in time, so that

Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)

1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)

2. Substitute the x x and t t values of the chosen points into the equation. Remember in calculating change ( Δ ) ( Δ ) we always use final value minus initial value.

This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.

Graphs of Motion when a a is constant but a ≠ 0 a ≠ 0

The graphs in Figure 2.46 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the position and velocity are initially 200 m and 15 m/s, respectively.

The graph of position versus time in Figure 2.46 (a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a position-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 2.46 (a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 2.46 (b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 2.46 (c).

Example 2.18

Determining instantaneous velocity from the slope at a point: jet car.

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the x x vs. t t graph in the graph below.

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 2.48 , where Q is the point at t = 25 s t = 25 s .

1. Find the tangent line to the curve at t = 25 s t = 25 s .

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope, v v .

This is the value given in this figure’s table for v v at t = 25 s t = 25 s . The value of 140 m/s for v Q v Q is plotted in Figure 2.48 . The entire graph of v v vs. t t can be obtained in this fashion.

Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a v v vs. t t graph, rise = change in velocity Δ v Δ v and run = change in time Δ t Δ t .

The Slope of v vs. t

The slope of a graph of velocity v v vs. time t t is acceleration a a .

Since the velocity versus time graph in Figure 2.46 (b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 2.46 (c).

Additional general information can be obtained from Figure 2.48 and the expression for a straight line, y = mx + b y = mx + b .

In this case, the vertical axis y y is V V , the intercept b b is v 0 v 0 , the slope m m is a a , and the horizontal axis x x is t t . Substituting these symbols yields

A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension .

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.

Graphs of Motion Where Acceleration is Not Constant

Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 2.49 . Time again starts at zero, and the initial velocity is 165 m/s. (This was the final velocity of the car in the motion graphed in Figure 2.46 .) Acceleration gradually decreases from 5 . 0 m/s 2 5 . 0 m/s 2 to zero when the car hits 250 m/s. The velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.

Example 2.19

Calculating acceleration from a graph of velocity versus time.

Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the v v vs. t t graph in Figure 2.49 (a).

The slope of the curve at t = 25 s t = 25 s is equal to the slope of the line tangent at that point, as illustrated in Figure 2.49 (a).

Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, a a .

Note that this value for a a is consistent with the value plotted in Figure 2.49 (b) at t = 25 s t = 25 s .

A graph of position versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.

Check Your Understanding

A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?

(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.

(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

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• Authors: Paul Peter Urone, Roger Hinrichs
• Publisher/website: OpenStax
• Book title: College Physics 2e
• Publication date: Jul 13, 2022
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Graphs of Motion

Introduction.

Modern mathematical notation is a highly compact way to encode ideas. Equations can easily contain the information equivalent of several sentences. Galileo's description of an object moving with constant speed (perhaps the first application of mathematics to motion) required one definition, four axioms, and six theorems . All of these relationships can now be written in a single equation.

When it comes to depth, nothing beats an equation.

Well, almost nothing. Think back to the previous section on the equations of motion. You should recall that the three (or four) equations presented in that section were only valid for motion with constant acceleration along a straight line. Since, as I rightly pointed out, "no object has ever traveled in a straight line with constant acceleration anywhere in the universe at any time" these equations are only approximately true, only once in a while.

Equations are great for describing idealized situations, but they don't always cut it. Sometimes you need a picture to show what's going on — a mathematical picture called a graph. Graphs are often the best way to convey descriptions of real world events in a compact form. Graphs of motion come in several types depending on which of the kinematic quantities (time, position, velocity, acceleration) are assigned to which axis.

position-time

Let's begin by graphing some examples of motion at a constant velocity. Three different curves are included on the graph to the right, each with an initial position of zero. Note first that the graphs are all straight. (Any kind of line drawn on a graph is called a curve. Even a straight line is called a curve in mathematics.) This is to be expected given the linear nature of the appropriate equation. (The independent variable of a linear function is raised no higher than the first power.)

Compare the position-time equation for constant velocity with the classic slope-intercept equation taught in introductory algebra.

Thus velocity corresponds to slope and initial position to the intercept on the vertical axis (commonly thought of as the "y" axis). Since each of these graphs has its intercept at the origin, each of these objects had the same initial position. This graph could represent a race of some sort where the contestants were all lined up at the starting line (although, at these speeds it must have been a race between tortoises). If it were a race, then the contestants were already moving when the race began, since each curve has a non-zero slope at the start. Note that the initial position being zero does not necessarily imply that the initial velocity is also zero. The height of a curve tells you nothing about its slope.

• slope is velocity
• the "y" intercept is the initial position
• when two curves coincide, the two objects have the same position at that time

In contrast to the previous examples, let's graph the position of an object with a constant, non-zero acceleration starting from rest at the origin. The primary difference between this curve and those on the previous graph is that this curve actually curves. The relation between position and time is quadratic when the acceleration is constant and therefore this curve is a parabola . (The variable of a quadratic function is raised no higher than the second power.)

As an exercise, let's calculate the acceleration of this object from its graph. It intercepts the origin, so its initial position is zero, the example states that the initial velocity is zero, and the graph shows that the object has traveled 9 m in 10 s. These numbers can then be entered into the equation.

When a position-time graph is curved, it is not possible to calculate the velocity from it's slope. Slope is a property of straight lines only. Such an object doesn't have a velocity because it doesn't have a slope. The words "the" and "a" are underlined here to stress the idea that there is no single velocity under these circumstances. The velocity of such an object must be changing. It's accelerating.

• straight segments imply constant velocity
• curve segments imply acceleration
• an object undergoing constant acceleration traces a portion of a parabola

Although our hypothetical object has no single velocity, it still does have an average velocity and a continuous collection of instantaneous velocities. The average velocity of any object can be found by dividing the overall change in position (a.k.a. the displacement) by the change in time.

This is the same as calculating the slope of the straight line connecting the first and last points on the curve as shown in the diagram to the right. In this abstract example, the average velocity of the object was…

Instantaneous velocity is the limit of average velocity as the time interval shrinks to zero.

As the endpoints of the line of average velocity get closer together, they become a better indicator of the actual velocity. When the two points coincide, the line is tangent to the curve. This limit process is represented in the animation to the right.

