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If a particle having an initial velocity of $20 {ms}^{- 1}$ , moves with a uniform acceleration of $10 {ms}^{- 2}$ , then the distance covered by it when it attains a final velocity of $80 {ms}^{- 1}$ is

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300 m

200 m

150 m

250 m

answer is D.

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Detailed Solution

Initial velocity

u = 20 {ms}^{- 1}

, Acceleration

a = 1 {0ms}^{- 2}

, Final velocity

v = 80 {ms}^{- 1}

s =?

We know $v^{2} - u^{2} = 2 as$

$s = \frac{v^{2} - u^{2}}{2 a} = \frac{{(80)}^{2} - {(20)}^{2}}{2 (10)} = 300 m$

$s = 300 m$

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