Question

A truck travelling with uniform acceleration crosses two points $A \& B$ with velocities $60\, m / s$ and $40\, m / s$ respectively. The speed of the body at the mid point of $A$ and $B$ is nearest to:

• A. 17 m / s
• B. 20 m / s
• C. 19.49 m / s
• D. 50.9 m / s

Answer 1

Answer: (D)

As we know the relation $V_{\text {mid }}=\sqrt{\frac{V_{A}^{2}+V_{B}^{2}}{2}}$
$V _{ mid }=\sqrt{\frac{(60)^{2}+(40)^{2}}{2}}=\sqrt{\frac{3600+1600}{2}}$
$V _{\text {mid }}=\sqrt{\frac{5200}{2}}=\sqrt{2600}=50.9 m / s$