• average velocity is the slope of the straight line connecting the endpoints of a curve
• instantaneous velocity is the slope of the line tangent to a curve at any point

Seven tangents were added to our generic position-time graph in the animation shown above. Note that the slope is zero twice — once at the top of the bump at 3.0 s and again in the bottom of the dent at 6.5 s. (The bump is a local maximum , while the dent is a local minimum . Collectively such points are known as local extrema .) The slope of a horizontal line is zero, meaning that the object was motionless at those times. Since the graph is not flat, the object was only at rest for an instant before it began moving again. Although its position was not changing at that time, its velocity was. This is a notion that many people have difficulty with. It is possible to be accelerating and yet not be moving, but only for an instant.

Note also that the slope is negative in the interval between the bump at 3.0 s and the dent at 6.5 s. Some interpret this as motion in reverse, but is this generally the case? Well, this is an abstract example. It's not accompanied by any text. Graphs contain a lot of information, but without a title or other form of description they have no meaning. What does this graph represent? A person? A car? An elevator? A rhinoceros? An asteroid? A mote of dust? About all we can say is that this object was moving at first, slowed to a stop, reversed direction, stopped again, and then resumed moving in the direction it started with (whatever direction that was). Negative slope does not automatically mean driving backward, or walking left, or falling down. The choice of signs is always arbitrary. About all we can say in general, is that when the slope is negative, the object is traveling in the negative direction.

• positive slope implies motion in the positive direction
• negative slope implies motion in the negative direction
• zero slope implies a state of rest

velocity-time

The most important thing to remember about velocity-time graphs is that they are velocity-time graphs, not position-time graphs. There is something about a line graph that makes people think they're looking at the path of an object. A common beginner's mistake is to look at the graph to the right and think that the the v  = 9.0 m/s line corresponds to an object that is "higher" than the other objects. Don't think like this. It's wrong.

Don't look at these graphs and think of them as a picture of a moving object. Instead, think of them as the record of an object's velocity. In these graphs, higher means faster not farther. The v  = 9.0 m/s line is higher because that object is moving faster than the others.

These particular graphs are all horizontal. The initial velocity of each object is the same as the final velocity is the same as every velocity in between. The velocity of each of these objects is constant during this ten second interval.

In comparison, when the curve on a velocity-time graph is straight but not horizontal, the velocity is changing. The three curves to the right each have a different slope. The graph with the steepest slope experiences the greatest rate of change in velocity. That object has the greatest acceleration. Compare the velocity-time equation for constant acceleration with the classic slope-intercept equation taught in introductory algebra.

You should see that acceleration corresponds to slope and initial velocity to the intercept on the vertical axis. Since each of these graphs has its intercept at the origin, each of these objects was initially at rest. The initial velocity being zero does not mean that the initial position must also be zero, however. This graph tells us nothing about the initial position of these objects. For all we know they could be on different planets.

• slope is acceleration
• the "y" intercept is the initial velocity
• when two curves coincide, the two objects have the same velocity at that time

The curves on the previous graph were all straight lines. A straight line is a curve with constant slope. Since slope is acceleration on a velocity-time graph, each of the objects represented on this graph is moving with a constant acceleration. Were the graphs curved, the acceleration would have been not constant.

• straight lines imply constant acceleration
• curved lines imply non-constant acceleration
• an object undergoing constant acceleration traces a straight line

Since a curved line has no single slope we must decide what we mean when asked for the acceleration of an object. These descriptions follow directly from the definitions of average and instantaneous acceleration. If the average acceleration is desired, draw a line connecting the endpoints of the curve and calculate its slope. If the instantaneous acceleration is desired, take the limit of this slope as the time interval shrinks to zero, that is, take the slope of a tangent.

• average acceleration is the slope of the straight line connecting the endpoints of a curve
• instantaneous acceleration is the slope of the line tangent to a curve at any point

Seven tangents were added to our generic velocity-time graph in the animation shown above. Note that the slope is zero twice — once at the top of the bump at 3.0 s and again in the bottom of the dent at 6.5 s. The slope of a horizontal line is zero, meaning that the object stopped accelerating instantaneously at those times. The acceleration might have been zero at those two times, but this does not mean that the object stopped. For that to occur, the curve would have to intercept the horizontal axis. This happened only once — at the start of the graph. At both times when the acceleration was zero, the object was still moving in the positive direction.

You should also notice that the slope was negative from 3.0 s to 6.5 s. During this time the speed was decreasing. This is not true in general, however. Speed decreases whenever the curve returns to the origin. Above the horizontal axis this would be a negative slope, but below it this would be a positive slope. About the only thing one can say about a negative slope on a velocity-time graph is that during such an interval, the velocity is becoming more negative (or less positive, if you prefer).

• positive slope implies an increase in velocity in the positive direction
• negative slope implies an increase in velocity in the negative direction
• zero slope implies motion with constant velocity

In kinematics, there are three quantities: position, velocity, and acceleration. Given a graph of any of these quantities, it is always possible in principle to determine the other two. Acceleration is the time rate of change of velocity, so that can be found from the slope of a tangent to the curve on a velocity-time graph. But how could position be determined? Let's explore some simple examples and then derive the relationship.

Start with the simple velocity-time graph shown to the right. (For the sake of simplicity, let's assume that the initial position is zero.) There are three important intervals on this graph. During each interval, the acceleration is constant as the straight line segments show. When acceleration is constant, the average velocity is just the average of the initial and final values in an interval.

0–4 s: This segment is triangular. The area of a triangle is one-half the base times the height. Essentially, we have just calculated the area of the triangular segment on this graph.

The cumulative distance traveled at the end of this interval is…

4–8 s: This segment is trapezoidal. The area of a trapezoid (or trapezium ) is the average of the two bases times the altitude. Essentially, we have just calculated the area of the trapezoidal segment on this graph.

16 m + 36 m = 52 m

8–10 s: This segment is rectangular. The area of a rectangle is just its height times its width. Essentially, we have just calculated the area of the rectangular segment on this graph.

16 m + 36 m + 20 m = 72 m

I hope by now that you see the trend. The area under each segment is the change in position of the object during that interval. This is true even when the acceleration is not constant.

Anyone who has taken a calculus course should have known this before they read it here (or at least when they read it they should have said, "Oh yeah, I remember that"). The first derivative of position with respect to time is velocity. The derivative of a function is the slope of a line tangent to its curve at a given point. The inverse operation of the derivative is called the integral. The integral of a function is the cumulative area between the curve and the horizontal axis over some interval. This inverse relation between the actions of derivative (slope) and integral (area) is so important that it's called the fundamental theorem of calculus . This means that it's an important relationship. Learn it! It's "fundamental". You haven't seen the last of it.

