Interpreting graphs of motion

A key element of the California content standards for forces and motion, and of the CSET Science I test, is that students should know how to interpret graphs of position versus time and graphs of speed versus time, for motion in a single direction. In this topic we will survey the various forms of such graphs, and link to websites with more examples.

Position-time or P-T graphs

A typical position-time graph plots time along the horizontal or x-axis and position along the vertical or y-axis.

Consider graphs of position versus time for an object that is

not moving
moving in the positive direction with velocity V1
moving in the positive direction with velocity V2, which is higher than V1

In the first case, the position of the object does not change. Say the object is at position 12 in some arbitrary units. It is at 12 at time t = 0, and it is at 12 at any time later, say t = 35. The P-T graph for a stationary object is thus a straight horizontal line, indicating a constant position with time:

If the object is moving in the positive direction, its position value will increase with time. Say the object is at position 12 (in some arbitrary units) from t = 0 to t = 5, then starts to move uniformly in the positive direction. At t = 20 it reaches position 19. The P-T graph looks like this:

We can use the graph to find the position of the object at any time from t = 0 to t = 20. For example, at t = 10 (x-axis value of 10) the purple line showing the position of the object is at 14 on the y-axis.

The velocity of the object can be found from the slope of the graph. (Some students remember how to calculate the slope as "the rise over the run," i.e., the change in y divided by the change in x.) Between t = 0 and t = 5, the change in time is 5, and the change in y is 0 (no change — the value stays constant at 12). The slope is 0 (because $\frac{0}{5} = 0$ ) and the velocity is 0.

The slope of the line is constant between t = 5 and t = 20, indicating a constant speed or velocity. We can use any part of the purple line between t = 5 and t = 20 to find the slope. For example, between t = 10 and t = 20, the position goes from 14 to 19. The change in position is (19 - 14 = 5) and the change in time is 10. The slope is then $\frac{5}{10}$ or $\frac{1}{2}$ . If the units of position were meters and the units of time were seconds, we would give the velocity (or the slope of the position-time graph) as $\frac{1}{2} {ms}^{- 1}$ .

Now the object changes to a higher uniform speed or velocity. At t = 20, it is at position 19, and at t = 22, it is at position 25. The position-time graph looks like this:

Again, we can find the new velocity by measuring the slope of the graph. To find the new, higher velocity, we compute the slope somewhere between $t = 20$ and $t = 22$ .

Change in position is $25 - 19 = 6$
Change in time corresponding to this change of position: $22 - 20 = 2$
Slope is $\frac{6}{2} = 3$

If we computed the slope over the whole time period, t=0 to t=22, we would be finding the average velocity over that time interval.

Change in position would be $25 - 12 = 13$
Change in time would be $22 - 0 = 22$
Slope would be $\frac{13}{22}$ or $0.6$

Notice that when the object went from a velocity of $\frac{1}{2}$ (in arbitrary units) to a velocity of 3 (same arbitrary units) the purple line showed a change of direction as the slope became steeper.

If an object is accelerating it will produce a line on a position-time graph that has an ever-increasing slope. If the object is accelerating uniformly, the slope will increase uniformly in each time interval. Here is a position-time graph for an object that accelerates uniformly from t = 5 onward, starting at position 6:

What is the meaning of the following P-T graph?

From t = 0 to t = 7, the object is at rest: its position is not changing with time.
From t = 7 to t = 13, the object is moving in the positive direction: its position is increasing with time. Its velocity can be found from the slope of the line: a speed of (15 - 5)/(13 - 7) = 1.7 in the positive-x direction.
From t = 13 to t = 26, the object is moving in the negative direction. It starts at position 15, and ends at position 12. Again, the slope gives the velocity: (12 - 15)/(26 - 13) = -3/13 or -0.23. This is a speed of 0.23 in the negative-x direction.

In summary, the slope of the line on a P-T graph gives the velocity of the object. A horizontal line, with a slope of 0, implies a velocity of zero, i.e., an object at rest. A positive slope implies an object whose position is increasing with time, i.e., an object moving in the positive-x direction. A negative slope implies an object moving in the negative-x direction. The steeper the slope, the higher the speed (the magnitude of the velocity).

For more practice interpreting position-time graphs, see this lesson from The Physics Classroom tutorial.

Velocity-time graphs

A velocity-time graph plots time along the x-axis and velocity along the y-axis.

Let's consider again an object at rest and an object moving with different positive velocities.

An object at rest has zero velocity, which is represented on the V-T graph as a horizontal line at `y = 0`:

An object — which initially is at rest for 3 seconds, and then moves with a constant, positive velocity, say 7 meters per second — would have a velocity-time graph like this:

At $t = 3$ in this graph, the velocity instantaneously changes from 0 to 7 meters per second, then stays constant at $7 {ms}^{- 1}$ .

A real object would require some time to go from 0 to 7 meters per second, but this might happen over a time scale too short to appear on the graph.

If the object reversed its course after $t = 13$ and traveled in the negative-x direction at a speed of 3 meters per second, the V-T graph would appear like this:

Notice that the graph shown here was extended in the negative-y direction to show a negative velocity — that is, a velocity in the opposite direction from that in the earlier time period.

What does a sloped line indicate on a V-T graph?

At $t = 0$ the velocity is $0$
At $t = 10$ , the velocity is $5 {ms}^{- 1}$
At $t = 20$ , the velocity is $10 {ms}^{- 1}$
At $t = 30$ , the velocity is $15 {ms}^{- 1}$

These data points indicate that the object is accelerating. In each 10-second interval, it increases its velocity by $5 {ms}^{- 1}$ . (That means that in each second, it increases its velocity by $0.5 {ms}^{- 1}$ ; the acceleration is 0.5 meters per second per second, or $0.5 {ms}^{- 2}$ .)

In general, the slope of a line on a V-T graph gives the acceleration. If the slope is, say, 7.3 meters per second "rise" or change in y divided by 3 seconds "run" or change in x, then the acceleration is $\frac{7.3}{3}$ or $2.4 {ms}^{- 2}$ . Any two points on the line can be chosen for the calculation (assuming the line does not change slope between those two points). The answer should come out the same, no matter which two points are chosen.

There is something special and important to remember about velocity-time graphs: The area "under the curve" on a V-T graph gives the displacement of the object. To put it more precisely, the area of the space between the plotted line and the axes is the displacement (in units of distance).

In the example shown here, the object accelerates from rest for 10 seconds, then maintains a constant velocity of $17 {ms}^{- 1}$ from $t = 10$ to $t = 13$ , and comes to rest.

The area we want to calculate is shown in orange:

We can find the area simply by counting squares and partial squares, or by dividing up the orange area into shapes for which we can calculate the area:

Area 1 is the area of a triangle with base 10 and height 17. The area is then
$\frac{10 x 17}{2} = 85$
Area 2 is the area of a rectangle with one side 3 and another side 17. The area is then $17 x 3 = 51$ .
The total area is $85 + 51 = 136$
What are the units? A unit square on this graph is 1 second across by $1 {ms}^{- 1}$ high. So the units are $[seconds] x [\frac{meters}{second}]$ and the seconds cancel, leaving just meters — the units of displacement.

This means that between time $t = 0$ and $t = 13$ , the object traveled 136 meters.

For more information and practice interpreting V-T graphs, see the following pages of the Physics Classroom tutorial: