Question

driver is traveling at 31.0 m/s when he sees a moose crossing the road 92.0 m ahead. The moose becomes distracted and stops in the middle of the road. The driver of the car slams on the brakes? (In British Columbia, Canada, alone, there are over 4000 moose-car accidents each year.)
1)
What is the minimum constant acceleration he must undergo to stop short of the moose and avert an accident?

Answer 1

Explanation:

First of all it is not acceleration it is retardation

now ,

distance S = 92m

initial velocity u = 31m/s

final velocity v = 0 m/s

now according to third rule of speed

${v}^{2} = {u}^{2} + 2as \\ 0 = {31}^{2} + 2 \times a \times 92 \\ 0 = 961 + 184a \\ - a = 961 \div 184 \\ - a = 5.22 \frac{m}{ {s}^{2} }$

thus the answer is -5.22 m/s²