## In This post you will get answer of Q.

The velocity-displacement graph of a particle is as shown in the figure. Which of the following graphs correctly represents the variation of acceleration with displacement ?

Solution:

## For the given velocity-displacement graph, intercept $=v_{0}$ and slope $=-frac{v_{0}}{x_{0}}$

Thus, the equation of given line of velocity-displacementgraph is

$v=frac{v_{0}}{x_{0}}x+v_{0}quadldotsleft(iright)$

Acceleration, $a=frac{dv}{dt}=frac{dv}{dx} frac{dx}{dt}=frac{dv}{dx}v$

$because frac{dv}{dx}=-frac{v_{0}}{x_{0}}$

$therefore a=frac{v_{0}}{x_{0}}left(-frac{v_{0}}{x_{0}}x+v_{0}right)quad$ (Using $(i)$)

$=frac{v^{2}_{0}}{x^{2}_{0}}x-frac{v^{2}_{0}}{x_{0}}$

It is a straight line with positive slope and a negative intercept.

The variation of $a$ with $x$ is as shown in the above figure.