A car travels with a speed of 40 text{ km h}^ | vedclass

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(C) Let $v_{CE}$ be the velocity of the car with respect to the Earth,$v_{RE}$ be the velocity of the rain with respect to the Earth,and $v_{RC}$ be t

Vedclass · Accepted Answer

$80 	ext{ km h}^{-1}$

Vedclass · Answer

(C) Let $v_{CE}$ be the velocity of the car with respect to the Earth,$v_{RE}$ be the velocity of the rain with respect to the Earth,and $v_{RC}$ be the velocity of the rain with respect to the car.Given: $v_{CE} = 40 	ext{ km h}^{-1}$.The rain falls vertically,so $v_{RE}$ is along the vertical direction.The trace of the rain on the side window makes an angle $	heta = 30^{\circ}$ with the vertical.From the vector triangle of relative velocity,we have $\vec{v}_{RC} = \vec{v}_{RE} - \vec{v}_{CE}$.In the right-angled triangle formed by these vectors,the side representing the car's velocity $v_{CE}$ is opposite to the angle $	heta = 30^{\circ}$.Therefore,$\sin 	heta = \frac{v_{CE}}{v_{RC}}$.Rearranging for $v_{RC}$,we get $v_{RC} = \frac{v_{CE}}{\sin 30^{\circ}}$.Substituting the values,$v_{RC} = \frac{40}{0.5} = 80 	ext{ km h}^{-1}$.

$A$ car travels with a speed of $40 \text{ km h}^{-1}$. Rain drops are falling at a constant speed vertically. The traces of the rain on the side windows of the car make an angle of $30^{\circ}$ with the vertical. The magnitude of the velocity of the rain with respect to the car is

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