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

Answer:

67.4 m/s

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:

$v^2-u^2 = -2\mu g s\\u=\sqrt{2\mu g s}=\sqrt{2(0.80)(9.8)(290)}=67.4 m/s$