Answer to Question #259606 in Physics for teph

Question

The brakes of a car whose initial velocity is 30 m/s are applied, and the car receives an acceleration of −2 m/s2. How far will it have gone (a) when its velocity has decreased to 15 m/s and (b) when it has come to a stop?

Answer 1

(a)

$v^2=v_0^2+2ad,$ $d=\dfrac{v^2-v_0^2}{2a},$ $d=\dfrac{(15\ \dfrac{m}{s})^2-(30\ \dfrac{m}{s})^2}{2\times(-2.0\ \dfrac{m}{s^2})}=169\ m.$

(b)

$v^2=v_0^2+2ad,$ $d=\dfrac{v^2-v_0^2}{2a}$ $d=\dfrac{0-(30\ \dfrac{m}{s})^2}{2\times(-2.0\ \dfrac{m}{s^2})}=225\ m.$