Rain is falling vertically with a speed of 12 | vedclass

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Question

(D) The velocity of rain is $\vec{v}_r = -12 \hat{j} \ ms^{-1}$.The velocity of the woman is $\vec{v}_w = -12 \hat{i} \ ms^{-1}$ (since she is moving

Vedclass · Accepted Answer

$45^{\circ}$,towards west

Vedclass · Answer

(D) The velocity of rain is $\vec{v}_r = -12 \hat{j} \ ms^{-1}$.The velocity of the woman is $\vec{v}_w = -12 \hat{i} \ ms^{-1}$ (since she is moving from east to west).The relative velocity of rain with respect to the woman is $\vec{v}_{rw} = \vec{v}_r - \vec{v}_w = -12 \hat{j} - (-12 \hat{i}) = 12 \hat{i} - 12 \hat{j} \ ms^{-1}$.To protect herself from rain,the woman must hold her umbrella in the direction of the relative velocity of the rain,which is opposite to $\vec{v}_{rw}$.The angle $	heta$ with the vertical is given by $	an 	heta = \frac{|v_w|}{|v_r|} = \frac{12}{12} = 1$ .Therefore,$	heta = 	an^{-1}(1) = 45^{\circ}$.Since the woman is moving towards the west,the relative velocity vector points towards the west,so she must hold the umbrella at an angle of $45^{\circ}$ towards the west to counteract the rain.

Rain is falling vertically with a speed of $12 \ ms^{-1}$. $A$ woman rides a bicycle with a speed of $12 \ ms^{-1}$ in the east to west direction. What is the direction in which she should hold her umbrella?

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