### Video Transcript

Two cars travel along a straight
road. Both cars accelerate at five meters
per second squared as they travel. At point A, car I is at rest and
car II has a velocity of 20 meters per second. At point B, car II has a velocity
of 30 meters per second. Car I has a velocity of 30 meters
per second when it is at point C. What is the distance between point
B and point C to the nearest meter?

The question is telling us that two
cars are traveling along a straight road. Both cars pass point A, and the
question is asking us to calculate the distance between point B and point C. We’re told that both cars
accelerate at five meters per second squared along a straight road. This means that the cars don’t
change direction, so we can assume that their acceleration is constant.

At point A, car I is at rest. In this context, the phrase “at
rest” means “not moving.” Therefore, car I’s initial velocity
at point A is zero meters per second. As for car II, we’re told that it
has an initial velocity of 20 meters per second at point A. Both cars accelerate and reach a
velocity of 30 meters per second, but at different points. Car I reaches 30 meters per second
at point C, while car II reaches 30 meters per second at point B.

The simplest way to calculate the
distance between point B and point C using the information given will be in three
steps. First, we will calculate the
distance traveled by car I to get from point A to point C. Then, we’ll calculate the distance
traveled by car II to get from point A to point B. After that, we can calculate the
difference between these two distances to find the distance between point B and
point C.

Since this question has given us
the initial velocity, final velocity, and acceleration of each car and is asking us
to find a distance, we need to use an equation of motion that combines all of these
quantities to help us answer the question. This type of equation is known as a
kinematic equation. The particular kinematic equation
that we want here is 𝑣 squared equals 𝑢 squared plus two times 𝑎 times 𝑠, where
𝑣 is the final velocity, 𝑢 is the initial velocity, 𝑎 is the acceleration, and 𝑠
is the distance moved.

In all the cases we’ll need this
equation, we will be solving for the distance 𝑠. So, we need to rearrange the
equation in order to make 𝑠 the subject. First, we can subtract 𝑢 squared
from both sides. After canceling out the positive
and negative 𝑢 squared terms on the right-hand side, we are left with 𝑣 squared
minus 𝑢 squared equals two times 𝑎 times 𝑠. Let’s now divide both sides by two
𝑎. Canceling two 𝑎 from the numerator
and denominator on the right, we have that 𝑠 is equal to 𝑣 squared minus 𝑢
squared all divided by two 𝑎.

Let’s use this equation to
calculate the distance traveled by car I to get from point A to point C. Car I’s initial velocity at point A
is zero meters per second, and we’ll label this as 𝑢 one. Its final velocity at point C is 30
meters per second, which we’ll label as 𝑣 one. Finally, the car’s acceleration is
five meters per second squared, which is our value of 𝑎.

After substituting these values
into this equation, we have that the distance from A to C, which we’ll call 𝑠 one,
is equal to the square of 30 meters per second minus the square of zero meters per
second all divided by two times five meters per second squared. This works out as 90 meters. So, we’ve found that the distance
car I travels to get from point A to point C is 90 meters, thus completing step
one.

Let’s now calculate the distance
traveled by car II to get from point A to point B, using the same method as we did
for car I. Car II’s initial velocity is 20
meters per second, and we’ll label this as 𝑢 two. Its final velocity is 30 meters per
second, and we’ll call this 𝑣 two. Its acceleration is the same value,
𝑎 equals five meters per second squared, that we had for car I.

Substituting these values into this
equation, we have that the distance from A to B, which we’ll call 𝑠 two, is equal
to the square of 30 meters per second minus the square of 20 meters per second all
divided by two times five meters per second squared. Evaluating this gives a result for
𝑠 two of 50 meters. This is our value for the distance
from point A to point B.

Now that we’ve calculated the
distance from A to B as well as that from A to C, the last step that we need to take
is to calculate the difference between these two distances that we have just
found. This will be the distance between
point B and point C. This distance will be equal to 𝑠
one minus 𝑠 two, which is 90 meters minus 50 meters. This is equal to 40 meters. This completes our third step. The question asks us to give the
answer to the nearest meter. Our answer is 40 meters, which is
already an integer number of meters. Therefore, we don’t need to round
this result; we already have our answer. The distance between point B and
point C is equal to 40 meters.