The displacement x of a body moving in a straight line is related to time t as $2 x^{2} + 3 x = t$ . If v is the velocity of the body at a certain instant 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
$- 4 v^{3}$

Differentiating $2 x^{2} + 3 x = t$ with respect to t, $4 x \frac{d x}{d t} + \frac{3 d x}{d t} = 1$
Now, $\frac{d x}{d t} = v$
$∴ 4 x v + 3 v = 1$
$\Rightarrow 4 x + 3 = \frac{1}{v}$
Diffrentiating with respect to time t,
$4 \frac{d x}{d t} = \frac{- 1}{v^{2}} \frac{d v}{d t}$
$\Rightarrow 4 v = \frac{- 1}{v^{2}} \frac{d v}{d t}$
$\Rightarrow a = \frac{d v}{d t} = - 4 v^{3}$
So, the correct choice is (d).

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The displacement x of a body moving in a straight line is related to time t as 2x2+3x=t. If v is the velocity of the body at a certain instant of time, its acceleration will be

The correct option is D −4v3 Differentiating 2x2+3x=t with respect to t, 4xdxdt+3dxdt=1 Now, dxdt=v ∴4xv+3v=1 ⇒4x+3=1v Diffrentiating with respect to time t, 4dxdt=−1v2dvdt ⇒4v=−1v2dvdt ⇒a=dvdt=−4v3 So, the correct choice is (d).

The displacement x of a body moving in a straight line is related to time t as $2 x^{2} + 3 x = t$ . If v is the velocity of the body at a certain instant of time, its acceleration will be