Question

A car is going at a speed of $6\text{m/s}$ when it encounters a $15\text{m}$ slope of angle ${30}^{\circ }$. The friction coefficient between the road and tyre is $0.5$. The minimum speed of car with which it can reach the bottom is $(g=10{\text{m/s}}^2)$

• A. $4\text{m/s}$
• B. $3\text{m/s}$
• C. $7.49\text{m/s}$
• D. $8.45\text{m/s}$

Answer 1

The correct option is C $7.49\text{m/s}$

Acceleration of car along slope
$=g\sin \theta -\mu g\cos \theta$
$=10\frac{1}{2}-\left(0.5\right)\left(10\right)\left(\frac{\sqrt{3}}{2}\right)=5-4.33$
$=0.67{\text{ms}}^{-2}$
Now from $v^2=u^2+2as$
$v=\sqrt{6^2+2\left(0.67\right)\left(15\right)}$
$=\sqrt{36+20.1}=\sqrt{56.1}$
$=7.49{\text{ms}}^{-1}$