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](https://tex.z-dn.net/?f=2aS%20%3D%20v_f%5E2-v_i%5E2%20%3D%20-v_i%5E2)
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](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B-v_i%5E2%7D%7B2S%7D%3D%20%5Cfrac%7B-%2815m%2Fs%29%5E2%7D%7B2%5Ccdot%2067%20m%7D%3D-1.68%20m%2Fs%5E2%20%20)
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](https://tex.z-dn.net/?f=F%3Dma)
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](https://tex.z-dn.net/?f=2aS%3D-v_i%5E2)