A particle starts from rest and moves along X-axis. The acceleration a varies with its displacement x as a = (2 - x) ms2 . Find magnitude of acceleration of particle when its displacement along positive X-axis is maximum.

a=2−x Vdvdx=2−x ∫Vdv=∫(2−x)dx [V22]0V=[2x−x22]0x V2=4x−x2 V=4x−x2For Max displacement dxdt=V=0∴ x=4⇒a=2−4=−2ms2|a¯|=2ms2

A particle starts from rest and moves along X-axis. The acceleration a varies with its displacement x as a = (2 - x) ms2 . Find magnitude of acceleration of particle when its displacement along positive X-axis is maximum.

a=2−x Vdvdx=2−x ∫Vdv=∫(2−x)dx [V22]0V=[2x−x22]0x V2=4x−x2 V=4x−x2For Max displacement dxdt=V=0∴ x=4⇒a=2−4=−2ms2|a¯|=2ms2

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A particle starts from rest and moves along X-axis. The acceleration a varies with its displacement x as a = (2 - x) $\frac{m}{s^{2}}$ . Find magnitude of acceleration of particle when its displacement along positive X-axis is maximum.

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$3 \frac{m}{s^{2}}$

$1 \frac{m}{s^{2}}$

$4 \frac{m}{s^{2}}$

$2 \frac{m}{s^{2}}$

answer is B.

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Detailed Solution

$a = 2 - x V \frac{d v}{d x} = 2 - x \int V d v = \int (2 - x) d x {[\frac{V^{2}}{2}]}_{0}^{V} = {[2 x - \frac{x^{2}}{2}]}_{0}^{x} V^{2} = 4 x - x^{2} V = \sqrt{4 x - x^{2}}$
For Max displacement
$\begin{array}{l} \frac{d x}{d t} = V = 0 \\ ∴ x = 4 \\ \Rightarrow a = 2 - 4 = - 2 \frac{m}{s^{2}} \\ | \bar{a} | = 2 \frac{m}{s^{2}} \end{array}$