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The initial velocity of a body moving along a straight line is $8u$ . It has a uniform acceleration of ${6ms}^{- 2}$ . The distance covered by the body in the $6^{th}$ second of its motion is $41m$ . The value of u is

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{4ms}^{- 1}

{16ms}^{- 1}

{8ms}^{- 1}

{12ms}^{- 1}

answer is C.

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Detailed Solution

Initial velocity

u = 8u, a = {6ms}^{- 2}

$S_{6} = 41m$

$S_{n} = u + a [n - \frac{1}{2}]$

$41 = 8u+ 6 [6 - \frac{1}{2}]$

$= 8 u + 6^{3} \times \frac{11}{2_{1}}$

$8u = 41 - 33$

$8u = 8$

$u = 1 m s^{- 1}$

Here $u = 8u$

so $u = 8 (1) = 8m / s .$

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