Question

A driver in a car,originally moving 12.2 m/s, applies the brakes until the car comes to a stop. The car moves a distance of 36.5 m while braking. How much time did it take for the car to stop? Assume constant acceleration during braking.

Answer 1

First we have to calculate the acceleration of the car,

$a =\frac{u-v}{t}$

Here, u is initial velocity of the car and its value is given 12.2 m/s and v is final velocity of the car and it comes to stop, so its value zero.

$a=\frac{0-12.2 m/s}{t} =\frac{-12.2 \ m/s}{t}$.

As during the braking the acceleration is constant, from the kinematic equation,

$s=ut + \frac{1}{2} a t^2$

Here, s is the distance traveled by the car during braking and its value is given 36. 5 m.

Substituting all the values in kinematic equation, we get

$36.5 m =(12.2 m/s) t + \frac{1}{2} (- \frac{12.2 m/s}{t}) t^2 \\\\ 36.5 \ m = (12.2 \ m/s) t -(6.1 \ m/s) t \\\\\ t = \frac{36.5 m}{6.1 m/s} = 5.98 s$

Therefore, car will stop after 5.98 s.