Question

A sports car skids to a stop, leaving skid marks 290 m long.

If the coefficient of kinetic friction between tires and pavement is 0.80, how fast was the car going before the skid?

Answer 1

67.4 m/s

Step-by-step explanation:

The force acting on the car, and that causes the car to slow down, is the force of friction, which is given by:

`F_f = -\mu mg`

where

`\mu_k = 0.80` is the coefficient of kinetic friction

m is the mass of the car

`g=9.8 m/s^2` is the acceleration of gravity

According to Newton's second law:

`F=ma`

where F is the net force on the car and a its acceleration. Comparing the two equations, we find an expression for the acceleration:

`ma=-\mu mg\\a=-\mu g` (1)

Since the motion of the car is a uniformly accelerated motion, we can use the following equation:

`v^2-u^2=2as`

where

v = 0 is the final velocity of the car

u is the initial velocity

a is the acceleration

s = 290 m is the distance covered by the car while slowing down

Using (1) and solving for u, we find the initial velocity:

`v^2-u^2 = -2\mu g s\\u=√(2\mu g s)=√(2(0.80)(9.8)(290))=67.4 m/s`