# Question Video: Comparing the Distance Travelled by Accelerating Cars Physics • 9th Grade

Two cars travel along a straight road. Both cars accelerate at 5 m/s² as they travel. At point A, car I is at rest and car II has a velocity of 20 m/s. At point B, car II has a velocity of 30 m/s. Car I has a velocity of 30 m/s when it is at point C. What is the distance between point B and point C, to the nearest meter?

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### 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.