# Video: Calculating Acceleration from a Velocity-Time Graph

A car travels along a straight road. The velocity of the car is plotted against time in the following graph. Between which points on the graph is the magnitude of the acceleration of the car smallest?

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### Video Transcript

A car travels along a straight road. The velocity of the car is plotted against time in the following graph. Between which points on the graph is the magnitude of the acceleration of the car smallest? a) A and B only, b) B and C only, c) A and C, d) C and D, e) D and E.

The question references This graph on the right-hand side, with 𝑡, time, on the horizontal axis and 𝑣𝑥, velocity, on the vertical axis. And the question asks us to use this graph to determine where the magnitude of the acceleration of the car is smallest. So, to do this, we need to be able to interpret a velocity–versus–time graph to find acceleration.

Let’s start by recalling the definition of acceleration. Acceleration is the rate of change of velocity, which is another way of saying the change of velocity with time. Symbolically, we’d write 𝑎, the acceleration, is equal to Δ𝑣, the change in velocity, divided by Δ𝑡, the time over which that change took place. Let’s take a closer look at the right-hand side of this equation in the context of a velocity–time graph like the one pictured.

Δ𝑣, the change in velocity, is a change in the vertical coordinate of the graph. In other words, it’s what we call the rise of the graph. And Δ𝑡, the time in which that rise occurred, is just the change in the horizontal coordinate of the graph. In other words, it’s what we call the run. So, in the context of a velocity–versus–time graph, Δ𝑣 over Δ𝑡 is the rise of the graph divided by the run of the graph. And rise over run is the quantity that we know as slope. So, in the context of a velocity–versus–time graph, the acceleration at any given time is the slope of the graph at that point. We can use this to figure out what region of the graph has the smallest magnitude of acceleration.

First, let’s note that the graph consists of three straight lines. Straight lines have constant slope. And since the slope is the acceleration, each of these regions represents a time when the car had a different acceleration. So, let’s figure out the acceleration in each of the regions.

The first region is between 𝑡 equals zero and 𝑡 equals three 𝑡. Zero to three 𝑡 is a run of three 𝑡. And the associated rise is the rise from negative two 𝑣 to 𝑣, which is just a total change of three 𝑣. So, the rise is three 𝑣, and the acceleration is three 𝑣 divided by three 𝑡. Three divided by three is one. And so, the acceleration in this first region is just 𝑣 over 𝑡.

In the second region, the run is from three 𝑡 to four 𝑡, which is just a run of 𝑡. In this region, the velocity is constant. It’s 𝑣 all the way between C and D. So, the total rise is zero. And the acceleration is zero divided by 𝑡, which is just zero. This is actually what it means for an object to have zero acceleration; the velocity doesn’t change.

In this last region, the run is from four 𝑡 to five 𝑡, which is again a total run of 𝑡. But this time, the velocity does change. It drops from an initial value of 𝑣 to a final value of zero, which is a rise of negative 𝑣. So, the acceleration in this third region is negative 𝑣 divided by 𝑡. Since we’re looking only for the magnitude of the acceleration, we can drop the negative sign and say the magnitude of the acceleration in this third region is also 𝑣 over 𝑡, just like in the first region.

Now that we know the values for acceleration, we can evaluate the answer choices. Even without knowing the values of acceleration, we could determine the choices a and b are not the right answer. This is because the region specified in choice a between points A and B and the region specified in choice b between points B and C lie along the same straight line, which means they have the same acceleration.

Choices a and b also each individually claim to represent the only such region with the smallest acceleration. But if between points A and B has the smallest acceleration, then so does the region between points B and C since they have the same acceleration. And neither choice a nor choice b is the only such region. So, both cannot possibly be correct.

So, we’re left with choices c, d, and e. Choice c is the region between points A and C; choice d, the region between points C and D; and choice e, the region between point D and E. But these are the three regions for which we’ve already worked out the acceleration. Between points A and C, the acceleration is that of the first region, 𝑣 over 𝑡. Between points C and D, the acceleration is zero. And between points D and E, the acceleration is negative 𝑣 over 𝑡, which has magnitude 𝑣 over 𝑡. So, of 𝑣 over 𝑡, zero, and 𝑣 over 𝑡, the smallest magnitude is clearly zero. So, the magnitude of the acceleration of the car is smallest between points C and D. And this is answer choice d.