# Kinematics of two cars

In summary, the blue car accelerates uniformly at a rate of 4.2 m/s2 for 3.2 seconds, then travels at a constant speed for 13.2 seconds before applying the brakes and coming to a stop 220.0 meters from its starting point. The yellow car also accelerates uniformly for the entire distance, catching the blue car just as it comes to a stop. To find the acceleration of the blue car once the brakes are applied, the equation Vf^2=Vi^2+2ax is used, resulting in a value of -3.8 or -3.84 for "a".

## Homework Statement

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 4.2 m/s2 for 3.2 seconds. It then continues at a constant speed for 13.2 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 220.0 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

What is the acceleration of the blue car once the brakes are applied

## The Attempt at a Solution

I found the distance traveled before the brakes were applied which was 198 m. I also found the speed at which it was constant which came out to be 13.44 m/s. These two values i have been confirmed is correct. I found the distance traveled after the brakes are applied which came out to be 22m (220.0-198).

So then i simply plugged the values into Vf^2=Vi^2+2ax → 0=(13.44)^2+2a(22)and solved for "a" but i got the wrong answer?

Look ok to me.

By the way, what is the yellow car doing in this problem.

I also agree that your work looks fine. However, I got 199 m instead of 198 m for the distance to where the blue car applies the brakes. Won't make much difference in the answer for the acceleration, but if you're plugging answers into on online grading system, then it might make a difference if they don't allow much tolerance.

Also, did you include the correct sign for the acceleration in your answer?

yes... very picky online grading system indeed. Thanks for your time guys :)

TRY this
0=(13)^2+2a(22)