• the area under the curve is the change in position

acceleration-time

The acceleration-time graph of any object traveling with a constant velocity is the same. This is true regardless of the velocity of the object. An airplane flying at a constant 270 m/s (600 mph), a sloth walking at a constant 0.4 m/s (1 mph), and a couch potato lying motionless in front of the TV for hours will all have the same acceleration-time graphs — a horizontal line collinear with the horizontal axis. That's because the velocity of each of these objects is constant. They're not accelerating. Their accelerations are zero. As with velocity-time graphs, the important thing to remember is that the height above the horizontal axis doesn't correspond to position or velocity, it corresponds to acceleration .

If you trip and fall on your way to school, your acceleration towards the ground is greater than you'd experience in all but a few high performance cars with the "pedal to the metal". Acceleration and velocity are different quantities. Going fast does not imply accelerating quickly. The two quantities are independent of one another. A large acceleration corresponds to a rapid change in velocity, but it tells you nothing about the values of the velocity itself.

When acceleration is constant, the acceleration-time curve is a horizontal line. The rate of change of acceleration with time is not often discussed, so the slope of the curve on this graph will be ignored for now. If you enjoy knowing the names of things, this quantity is called jerk . On the surface, the only information one can glean from an acceleration-time graph appears to be the acceleration at any given time.

• slope is jerk
• the "y" intercept equals the initial acceleration
• when two curves coincide, the two objects have the same acceleration at that time
• an object undergoing constant acceleration traces a horizontal line
• zero slope implies motion with constant acceleration

Acceleration is the rate of change of velocity with time. Transforming a velocity-time graph to an acceleration-time graph means calculating the slope of a line tangent to the curve at any point. (In calculus, this is called finding the derivative.) The reverse process entails calculating the cumulative area under the curve. (In calculus, this is called finding the integral.) This number is then the change of value on a velocity-time graph.

Given an initial velocity of zero (and assuming that down is positive), the final velocity of the person falling in the graph to the right is…

and the final velocity of the accelerating car is…

• the area under the curve equals the change in velocity

There are more things one can say about acceleration-time graphs, but they are trivial for the most part.

phase space

There is a fourth graph of motion that relates velocity to position. It is as important as the other three types, but it rarely gets any attention below the advanced undergraduate level. Some day I will write something about these graphs called phase space diagrams, but not today.

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2 Kinematics

14 2.8 Graphical Analysis of One-Dimensional Motion

• Describe a straight-line graph in terms of its slope and y-intercept.
• Determine average velocity or instantaneous velocity from a graph of position vs. time.
• Determine average or instantaneous acceleration from a graph of velocity vs. time.
• Derive a graph of velocity vs. time from a graph of position vs. time.
• Derive a graph of acceleration vs. time from a graph of velocity vs. time.

A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.

Slopes and General Relationships

First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable . If we call the horizontal axis the[latex]\boldsymbol{x}\text{-axis}[/latex]and the vertical axis the[latex]\boldsymbol{y}\text{-axis}[/latex], as in Figure 1 , a straight-line graph has the general form

Here[latex]\boldsymbol{m}[/latex]is the slope , defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter[latex]\boldsymbol{b}[/latex]is used for the y -intercept , which is the point at which the line crosses the vertical axis.

Graph of Displacement vs. Time ( a = 0, so v is constant)

Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have[latex]\boldsymbol{x}[/latex]on the vertical axis and[latex]\boldsymbol{t}[/latex]on the horizontal axis. Figure 2 is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada .

Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity[latex]\boldsymbol{\bar{v}}[/latex]and the intercept is displacement at time zero—that is,[latex]\boldsymbol{x_0}.[/latex]Substituting these symbols into[latex]\boldsymbol{y}=\boldsymbol{mx+b}[/latex]gives

Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.

THE SLOPE OF X VS. T

The slope of the graph of displacement [latex]\boldsymbol{x}[/latex]vs. time[latex]\boldsymbol{t}[/latex]is velocity[latex]\boldsymbol{v}.[/latex]

Notice that this equation is the same as that derived algebraically from other motion equations in Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension .

From the figure we can see that the car has a displacement of 25 m at 0.50 s and 2000 m at 6.40 s. Its displacement at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.

Example 1: Determining Average Velocity from a Graph of Displacement versus Time: Jet Car

Find the average velocity of the car whose position is graphed in Figure 2 .

The slope of a graph of[latex]\boldsymbol{x}[/latex]vs.[latex]\boldsymbol{t}[/latex]is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that

Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)

1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)

2. Substitute the[latex]\boldsymbol{x}[/latex]and[latex]\boldsymbol{t}[/latex]values of the chosen points into the equation. Remember in calculating change[latex]\boldsymbol{(\Delta)}[/latex]we always use final value minus initial value.

This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.

Graphs of Motion when α is constant but α≠0

The graphs in Figure 3 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively.

The graph of displacement versus time in Figure 3 (a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3 (a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3 (b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3 (c).

Example 2: Determining Instantaneous Velocity from the Slope at a Point: Jet Car

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the[latex]\boldsymbol{x}[/latex]vs.[latex]\boldsymbol{t}[/latex]graph in the graph below.

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5 , where Q is the point at[latex]\boldsymbol{t=25\textbf{ s}}.[/latex]

1. Find the tangent line to the curve at[latex]\boldsymbol{t=25\textbf{ s}}.[/latex]

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope,[latex]\boldsymbol{v}.[/latex]

This is the value given in this figure’s table for[latex]\boldsymbol{v}[/latex]at[latex]\boldsymbol{t=25\textbf{ s}}.[/latex]The value of 140 m/s for[latex]\boldsymbol{v_Q}[/latex]is plotted in Figure 5 . The entire graph of[latex]\boldsymbol{v}[/latex]vs.[latex]\boldsymbol{t}[/latex]can be obtained in this fashion.

Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a[latex]\boldsymbol{v}[/latex]vs.[latex]\boldsymbol{t}[/latex]graph, rise = change in velocity[latex]\boldsymbol{\Delta{v}}[/latex]and run = change in time[latex]\boldsymbol{\Delta{t}}.[/latex]

THE SLOPE OF V VS. T

The slope of a graph of velocity[latex]\boldsymbol{v}[/latex]vs. time[latex]\boldsymbol{t}[/latex]is acceleration[latex]\boldsymbol{a}.[/latex]

Since the velocity versus time graph in Figure 3 (b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 3 (c).

