Question

A car is moving at a speed of $21.6km/hr$ when it encounters a $12.8m$ long slope of angle ${30}^o$ (figure). The friction coefficient between the road and the tyre is $1/2\sqrt{3}$. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than equal to $36km/hr$. Take $g=10{m/s}^2$.

Answer 1

The initial velocity is given as,

$u=21.6\times \frac{5}{18}$

$=6m/s$

Let the mass of the car is $m$.

The normal force on the car by the incline is given as,

$F=mg\cos {30}^{\circ }$

$=\frac{\sqrt{3}}{2}mg$

The maximum friction force is given as,

$F_f=\mu F$

$=\frac{1}{2\sqrt{3}}\times \frac{\sqrt{3}}{2}mg$

$=\frac{1}{4}mg$

The acceleration is given as,

$a=\frac{F_f}{m}$

$a=\frac{m\times 10}{4m}$

$=2.5m/s^2$

The final velocity is given as,

$v^2=u^2+2as$

$v^2={\left(6\right)}^2+2\times 2.5\times 12.8$

$v=10m/s$

$=36km/h$

Thus, it is shown that the minimum speed of the car when it reaches the bottom is $36km/h$.