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(C) Let the velocity of car $A$ be $v_A = 100 	ext{ km/h}$ and car $B$ be $v_B = 80 	ext{ km/h}$.Relative velocity of $A$ with respect to $B$ is $v_

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(C) Let the velocity of car $A$ be $v_A = 100 	ext{ km/h}$ and car $B$ be $v_B = 80 	ext{ km/h}$.Relative velocity of $A$ with respect to $B$ is $v_{AB} = v_A - v_B = 100 - 80 = 20 	ext{ km/h}$.Converting this to $	ext{m/s}$,we get $20 	imes (5/18) = 5.55 	ext{ m/s}$.Let the velocity of the stone relative to car $B$ be $v$. Since the stone is thrown towards $A$,its velocity relative to the ground is $v_{sg} = v + v_B$.The velocity of the stone relative to car $A$ is $v_{sa} = v_{sg} - v_A = (v + v_B) - v_A = v - (v_A - v_B) = v - 20 	ext{ km/h}$.Given that the speed of the stone relative to $A$ is $5 	ext{ m/s}$,which is $5 	imes (18/5) = 18 	ext{ km/h}$.Thus,$|v - 20| = 18$.Since the stone must hit car $A$ (which is ahead),$v$ must be greater than $20 	ext{ km/h}$.Therefore,$v - 20 = 18$,which gives $v = 38 	ext{ km/h}$.

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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