Answer to Question #316296 in Mechanics | Relativity for Daniel

Question

A car is preceding down a 1 in 10 incline at 65KPH when it's brake are abruptly applied. How far does it skid along the road before stopping if the wheels lock and the coefficient of friction between the tyres and the road is 0.7 ?

Answer 1

From Newton's second law:

$mgsin\alpha-F_f=ma,$ (1)

where $\alpha=arctan(\frac{1}{10})=6\degree$- the angle of road incline;

$F_f=\mu mgcos\alpha$ - the force of friction;

$\mu$ - the coefficient of friction;

$a$ - the acceleration along the road.

From (1)

$a=g(sin\alpha-\mu cos\alpha)=9.8(sin6\degree-0.7cos6\degree)=-5.79\space m/s^2.$

Let find the time t from applaing the brake to stop:

$v_0+at=0$, (2)

where $v_0=65\space KPH=18.1\space m/s$ - the initial speed of the car.

From (2)

$t=-\frac{v_0}{a}=-\frac{18.1}{-5.79}=3.13\space s$

The distance, that car skid along the road before stopping is

$s=v_0t+\frac{at^2}{2}=18.1\cdot 3.13+\frac{(-5.79)\cdot 3.13^2}{2}=28.3\space m.$