A body starts with an initial velocity of $10 {ms}^{- 1}$ and is moving along a straight line with constant acceleration. When the velocity of the particle is $50 {ms}^{- 1}$ , the acceleration is reversed in direction. The velocity of the particle when it again reaches the starting point is

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$70 {ms}^{- 1}$

$10 {ms}^{- 1}$

$60 {ms}^{- 1}$

$30 {ms}^{- 1}$

answer is A.

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Detailed Solution

${(50)}^{2} = {(10)}^{2} + 2 a s$ ….(i)

$v^{2} = (50)^{2} + 2 (- a) (- s) = (50)^{2} + 2 a s$ ….(ii)

Solving these two equations, we get $v = 70 m / s$

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