The distance $x$ covered by a body moving in a straight line in $t$ is given by the relation $2 x^{2} + 3 x = t .$ If $v$ is the velocity of the body at a certain of time, its acceleration will be :

- v^{3}

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- 2 v^{3}

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- 3 v^{3}

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- 4 v^{3}

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Solution

The correct option is D $- 4 v^{3}$
The time is given as,

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

Differentiate above expression with respect to time, we get

$1 = (4 x + 3) \frac{d x}{d t}$

$1 = (4 x + 3) v$

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

Again differentiate both sides with respect to time, we get

$(\frac{- 1}{v^{2}}) \frac{d v}{d t} = (4) \frac{d x}{d t}$

$(\frac{- 1}{v^{2}}) a = (4) v$

$a = - 4 v^{3}$

Thus, the acceleration is $- 4 v^{3}$ .

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