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 ?

Updated On: Jul 7, 2022
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The Correct Option is A

Solution and Explanation

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.
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Concepts Used:

Acceleration

In the real world, everything is always in motion. Objects move at a variable or a constant speed. When someone steps on the accelerator or applies brakes on a car, the speed of the car increases or decreases and the direction of the car changes. In physics, these changes in velocity or directional magnitude of a moving object are represented by acceleration

acceleration