### 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.