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

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Solution

The correct option is A
4m

Use, a = v $\frac{d v}{d x}$

a = $x^{3}$ - 3x = v $\frac{d v}{d x}$ $\Rightarrow$ $\int_{0}^{v}$ vdv= $\int_{1}^{3}$ ( $x^{3}$ -3x) dx

$\Rightarrow v = 4 m s^{- 1}$

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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 4m Use, a = vdvdx a = x3 - 3x = vdvdx ⇒ ∫v0 vdv= ∫31(x3-3x) dx ⇒v=4ms−1

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

The correct option is A
4m

Use, a = v $\frac{d v}{d x}$

a = $x^{3}$ - 3x = v $\frac{d v}{d x}$ $\Rightarrow$ $\int_{0}^{v}$ vdv= $\int_{1}^{3}$ ( $x^{3}$ -3x) dx

$\Rightarrow v = 4 m s^{- 1}$