Additional general information can be obtained from Figure 5 and the expression for a straight line,[latex]\boldsymbol{y=mx+b}.[/latex]

In this case, the vertical axis[latex]\boldsymbol{y}[/latex]is[latex]\textbf{V}[/latex], the intercept[latex]\boldsymbol{b}[/latex]is[latex]\boldsymbol{v_0},[/latex]the slope[latex]\boldsymbol{m}[/latex]is[latex]\boldsymbol{a},[/latex]and the horizontal axis[latex]\boldsymbol{x}[/latex] is [latex]\boldsymbol{t}.[/latex]Substituting these symbols yields

A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension .

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.

Graphs of Motion Where Acceleration is Not Constant

Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 6 . Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure 3 .) Acceleration gradually decreases from[latex]\boldsymbol{5.0\textbf{ m/s}^2}[/latex]to zero when the car hits 250 m/s. The slope of the[latex]\boldsymbol{x}[/latex]vs.[latex]\boldsymbol{t}[/latex]graph increases until[latex]\boldsymbol{t=55\textbf{ s}},[/latex]after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.

Example 3: Calculating Acceleration from a Graph of Velocity versus Time

Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the[latex]\boldsymbol{v}[/latex]vs.[latex]\boldsymbol{t}[/latex]graph in Figure 6 (b).

The slope of the curve at[latex]\boldsymbol{t=25\textbf{ s}}[/latex]is equal to the slope of the line tangent at that point, as illustrated in Figure 6 (b).

Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope,[latex]\boldsymbol{a}.[/latex]

Note that this value for[latex]\boldsymbol{a}[/latex]is consistent with the value plotted in Figure 6 (c) at[latex]\boldsymbol{t=25\textbf{ s}}.[/latex]

A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.

Check Your Understanding

1: A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?

Section Summary

• Graphs of motion can be used to analyze motion.
• Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
• The slope of a graph of displacement[latex]\boldsymbol{x}[/latex]vs. time[latex]\boldsymbol{t}[/latex]is velocity[latex]\boldsymbol{v}.[/latex]
• The slope of a graph of velocity[latex]\boldsymbol{v}[/latex]vs. time[latex]\boldsymbol{t}[/latex]graph is acceleration[latex]\boldsymbol{a}.[/latex]
• Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.

Conceptual Questions

1: (a) Explain how you can use the graph of position versus time in Figure 8 to describe the change in velocity over time. Identify (b) the time ([latex]\boldsymbol{t_a, t_b,t_c,t_d,}\text{ or }\boldsymbol{t_e}[/latex]) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative.

2: (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 9 . (b) Identify the time or times ([latex]\boldsymbol{t_a,t_b,t_c},[/latex]etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative?

3: (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 10 . (b) Based on the graph, how does acceleration change over time?

4: (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 11 . (b) Identify the time or times ([latex]\boldsymbol{t_a,t_b,t_c},[/latex]etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?

5: Consider the velocity vs. time graph of a person in an elevator shown in Figure 12 . Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.

6: A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.

Problems & Exercises

Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.

1: (a) By taking the slope of the curve in Figure 13 , verify that the velocity of the jet car is 115 m/s at[latex]\boldsymbol{t=20\textbf{ s}}.[/latex](b) By taking the slope of the curve at any point in Figure 14 , verify that the jet car’s acceleration is[latex]\boldsymbol{5.0\textbf{ m/s}^2}.[/latex]

2: Using approximate values, calculate the slope of the curve in Figure 15 to verify that the velocity at[latex]\boldsymbol{t=10.0\textbf{ s}}[/latex]is 0.208 m/s. Assume all values are known to 3 significant figures.

3: Using approximate values, calculate the slope of the curve in Figure 15 to verify that the velocity at[latex]\boldsymbol{t=30.0\textbf{ s}}[/latex]is 0.238 m/s. Assume all values are known to 3 significant figures.

4: By taking the slope of the curve in Figure 16 , verify that the acceleration is[latex]\boldsymbol{3.2\textbf{ m/s}^2}[/latex]at[latex]\boldsymbol{t=10\textbf{ s}}.[/latex]

5: Construct the displacement graph for the subway shuttle train as shown in Chapter 2.4 Figure 7 (a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

6: (a) Take the slope of the curve in Figure 17 to find the jogger’s velocity at[latex]\boldsymbol{t=2.5\textbf{ s}}.[/latex](b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 18 .

7: A graph of[latex]\boldsymbol{v(t)}[/latex]is shown for a world-class track sprinter in a 100-m race. (See Figure 20 ). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at[latex]\boldsymbol{t=5\textbf{ s}}?[/latex](c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

8: Figure 21 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.

1: (a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.

(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.

(a)[latex]\boldsymbol{115\textbf{ m/s}}[/latex]

(b)[latex]\boldsymbol{5.0\textbf{ m/s}^2}[/latex]

[latex]\boldsymbol{v=}[/latex][latex]\boldsymbol{\frac{(11.7\:-\:6.95)\times10^3\textbf{ m}}{(40.0\:-\:20.0)\textbf{ s}}}[/latex][latex]\boldsymbol{=238\textbf{ m/s}}[/latex]

(c)[latex]\boldsymbol{3\textbf{ m/s}^2}[/latex]

College Physics chapters 1-17 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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• Graphical Representation of Motion

Graphical Representation makes it simpler for us to understand data. When analyzing motion, graphs representing values of various parameters of motion make it simpler to solve problems. Let us understand the concept of motion and the other entities related to it using the graphical method.

Suggested Videos

Using a graph for a pictorial representation of two sets of data is called a graphical representation of data . One entity is represented on the x-axis of the graph while the other is represented on the y-axis. Out of the two entities, one is a dependent set of variables while the other is independent an independent set of variables.

We use line graphs to describe the motion of an object. This graph shows the dependency of a physical quantity speed or distance on another quantity, for example, time.

Browse more Topics under Motion

• Introduction to Motion and its Parameters
• Equations of Motion
• Uniform Circular Motion

Distance Time Graph

The distance-time graph determines the change in the position of the object. The speed of the object as well can be determined using the line graph. Here the time lies on the x-axis while the distance on the y-axis. Remember, the line graph of uniform motion is always a straight line .

Why? Because as the definition goes, uniform motion is when an object covers the equal amount of distance at equal intervals of time.  Hence the straight line. While the graph of a non-uniform motion is a curved graph.

Velocity and  Time Graph

A velocity-time graph is also a straight line. Here the time is on the x-axis while the velocity is on the y-axis. The product of time and velocity gives the displacement of an object moving at a uniform speed. The velocity of time and graph of a velocity that changes uniformly is a straight line. We can use this graph to calculate the acceleration of the object.

