Question

A car is traveling at a velocity of 20 m/s. the driver applies the brakes, and the car slows with an acceleration of a = -4.0 meters per second squared. what is the stopping difference of the car?

Answer 1

   
$\displaystyle\\ V = a\times t\\\\ t= \frac{V}{a} = \frac{20}{4} = 5~s\\\\ D = \frac{a\times t^2}{2} = \frac{4\times 5^2}{2} = 2 \times 25 = 50~m$