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A particle initially at rest moves with a uniform acceleration of ${10 ms}^{- 2}$ and attains a velocity of ${90 ms}^{- 1}$ after some time. The time taken to travel a distance of 405 m is

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3 s

10 s

8 s

9 s

answer is C.

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Detailed Solution

$\to u = 0$ $\to a = {10 m/s}^{2}$ $\to v = 9 0 m/s$

$S = 405m$

$S = ut + {1/2 at}^{2}$

$\Rightarrow 405 = 1 / 2 (10) {(t)}^{2}$

$t^{2} = \frac{405}{5} = 81$

$\Rightarrow t^{2} = 81$

$t = 9 \sec$

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