Acceleration =(Change in velocity)/time

For calculating acceleration draw a perpendicular on the x-axis from the graph point as shown in the figure. Here the acceleration will be equal to the slope of the velocity-time graph. Distance travelled will be equal to the area of the triangle, Therefore,

Distance traveled= (Base × Height)/2

Just like in the distance-time graph, when the velocity is non-uniform the velocity-time graph is a curved line.

Solved Examples for You

Question: The graph shows position as a function of time for an object moving along a straight line. During which time(s) is the object at rest?

• 0.5 seconds
• From 1 to 2 seconds
• 2.5 seconds

Solution: Option B. Slope of the curve under the position-time graph gives the instantaneous velocity of the object. The slope of the curve is zero only in the time interval 1 < t < 2 s. Thus the object is at rest (or velocity is zero) only from 1 to 2 s. Hence option B is correct.

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9A: One-Dimensional Motion Graphs

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• Page ID 2462

• Jeffrey W. Schnick
• Saint Anselm College

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Consider an object undergoing motion along a straight-line path, where the motion is characterized by a few consecutive time intervals during each of which the acceleration is constant but typically at a different constant value than it is for the adjacent specified time intervals. The acceleration undergoes abrupt changes in value at the end of each specified time interval. The abrupt change leads to a jump discontinuity in the Acceleration vs. Time Graph and a discontinuity in the slope (but not in the value) of the Velocity vs. Time Graph (thus, there is a “corner” or a “kink” in the trace of the Velocity vs. Time graph). The thing is, the trace of the Position vs. Time graph extends smoothly through those instants of time at which the acceleration changes. Even folks that get quite proficient at generating the graphs have a tendency to erroneously include a kink in the Position vs. Time graph at a point on the graph corresponding to an instant when the acceleration undergoes an abrupt change.

Your goals here all pertain to the motion of an object that moves along a straight line path at a constant acceleration during each of several time intervals but with an abrupt change in the value of the acceleration at the end of each time interval (except for the last one) to the new value of acceleration that pertains to the next time interval. Your goals for such motion are:

• Given a description (in words) of the motion of the object; produce a graph of position vs. time, a graph of velocity vs. time, and a graph of acceleration vs. time, for that motion.
• Given a graph of velocity vs. time, and the initial position of the object; produce a description of the motion, produce a graph of position vs. time, and produce a graph of acceleration vs. time.
• Given a graph of acceleration vs. time, the initial position of the object, and the initial velocity of the object; produce a description of the motion, produce a graph of position vs. time, and produce a graph of velocity vs. time.

The following example is provided to more clearly communicate what is expected of you and what you have to do to meet those expectations:

A car moves along a straight stretch of road upon which a start line has been painted. At the start of observations, the car is already \(225 m\) ahead of the start line and is moving forward at a steady \(15 m/s\). The car continues to move forward at \(15 m/s\) for \(5.0\) seconds. Then it begins to speed up. It speeds up steadily, obtaining a speed of \(35 m/s\) after another \(5.0\) seconds. As soon as its speed gets up to \(35 m/s\), the car begins to slow down. It slows steadily, coming to rest after another \(10.0\) seconds. Sketch the graphs of position vs. time, velocity vs. time, and acceleration vs. time pertaining to the motion of the car during the period of time addressed in the description of the motion. Label the key values on your graphs of velocity vs. time and acceleration vs. time. Solution Okay, we are asked to draw three graphs, each of which has the time, the same “stopwatch readings” plotted along the horizontal axis. The first thing I do is to ask myself whether the plotted lines/curves are going to extend both above and below the time axis. This helps to determine how long to draw the axes. Reading the description of motion in the case at hand, it is evident that: The car goes forward of the start line but it never goes behind the start line. So, the \(x\) vs. \(t\) graph will extend above the time axis (positive values of \(x\)) but not below it (negative values of \(x\)). The car does take on positive values of velocity, but it never backs up, that is, it never takes on negative values of velocity. So, the \(v\) vs. \(t\) graph will extend above the time axis but not below it. The car speeds up while it is moving forward (positive acceleration), and it slows down while it is moving forward (negative acceleration). So, the \(a\) vs. \(t\) graph will extend both above and below the time axis. Next, I draw the axes, first for \(x\) vs. \(t\), then directly below that set of axes, the axes for \(v\) vs. \(t\), and finally, directly below that, the axes for \(a\) vs. \(t\). Then I label the axes, both with the symbol used to represent the physical quantity being plotted along the axis and, in brackets, the units for that quantity. Now I need to put some tick marks on the time axis. To do so, I have to go back to the question to find the relevant time intervals. I’ve already read the question twice and I’m getting tired of reading it over and over again. This time I’ll take some notes: From my notes it is evident that the times run from \(0\) to \(20\) seconds and that labeling every \(5\) seconds would be convenient. So I put four tick marks on the time axis of \(x\) vs. \(t\). I label the origin \(0\), \(0\) and label the tick marks on the time axis \(5\), \(10\), \(15\), and \(20\) respectively. Then I draw vertical dotted lines, extending my time axis tick marks up and down the page through all the graphs. They all share the same times and this helps me ensure that the graphs relate properly to each other. In the following diagram we have the axes and the graph. Except for the labeling of key values I have described my work in a series of notes. To follow my work, please read the numbered notes, in order, from \(1\) to \(10\).

The key values on the \(v\) vs. \(t\) graph are givens so the only “mystery,” about the diagram above, that remains is, “How were the key values on \(a\) vs. \(t\) obtained?” Here are the answers:

On the time interval from \(t=5\) seconds to \(t=10\) seconds, the velocity changes from \(15\dfrac{m}{s}\) to \(35\dfrac{m}{s}\). Thus, on that time interval the acceleration is given by:

\[a=\dfrac{\triangle v}{\triangle t}=\dfrac{v_f-v_i}{t_f-t_i}=\dfrac{35\dfrac{m}{s}-15\dfrac{m}{s}}{10s-5s}=4\dfrac{m}{s^2} \nonumber \]

On the time interval from \(t=10\) seconds to \(t=20\) seconds, the velocity changes from \(35\dfrac{m}{s}\) to \(0\dfrac{m}{s}\). Thus, on that time interval the acceleration is given by:

\[a=\dfrac{\triangle v}{\triangle t}=\dfrac{v_f-v_i}{t_f-t_i}=\dfrac{0\dfrac{m}{s}-35\dfrac{m}{s}}{20s-10s}=-3.5\dfrac{m}{s^2} \nonumber \]

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1.2 Representations of Motion

13 min read • december 22, 2022

Daniella Garcia-Loos

Understanding Representations of Motion

In AP Physics 1, we study different representations of motion to understand and analyze the movement of objects. Some common types of representations include:

Graphical representations: These include position-time graphs , velocity-time graphs , and acceleration-time graphs . These graphs can be used to represent the motion of an object and to understand its characteristics, such as its speed and acceleration .

