Question

please solve this.........
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Answer 1

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!

Solution :

For the given velocity-displacement graph,

• Intercept = $\sf{v_0}$
• Slope = $\sf{-\dfrac{v_0}{x_0}}$

Thus, the equation of given line of velocity-displacement graph is

$\dag\:\underline{\boxed{\bf{\orange{v=-\dfrac{v_0}{x_0}x+v_0}}}}$ .......... (I)

In order to find acceleration of the particle, we need to differentiate the velocity equation with respect to time.

$:\implies\sf{a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=\dfrac{dv}{dx}\cdot v}$

$\sf\because\:\dfrac{dv}{dx}=-\dfrac{v_0}{x_0}$

$\sf\therefore\:a=-\dfrac{v_0}{x_0}\left(-\dfrac{v_0}{x_0}x+v_0\right)\:\dots\:(Using\:Eq.\:I)$

$:\implies\:\underline{\boxed{\bf{\purple{a=\dfrac{v_0^2}{x_0^2}x-\dfrac{v_0^2}{x_0}}}}}$

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

Option - A shows the correct variation of a with x.