A glass windscreen,whose inclination with the | vedclass

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(A) Let the velocity of the car be $\vec{v}_c = 2\hat{i}$ and the velocity of the rain drops be $\vec{v}_r = -6\hat{j}$.The velocity of the rain drops

Vedclass · Accepted Answer

$	an^{-1}(3)$

Vedclass · Answer

(A) Let the velocity of the car be $\vec{v}_c = 2\hat{i}$ and the velocity of the rain drops be $\vec{v}_r = -6\hat{j}$.The velocity of the rain drops relative to the car is $\vec{v}_{rc} = \vec{v}_r - \vec{v}_c = -2\hat{i} - 6\hat{j}$.For the rain drops to strike the windscreen perpendicularly,the windscreen must be perpendicular to the relative velocity vector $\vec{v}_{rc}$.The angle $\alpha$ that the relative velocity vector $\vec{v}_{rc}$ makes with the vertical ($-y$ axis) is given by $	an \alpha = \frac{|v_x|}{|v_y|} = \frac{2}{6} = \frac{1}{3}$.However,the windscreen is inclined at an angle $\alpha$ with the vertical such that its normal is parallel to the relative velocity vector. The angle of the windscreen with the vertical is the same as the angle the relative velocity vector makes with the horizontal,which is $	heta = 	an^{-1}(\frac{6}{2}) = 	an^{-1}(3)$.

Column-$I$	Column-$II$
$(1)$ Velocity of $A$ relative to $B$ if $A$ and $B$ are moving perpendicular to each other	$(a)$ $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$
$(2)$ Velocity of rain drops relative to a man	$(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
	$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

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