Numerical representations: These include tables or lists of numerical data that describe the motion of an object. These data may include position, velocity , and acceleration at different times.

Analytical representations: These include mathematical equations that describe the motion of an object. These equations may involve variables such as position, velocity , and acceleration , and can be used to make predictions about the motion of an object.

Diagrammatic representations: These include sketches or diagrams that show the position, velocity , and acceleration of an object at different times. These representations can help visualize and understand the motion of an object.

Overall, different representations of motion can be used to help understand the characteristics and behavior of moving objects, and can be useful tools for predicting and analyzing the motion of objects in different situations.

Center of Mass

The acceleration of the center of mass of a system is related to the net force exerted on the system, where a = F/m.

Key Vocabulary: Center of mass - a point on an object or system that is the mean position of the matter.

⟶ A force may be applied to this point to cause a linear acceleration without angular acceleration occurring.

4.A.1 Essential Knowledge

The linear motion of a system can be described by the displacement , velocity , and acceleration of its center of mass .

4.A.2 Essential Knowledge

The acceleration is equal to the rate of change of velocity with time, and velocity is equal to the rate of change of position with time.

Graphical Representations of Motion

As we covered in 1.1 Position, Velocity and Acceleration there are ways to represent all three quantities graphically. We also reviewed how to interpret these graphs,  but it is imperative to understand the relationships each graph has to one another. 

Image Courtesy of geogebra

From the image above, we can use position to find velocity for any given period of time by looking at the slope of the Position vs. Time Graph. Working from velocity to position we can look at the area underneath the curve to find displacement , but it is not possible to determine how far from the detector the object is located from a Velocity vs. Time Graph.  

When working from velocity to acceleration we look at the slope of the Velocity vs. Time Graph. Similarly, when working from acceleration to velocity we look at the area under the curve to find velocity . 

Linearization

Many graphs in physics will not be perfectly straight lines, but we can turn the curve into a straight sloped line! We accomplish this by squaring the x-axis value. This is called Linearization. This is a concept that will be used throughout the course, so get comfortable with it now!

Image Courtesy of x-engineer.org

Linearization of a graph is a method of approximating the behavior of a nonlinear function with a straight line. This can be useful for making predictions or understanding the general trend of the function. Here are some key points to consider when linearizing a graph:

A linear function is defined by a straight line, which can be represented by the equation y = mx + b, where m is the slope and b is the y-intercept.

A nonlinear function is any function that is not a straight line. Nonlinear functions can take many different shapes, including curves, loops, and jumps.

To linearize a nonlinear function, we need to find a straight line that is a good approximation of the function. This can be done by finding two points on the nonlinear function and drawing a straight line through them.

The equation of the tangent line can be found using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is a point on the function and m is the slope of the tangent line at that point.

The accuracy of the linear approximation depends on how closely the straight line follows the shape of the nonlinear function. The closer the line is to the function, the more accurate the approximation will be.

Linearization can be useful in physics because many physical systems can be modeled with nonlinear functions. By linearizing these functions, we can make predictions and understand the general trend of the system's behavior.

Example Problem

A ball is dropped from a height of 10 meters, and we want to find the time it takes to hit the ground. We know that the equation for the position of the ball as a function of time is:

y = 10 - 4.9t^2

However, this is a nonlinear equation and it is difficult to solve for t. Instead, we can use linearization to find an approximate solution.

To linearize the equation, we need to find the equation of the tangent line at a specific point on the curve. Let's choose the point where t = 1 second. At this point, the position of the ball is:

y = 10 - 4.9(1^2) = 5.1

Now we need to find the equation of the tangent line at this point. We can do this by using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope of the tangent line at that point.

Since the point on the curve is (1, 5.1), we can set y1 = 5.1 and x1 = 1. To find the slope of the tangent line, we can plot a second point on the curve that is close to (1, 5.1) and find the slope between the two points. For example, we can choose the point where t = 1.1 seconds:

y = 10 - 4.9(1.1^2) = 4.61

The slope between the two points is (4.61 - 5.1)/(1.1 - 1) = -0.49.

Now we can plug the values into the point-slope form of the linear equation to find the equation of the tangent line:

y - 5.1 = -0.49(x - 1)

y = -0.49x + 5.6

We can use this linear equation to approximate the position of the ball as it falls. For example, if we want to find the position of the ball after 0.5 seconds, we can substitute t = 0.5 into the linear equation to get:

y = -0.49*0.5 + 5.6 = 5.355

This means that the ball has fallen about 5.355 meters after 0.5 seconds.

This is just an approximate solution, but it is a much simpler calculation than solving the nonlinear equation for the position of the ball. Linearization can be useful for finding approximate solutions to problems involving nonlinear functions, especially when the nonlinear function is difficult to work with or when we only need an approximate solution.

⟶ Still feeling a little confused on Linearization? Don’t worry! Check out this video from AP Physics 1 Online for more practice!

Mathematical Representations of Motion

In Kinematics there are four major equations you must understand to begin calculations. They relate acceleration , displacement , initial and final velocity , and time together . 

Variable Interpretation: Δx is horizontal displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and a is acceleration in m/s/s. 

⟶ In order to solve for a variable without having all four other quantities known, we look at the ‘Variable Missing’ column to pick the equation that best suits our question. 

Key Vocabulary: Free Fall - an object only under the influence of gravity

Equation: velocity = force of gravity x time

Key Vocabulary: Acceleration due to Gravity - 9.8 m/s/s (it is acceptable to round up to 10 m/s/s on the AP Physics 1 exam)

Variable Interpretation: Δy is vertical displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and g is acceleration due to gravity in m/s/s. 

⟶ In free fall equations, we now replace Δx with Δy and a with g giving us a modified list of The Big Four as seen in the table above. 

Object Dropped (trip down)

Vo (initial velocity ) = 0 m/s

g works in the direction of motion

Object Tossed (trip up)

Vf (final velocity ) = 0 m/s

At maximum height of its trip, an object has a velocity of 0 m/s

g works against the direction of motion

Projectile Motion

Projectile motion is the motion of an object that is thrown, launched, or projected into the air and is then subject only to the force of gravity. Here are some key points to consider when solving projectile motion problems:

In projectile motion, the object moves in two dimensions: horizontally and vertically.

