Question

A car of mass 875 kg is traveling 30.0 m/s when the driver applies the brakes, which lock the wheels. The car skids for 5.60 s in the positive x - direction before coming to rest. (a) What is the car’s acceleration? (b) What magnitude force acted on the car during this time? (c) How far did the car travel?

Answer 1

Explanation:

It is given that,

Mass of the car, m = 875 kg

Initial speed of the car, u = 30 m/s

Brakes are applied i.e. v = 0

The car skids for 5.60 s in the positive x - direction before coming to rest, t = 5.6

(a) Acceleration of the car, $a=\dfrac{v-u}{t}$

$a=\dfrac{0-30}{5.6}$

$a=-5.35\ m/s^2$

(b) Force, F = ma

$F=875\ kg\times -5.35\ m/s^2$

F = -4681.25 N

So, the force of 4681.25 N is acting on the car.

(c) Let x is the distance covered by the car. So,

$v^2-u^2=2ax$

$0-u^2=2ax$

$x=\dfrac{-u^2}{2a}$

$x=\dfrac{-(30\ m/s)^2}{2\times -5.35\ m/s^2}$

x = 84.11 meters

So, the distance covered by the car is 84.11 meters. Hence, this is the required solution.