Question

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 ?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

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

$v=\frac{v_{0}}{x_{0}}x+v_{0}\quad\ldots\left(i\right)$
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.