The horizontal motion of a projectile is uniform, which means it moves at a constant speed in a straight line. This is because there are no forces acting on the projectile in the horizontal direction.

The vertical motion of a projectile is affected by the force of gravity, which pulls the projectile downward. The force of gravity causes the projectile to accelerate downward at a rate of 9.8 m/s^2.

The path of a projectile is called its trajectory , and it is a parabolic curve.

The initial velocity of a projectile is a vector that specifies both the magnitude and direction of the projectile's motion. It can be broken down into its horizontal and vertical components.

The range of a projectile is the horizontal distance it travels from the starting point to where it lands.

The maximum height of a projectile is the highest point it reaches during its motion.

The time of flight of a projectile is the total time it takes to complete its motion from the starting point to the landing point.

The angle at which a projectile is launched has an effect on its range, maximum height, and time of flight.

Air resistance is ignored

When dealing with horizontal projectile launches, launches that have an entirely horizontal initial velocity (ex. a ball rolling off a table), we break it up into two sets of equations: vertical and horizontal. 

Variable Interpretation: Δy is vertical displacement in meters, Δx is horizontal displacement in meters, Vfy is vertical final velocity in meters/second, Voy is vertical initial velocity in meters/second, Vx is horizontal velocity in m/s, t is time in seconds, and g is acceleration due to gravity in m/s/s. 

As shown in the chart above, there are three equations we use for the vertical component of launches and one for the horizontal. You cannot put an x-component into a formula without a y-component!

⟶ Still feeling a little confused on Projectile Launches? Don’t worry! Check out this live stream from Fiveable for more practice!

Angled Motion

Key Vocabulary: Angled Launches - launches at an angle that includes both a horizontal and vertical component of initial velocity

Key Vocabulary: Vector Components - the horizontal and vertical parts of a vector

Angled launches require you to find the Vox and Voy, or vector components , based on the initial velocity Vo and the angle Ө. Now you can solve for the following: Vo, Vox, Voy, Ө, t, X, Ymax, and Vf. 

If an object is shooting upward, Voy is (+)

If an object is shooting downward, Voy is (-)

Vertical velocity is 0 m/s at the top

Flight is symmetric if the projectile starts and ends at the same height

Variable Interpretation: Voy is vertical initial velocity in meters/second, Vox is initial horizontal velocity in m/s, t is time in seconds, t is time in seconds, and g is acceleration due to gravity in m/s/s. 

⟶ Still feeling a little confused on Angled Launches ? Don’t worry! Check out this video from Khan Academy for more practice!  Want more practice - Check out the Fiveable Live streams on this topic:

🎥 Watch AP Physics 1 - 2D Motion & Freefall

🎥 Watch AP Physics 1 - Horizontal Launch Problems

🎥 Watch AP Physics 1 - Angle Launch Problems

Key Terms to Review ( 15 )

Acceleration

Acceleration-Time Graphs

Angled Launches

Center of mass

Cosine Function (cos)

Displacement

Initial Velocity Vector Components

Numerical Data

Position-Time Graphs

Sine Function (sin)

Vector Components

Velocity-Time Graphs

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AP®︎/College Physics 1

Course: ap®︎/college physics 1   >   unit 1, representations of motion.

• Deriving displacement as a function of time, acceleration, and initial velocity
• Plotting projectile displacement, acceleration, and velocity
• Position vs. time graphs
• Why distance is area under velocity-time line
• (Choice A)   CD A CD
• (Choice B)   BC B BC
• (Choice C)   DE C DE
• (Choice D)   AB D AB

Uniform Motion

• March 24, 2023
• Kinematics , Mechanics

Table of Contents

Introduction

Uniform motion plays a significant role in understanding the movement of objects in our everyday lives. In this article, we will learn about the concept of uniform motion in physics. We will also learn about its graphical representation along with the initial instant of time. In the end, I will discuss some of the frequently asked questions about this concept.

We will also provide a quiz at the end of the article where you can check your understanding of uniform motion in physics.

Understanding Uniform Motion:

Uniform motion definition:.

Uniform motion is defined as the motion of a particle whose coordinate (position) is a linear function of time.

The equation for this type of motion is $x = vt + b$, where $v$ and $b$ are constants. Any motion that does not fit this definition is considered variable motion .

Equation of Motion:

The equation $x = vt + b$, represents uniform motion, where $x$ is the coordinate of the particle, $v$ is the constant velocity, $t$ is the time, and $b$ is the initial position.

Graphical Representation of Uniform Motion:

Importance of graphs.

Illustrations, such as graphs, are crucial for understanding complex concepts in physics, including uniform and variable motion. They provide a visual representation of the motion, allowing for easier interpretation and analysis of the motion’s characteristics.

Visualizing Motion – Plotting a Graph:

The motion equation can be visually represented by plotting a graph. This is done by constructing a coordinate system with time (t) on the x-axis and the particle’s coordinate (x) on the y-axis.

To better understand this, consider the data from our article on Rectilinear Motion . The table given below shows the measurements of the car’s position in relation to the markers at regular intervals of time.

By plotting the values of the variables from the table on both axes, we proceed to create perpendicular lines from the axes at these points. The intersection of these perpendicular lines forms a series of points.

Drawing a smooth line connecting these points results in a motion graph. As the coordinate in uniform motion is a linear function of time, the resulting graph appears as a straight line as shown below in the figure.

The Initial Instant of Time:

The initial instant of time is the point at which an experiment or observation begins. It is not necessarily the beginning of the motion but rather the starting point of the study.

About Initial Coordinate:

The initial coordinate, denoted by $x_0$, is the distance from the moving particle to the origin of coordinates at the initial instant of time. This can be determined by setting $t = 0$ in the equation of motion, which gives $x_0 = b$.

Example of uniform motion

List of few other examples of uniform motion.

• The revolution or orbital motion of the Earth around the Sun. 
• Earth’s rotation about its axis
• A man-made satellite orbiting the Earth.
• Walking at a steady speed.
• The hour hand of the clock 
• A car traveling at a constant speed along a straight, flat route.
• A sewing machine’s vibrating spring.

Challenges and Limitations:

Although the uniform motion is a fundamental concept in physics, real-world applications often involve variable motion due to factors such as friction, air resistance, and changing forces. Therefore, uniform motion serves as a basis for understanding more complex motion scenarios.

Frequently asked questions

1. what is the difference between uniform and non-uniform motion with examples.

Uniform motion occurs when an object moves at a constant speed in a straight line. In this case, the object covers equal distances in equal intervals of time. For example, a car traveling at a constant speed of 60 km/h on a straight highway exhibits uniform motion.

