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Q.

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 speed of the particle when it again reaches the starting point is

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a

$30 {ms}^{- 1}$

b

$10 {ms}^{- 1}$

c

$70 {ms}^{- 1}$

d

$60 {ms}^{- 1}$

answer is A.

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Detailed Solution

${(50)}^{2} - {(10)}^{2} = 2 \times a \times s$

$v^{2} - {(- 50)}^{2} = 2 \times a \times s$

$2500 - 100 = v^{2} - 2500$

$v^{2} = 4900 \Rightarrow v = 70 m / s$

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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 speed of the particle when it again reaches the starting point is