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 3.6 m/s2 for 4.5 seconds. It then continues at a constant speed for 7.3 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 201 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.

How fast is the blue car going 3.6 seconds after it starts? Answer: 12.96 m/s
How fast is the blue car going 8.8 seconds after it starts? Answer: 16.2 m/s
How far does the blue car travel before its brakes are applied to slow down? Answer: 154.71 m/s

These problems I had difficulty solving for. So any help is greatly appreciated. Thanks.

What is the acceleration of the blue car once the brakes are applied?
What is the total time the blue car is moving?
What is the acceleration of the yellow car?

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  1. at 3.6 seconds v = a t = 3.6*3.6 = 12.96 m/s

    at 4.5 s
    v = a t = 3.6*4.5 = 16.2 m/s
    x(position) = d (distance traveled)
    = average speed (16.2/2) * time (4.5)
    = 8.1 m/s * 4.5 s = 36.45 meters

    speed remains 16.2 until t = 4.5+7.3 = 11.8 seconds

    at 4.5 + 7.3 = 11.8 seconds
    v = 16.2 still
    d = 16.2 * 7.3 = 118.26 m
    x = 118.26 + 36.45 = 154.71 meters

    at end
    v = 0
    average v = (16.2+0)/2 = 8.1 m/s
    d = 201 - 154.71 = 46.29 meters to stop
    time to stop = 46.29/8.1 = 5.71
    acceleration while braking = (0-16.2)/5.71 = -2.84 m/s^2
    total time = 5.71 + 11.8 = 17.5 s

    yellow car goes 201 meters in 17.5 seconds
    201 = a t^2 = a (17.5)^2
    a yellow = .656 m/s^2

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  2. formula is wrong in the last problem. 1/2 is missing before a.

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