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Standard XII

Physics

Average Velocity in 2D Motion

The velocity ...

Question

The velocity of a particle moving in straight line depends on its position $x$ w.r.t. to fixed origin according to the relation $16 x^{2} + 4 v^{2} = 9$ . The acceleration of particle when it is at $x = 2 m$ is:

- 2 m / s^{2}

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- 8 m / s^{2}

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- 6 m / s^{2}

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Not defined

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Solution

The correct option is B $- 8 m / s^{2}$
The given relation is $16 x^{2} + 4 v^{2} = 9$
Differentiating it w.r.t. $x$ , we get
$32 x + 8 v \frac{d v}{d x} = 0$
Or $v \frac{d v}{d x} = \frac{- 32 x}{8}$
We know that acceleration $a = v \frac{d v}{d x}$
$⟹ a = \frac{- 32 x}{8} = - 4 x$
At $x = 2 m,$ $a = - 4 \times 2 = - 8 m / s^{2}$
Correct answer is option B.

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The velocity of a particle moving in straight line depends on its position x w.r.t. to fixed origin according to the relation 16x2+4v2=9. The acceleration of particle when it is at x=2 m is:

The correct option is B −8 m/s2The given relation is 16x2+4v2=9Differentiating it w.r.t. x, we get32x+8vdvdx=0Or vdvdx=−32x8We know that acceleration a=vdvdx⟹ a=−32x8=−4xAt x=2 m, a=−4×2=−8 m/s2Correct answer is option B.

The velocity of a particle moving in straight line depends on its position $x$ w.r.t. to fixed origin according to the relation $16 x^{2} + 4 v^{2} = 9$ . The acceleration of particle when it is at $x = 2 m$ is: