A particle is moving with uniform acceleration along a straight line ABC. Its velocity at A and B are $6$ $m s^{- 1}$ and $9$ $m s^{- 1}$ respectively. If $A B : B C = 5 : 16$ then its velocity at C is

9.6

m s^{- 1}

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m s^{- 1}

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m s^{- 1}

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21.5

m s^{- 1}

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Solution

The correct option is C 15 $m s^{- 1}$
$v_{A} = 6 m / s$
$v_{B} = 9 m / s$
$v_{B}^{2} = v_{A}^{2} + 2 a (5 x)$
$9^{2} = 6^{2} + 10 a x$
$\frac{45}{10 x} = a$
Now, B to C:
$v_{C}^{2} = v_{B}^{2} + 2 a (16 x)$
$v_{C}^{2} = 9^{2} + 2 (\frac{45}{10 x}) \times 16 x$
$v_{C}^{2} = 9^{2} + 144 = 225$
$v_{C} = 15 m / s$

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