Physics - Kinematics Question with Solution | TestHub

Distance along a line i.e., displacement ( $s$ ) $=t^3(\because s\propto t^3$ given $)$ By double differentiation of displacement, we get acceleration. $V=\frac{ds}{dt}=\frac{d{t}^3}{dt}=3t^2$ and $$\begin{array}{rl}& a=\frac{dv}{dt}=\frac{d3t^2}{dt}=6t\\ & a=6t\text{ or }a\propto t\end{array}$$ Hence graph (b) is correct.