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)

Answer 2

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.