Non-uniform motion , on the other hand, occurs when an object’s speed, direction, or both change over time. The object does not cover equal distances in equal intervals of time. For example, a roller coaster with varying speeds and changing directions demonstrates non-uniform motion.

2. What is the purpose of uniform motion in physics, and how is it used in everyday life?

Uniform motion is a fundamental concept in physics that helps us understand the motion of objects in the real world. It forms the basis for understanding more complex motion patterns and phenomena. In everyday life, uniform motion can be seen in various situations, such as a person jogging at a constant pace around a track or a conveyor belt moving at a steady speed in a factory.

3. How can one maintain uniform motion?

To maintain uniform motion, an object must maintain a constant speed and move in a straight line. This can be achieved by ensuring that no external forces act upon the object, or that the net force acting on the object is zero. In practice, this can be difficult due to factors such as friction, air resistance, and changes in terrain or environment.

4. What does the path of an object look like when it’s in uniform motion?

When an object is in uniform motion, its path is a straight line. This is because the object moves at a constant speed without changing direction. The graphical representation of uniform motion, with time on the horizontal axis and position on the vertical axis, would also appear as a straight line.

5. Does a body subjected to uniform motion always move in a straight line?

Yes, a body subjected to uniform motion always moves in a straight line. This is because uniform motion is characterized by constant speed and unchanging direction. Any deviation from a straight-line path would indicate a change in direction, which would mean the motion is no longer uniform.

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• Math Article

Graphical Representation

Graphical Representation is a way of analysing numerical data. It exhibits the relation between data, ideas, information and concepts in a diagram. It is easy to understand and it is one of the most important learning strategies. It always depends on the type of information in a particular domain. There are different types of graphical representation. Some of them are as follows:

• Line Graphs – Line graph or the linear graph is used to display the continuous data and it is useful for predicting future events over time.
• Bar Graphs – Bar Graph is used to display the category of data and it compares the data using solid bars to represent the quantities.
• Histograms – The graph that uses bars to represent the frequency of numerical data that are organised into intervals. Since all the intervals are equal and continuous, all the bars have the same width.
• Line Plot – It shows the frequency of data on a given number line. ‘ x ‘ is placed above a number line each time when that data occurs again.
• Frequency Table – The table shows the number of pieces of data that falls within the given interval.
• Circle Graph – Also known as the pie chart that shows the relationships of the parts of the whole. The circle is considered with 100% and the categories occupied is represented with that specific percentage like 15%, 56%, etc.
• Stem and Leaf Plot – In the stem and leaf plot, the data are organised from least value to the greatest value. The digits of the least place values from the leaves and the next place value digit forms the stems.
• Box and Whisker Plot – The plot diagram summarises the data by dividing into four parts. Box and whisker show the range (spread) and the middle ( median) of the data.

General Rules for Graphical Representation of Data

There are certain rules to effectively present the information in the graphical representation. They are:

• Suitable Title: Make sure that the appropriate title is given to the graph which indicates the subject of the presentation.
• Measurement Unit: Mention the measurement unit in the graph.
• Proper Scale: To represent the data in an accurate manner, choose a proper scale.
• Index: Index the appropriate colours, shades, lines, design in the graphs for better understanding.
• Data Sources: Include the source of information wherever it is necessary at the bottom of the graph.
• Keep it Simple: Construct a graph in an easy way that everyone can understand.
• Neat: Choose the correct size, fonts, colours etc in such a way that the graph should be a visual aid for the presentation of information.

Graphical Representation in Maths

In Mathematics, a graph is defined as a chart with statistical data, which are represented in the form of curves or lines drawn across the coordinate point plotted on its surface. It helps to study the relationship between two variables where it helps to measure the change in the variable amount with respect to another variable within a given interval of time. It helps to study the series distribution and frequency distribution for a given problem.  There are two types of graphs to visually depict the information. They are:

• Time Series Graphs – Example: Line Graph
• Frequency Distribution Graphs – Example: Frequency Polygon Graph

Principles of Graphical Representation

Algebraic principles are applied to all types of graphical representation of data. In graphs, it is represented using two lines called coordinate axes. The horizontal axis is denoted as the x-axis and the vertical axis is denoted as the y-axis. The point at which two lines intersect is called an origin ‘O’. Consider x-axis, the distance from the origin to the right side will take a positive value and the distance from the origin to the left side will take a negative value. Similarly, for the y-axis, the points above the origin will take a positive value, and the points below the origin will a negative value.

Generally, the frequency distribution is represented in four methods, namely

• Smoothed frequency graph
• Pie diagram
• Cumulative or ogive frequency graph
• Frequency Polygon

Merits of Using Graphs

Some of the merits of using graphs are as follows:

• The graph is easily understood by everyone without any prior knowledge.
• It saves time
• It allows us to relate and compare the data for different time periods
• It is used in statistics to determine the mean, median and mode for different data, as well as in the interpolation and the extrapolation of data.

Example for Frequency polygonGraph

Here are the steps to follow to find the frequency distribution of a frequency polygon and it is represented in a graphical way.

• Obtain the frequency distribution and find the midpoints of each class interval.
• Represent the midpoints along x-axis and frequencies along the y-axis.
• Plot the points corresponding to the frequency at each midpoint.
• Join these points, using lines in order.
• To complete the polygon, join the point at each end immediately to the lower or higher class marks on the x-axis.

Draw the frequency polygon for the following data

Mark the class interval along x-axis and frequencies along the y-axis.

Let assume that class interval 0-10 with frequency zero and 90-100 with frequency zero.

Now calculate the midpoint of the class interval.

Using the midpoint and the frequency value from the above table, plot the points A (5, 0), B (15, 4), C (25, 6), D (35, 8), E (45, 10), F (55, 12), G (65, 14), H (75, 7), I (85, 5) and J (95, 0).

To obtain the frequency polygon ABCDEFGHIJ, draw the line segments AB, BC, CD, DE, EF, FG, GH, HI, IJ, and connect all the points.

Frequently Asked Questions

What are the different types of graphical representation.

Some of the various types of graphical representation include:

• Line Graphs
• Frequency Table
• Circle Graph, etc.

Read More:  Types of Graphs

What are the Advantages of Graphical Method?

Some of the advantages of graphical representation are:

• It makes data more easily understandable.
• It saves time.
• It makes the comparison of data more efficient.

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Very useful for understand the basic concepts in simple and easy way. Its very useful to all students whether they are school students or college sudents

Thanks very much for the information

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