Question

A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to stop if it was instead traveling at 45 m/s?

Answer 1

1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
$2aS = v_f^2-v_i^2 = -v_i^2$
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
$a= \frac{-v_i^2}{2S}= \frac{-(15m/s)^2}{2\cdot 67 m}=-1.68 m/s^2$

2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
$F=ma$
This force will be constant, and since m is always the same, then a is the same even in the second situation.

3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
$2aS=-v_i^2$
$S= \frac{-v_i^2}{2a} = \frac{-(45 m/s)^2}{2\cdot (-1.68 m/s^2)}=603 m$