A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20ms^{- 1}$ and d$30\ ms^{- 1}$ respectively. The speed of the body at the mid-point of $A$ and $B\ $ is nearest to

Question

Dr. Abubacker · Accepted Answer

(a) [image:3_image159.emf]
$\because v^{''} = \sqrt{\frac{{\left( v_{1} \right)^{2} + \left( v_{2} \right)}^{2}}{2}} = \sqrt{\frac{900 + 400}{2}} = \sqrt{650} = 25.5\ ms^{- 1}$

Q. A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20ms^{- 1}$ and d$30\ ms^{- 1}$ respectively. The speed of the body at the mid-point of $A$ and $B\ $ is nearest to

Explanation

Reviewed by Dr. Abubacker

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