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(B) Let the acceleration of the car be $a$ and the distance between $P$ and $Q$ be $s$.Using the equation of motion $v^2 = u^2 + 2as$ for the path $PQ

Vedclass · Accepted Answer

$25\sqrt{2}\; km/h$

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(B) Let the acceleration of the car be $a$ and the distance between $P$ and $Q$ be $s$.Using the equation of motion $v^2 = u^2 + 2as$ for the path $PQ$:$40^2 = 30^2 + 2as$$1600 = 900 + 2as$$2as = 700$$as = 350$Let $V$ be the velocity at the midpoint of $PQ$. The distance from $P$ to the midpoint is $s/2$.Using the equation of motion $V^2 = u^2 + 2a(s/2)$:$V^2 = 30^2 + as$$V^2 = 900 + 350$$V^2 = 1250$$V = \sqrt{1250} = \sqrt{625 	imes 2} = 25\sqrt{2}\; km/h$.Alternatively,for uniform acceleration,the velocity at the midpoint $V_{mid}$ is given by:$V_{mid} = \sqrt{\frac{v_P^2 + v_Q^2}{2}}$$V_{mid} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{900 + 1600}{2}} = \sqrt{\frac{2500}{2}} = \sqrt{1250} = 25\sqrt{2}\; km/h$.

$A$ car is moving along a straight road with uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance $s$ with velocities $30\; km/h$ and $40\; km/h$ respectively. Find the velocity of the car midway between $P$ and $Q$.

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