Physics - Calculus Question with Solution | TestHub

The graph of $v^2$ versus $x$ is a straight line passing through the origin with a positive slope. This implies $v^2=kx$, where $k$ is a positive constant. To find the acceleration, we differentiate this equation with respect to time $t$: $2v\frac{dv}{dt}=k\frac{dx}{dt}$.

Since $v=\frac{dx}{dt}$ and $a=\frac{dv}{dt}$, we have $2va=kv$. For $v\ne 0$, we can cancel $v$ from both sides, giving $2a=k$, or $a=\frac{k}{2}$. Thus, the acceleration is constant and non-zero.