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(B) Let the acceleration of the body be $a$ and the distance between points $A$ and $B$ be $d$. Using the equation of motion $v^2 = u^2 + 2as$: For th

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$25.5 \,m/s$

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(B) Let the acceleration of the body be $a$ and the distance between points $A$ and $B$ be $d$. Using the equation of motion $v^2 = u^2 + 2as$: For the path from $A$ to $B$: $(30)^2 = (20)^2 + 2ad$ $900 = 400 + 2ad$ $2ad = 500$ $ad = 250$ Let $v_m$ be the velocity at the midpoint of $AB$. The distance from $A$ to the midpoint is $d/2$. Using the equation of motion for the path from $A$ to the midpoint: $v_m^2 = (20)^2 + 2a(d/2)$ $v_m^2 = 400 + ad$ Substituting $ad = 250$: $v_m^2 = 400 + 250 = 650$ $v_m = \sqrt{650} \approx 25.495 \,m/s \approx 25.5 \,m/s$.

$A$ body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 \,m/s$ and $30 \,m/s$ respectively. The speed of the body at the midpoint of $A$ and $B$ is (nearly)

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