The velocity-displacement graph of a particle | vedclass

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(C) From the given graph,the velocity $v$ as a function of displacement $x$ is a straight line with a negative slope.Using the equation of a line $y =

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(C) From the given graph,the velocity $v$ as a function of displacement $x$ is a straight line with a negative slope.Using the equation of a line $y = mx + c$,we have:$v = -\left(\frac{v_{0}}{x_{0}}\right)x + v_{0}$We know that acceleration $a$ is given by:$a = v \frac{dv}{dx}$First,find the derivative $\frac{dv}{dx}$:$\frac{dv}{dx} = -\frac{v_{0}}{x_{0}}$Now,substitute $v$ and $\frac{dv}{dx}$ into the expression for $a$:$a = \left[-\left(\frac{v_{0}}{x_{0}}\right)x + v_{0}\right] \left[-\frac{v_{0}}{x_{0}}\right]$$a = \left(\frac{v_{0}}{x_{0}}\right)^{2}x - \frac{v_{0}^{2}}{x_{0}}$This equation represents a straight line with a positive slope $\left(\frac{v_{0}}{x_{0}}\right)^{2}$ and a negative intercept $-\frac{v_{0}^{2}}{x_{0}}$. Looking at the options,graph $C$ represents a linear relationship where $a$ increases with $x$ starting from a negative value,which matches our derived equation.

The velocity-displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by:

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