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The distance $(x)$ travelled by a particle in time, t, is given by $t = 2 x^{2} + 3 x .$ If ‘u’ is the velocity, then acceleration will be:

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$- 4 u^{3}$

$- 3 u^{4}$

$4 u^{2}$

$- 3 u^{3}$

answer is A.

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Detailed Solution

$t = 2 x^{2} + 3 x$

Differentiating wrt t,

1= $4 x v + 3 v$

$v = \frac{1}{4 x + 3}$

Differentiating again,

$0 = 4 v^{2} + 4 x a + 3 a - 4 v^{2} = a / v a = - 4 v^{3}$

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