Force acting on a body of mass 1 kg is related to its position as, $F = x^{3} - 3 x$ newton. It is at rest at x=1m. Its velocity at x=3 m can be,

4 m s^{- 1}

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3 m s^{- 1}

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2 m s^{- 1}

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5 m s^{- 1}

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Solution

The correct option is A $4 m s^{- 1}$
Use, $a = v \frac{d v}{d x}$
$a = x^{3} - 3 x = v \frac{d v}{d x} \Rightarrow \int_{0}^{v} v d v = \int_{1}^{3} (x^{3} - 3 x) d x$
$\Rightarrow v = 4 m s^{- 1}$

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Force acting on a body of mass 1 kg is related to its position as, F=x3−3x newton. It is at rest at x=1m. Its velocity at x=3 m can be,

The correct option is A 4 ms−1Use, a=vdvdx a=x3−3x=vdvdx⇒∫v0v dv=∫31(x3−3x)dx ⇒v=4 ms−1

Force acting on a body of mass 1 kg is related to its position as, $F = x^{3} - 3 x$ newton. It is at rest at x=1m. Its velocity at x=3 m can be,

The correct option is A $4 m s^{- 1}$
Use, $a = v \frac{d v}{d x}$
$a = x^{3} - 3 x = v \frac{d v}{d x} \Rightarrow \int_{0}^{v} v d v = \int_{1}^{3} (x^{3} - 3 x) d x$
$\Rightarrow v = 4 m s^{- 1}$