Solve the following :


A car is going at a speed of $21.6 \mathrm{~km} / \mathrm{hr}$ when it encounters a $12.8 \mathrm{~m}$ long slope of angle $30^{\circ}$. The friction co-efficient between the road and the tyre is $\frac{\sqrt{3}}{2}$. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than $36 \mathrm{~km} / \mathrm{hr}$. Take $\mathrm{g}=10^{\mathrm{m} / \mathrm{s}^{2}}$


When driver applies brakes frictional force will be opposite to motion of car.

$a=g \sin 30^{\circ}-\mu g \cos 30^{\circ}$

$=\frac{g}{2}-\frac{1}{2 \sqrt{3}} \cdot g\left(\frac{\sqrt{3}}{2}\right)$

$a=\frac{g}{4}=\frac{10}{4}=2.5 \mathrm{~m} / \mathrm{s}^{2}$

$u=21.6 \frac{\mathrm{km}}{\mathrm{hr}}=21.6 \times \frac{1000}{60 \times 60}=6 \mathrm{~m} / \mathrm{s}$


We know that,

$v^{2}=u^{2}+2 a s$



$v=10 \times \frac{18}{5} \mathrm{Kmph}$

$\mathrm{y}=36 \mathrm{Kmph}$




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