Answer to Question #143705 in Physics for Sydney

Question

Mr. Sheppard is driving along a country road at 24.5 m/s when a deer jumps out in front of his vehicle (mass of car = 1510 kg). If the deer was 35.3 m in front of him when he applied the brakes, and the coefficient of friction between the tires and asphalt is 0.890, where will Mr. Sheppard’s car come to a stop in relation to the deer?

Answer 1

$$V^2=2as=2\mu gs\\24.5^2=2(0.89)(9.8)s\\s=34.4\ m$$

The distance from the deer is

$$S-s=35.3-34.4=0.9\ m$$