The velocity-displacement (v-x) graph for a particle moving along a straight line is shown in figure. The relation between acceleration(a) and displacement(x) is given by

a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}}

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a = \frac{{- v}_{0}^{2}}{x_{0}^{2}} x + \frac{v_{0}^{2}}{x_{0}}

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a = \frac{{- v}_{0}}{x_{0}} x + v_{0}

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a = \frac{v_{0}}{x_{0}} x - v_{0}

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Solution

The correct option is A $a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}}$
$a = \frac{d v}{d t}$
$a = v . \frac{d v}{d t}$ ${\frac{d x}{d t} = v$
$v = - \frac{v_{0}}{x_{0}} (x) + v_{0}$
$a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}}$

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The velocity-displacement (v-x) graph for a particle moving along a straight line is shown in figure. The relation between acceleration(a) and displacement(x) is given by

The correct option is A a=v20x20x−v20x0a=dvdta=v.dvdt {dxdt=vv=−v0x0(x)+v0a=v20x20x−v20x0

The correct option is A $a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}}$
$a = \frac{d v}{d t}$
$a = v . \frac{d v}{d t}$ ${\frac{d x}{d t} = v$
$v = - \frac{v_{0}}{x_{0}} (x) + v_{0}$
$a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}}$