The velocity-displacement graph of a particle is shown in the figure.
$(a)$ Write the relation between $v$ and $x$.
$(b)$ Obtain the relation between acceleration and displacement and plot it.

(N/A) From the graph,the initial velocity is $v_{0}$ at $x=0$ and the velocity is $0$ at $x=x_{0}$.
$(a)$ The equation of a straight line with intercept $v_{0}$ on the $v$-axis and intercept $x_{0}$ on the $x$-axis is given by the intercept form:
$\frac{v}{v_{0}} + \frac{x}{x_{0}} = 1$
$v = v_{0} \left(1 - \frac{x}{x_{0}}\right) = v_{0} - \frac{v_{0}}{x_{0}}x$
$(b)$ We know that acceleration $a = v \frac{dv}{dx}$.
From the relation $v = v_{0} - \frac{v_{0}}{x_{0}}x$,we differentiate with respect to $x$:
$\frac{dv}{dx} = -\frac{v_{0}}{x_{0}}$
Substituting these into the expression for $a$:
$a = \left(v_{0} - \frac{v_{0}}{x_{0}}x\right) \left(-\frac{v_{0}}{x_{0}}\right)$
$a = -\frac{v_{0}^{2}}{x_{0}} + \frac{v_{0}^{2}}{x_{0}^{2}}x$
This is a linear equation of the form $a = mx + c$,where the slope is $\frac{v_{0}^{2}}{x_{0}^{2}}$ and the intercept is $-\frac{v_{0}^{2}}{x_{0}}$. The graph is a straight line starting from $-\frac{v_{0}^{2}}{x_{0}}$ at $x=0$ and passing through $x=x_{0}$ at $a=0$.