Question

A 1200-kg car going 30 m/s applies its brakes and skids to rest. If the friction force between the tires and the pavement is 6000N, how far does the car skid before coming to rest

Answer 1

Mass of car (m) = 1200 kg
Initial velocity (u) = 30 m /s
Final velocity (v) = 0 m / s
Force (F) = -6000 N (Force of friction is applied opposite to motion)
Acceleration (a) = $\frac{F}{m}$
a = $\frac{-6000}{1200}$ = -5 m s^-2
Using the third equation of motion , v² - u² = 2as
(0)² - (30)² = 2$\times$ -5 $\times$ s
Distance (s) = $\frac{900}{10}$ = 90 m
Hence , the car skid will go 90 m before coming to rest.