A particle starts from rest and moves with an acceleration $a=kt$, where $k$ is a constant. What is the distance covered by the particle in time $t$?

Solution

1. Given the acceleration $$\begin{array}{rl}a& =\frac{dv}{dt}\\ & =kt\end{array}$$, integrate with respect to time to find velocity: $$\begin{array}{rl}v& =\int ktdt\\ & =\frac{k{t}^2}{2}+C\end{array}$$.
2. Since the particle starts from rest, $v(0)=0$, so the constant $C=0$ and $v=\frac{k{t}^2}{2}$.
3. Integrate the velocity to find the displacement $s$: $$\begin{array}{rl}s& =\int vdt\\ & =\int \frac{k{t}^2}{2}dt\\ & =\frac{k{t}^3}{6}\end{array}$$.
4. Hence the answer is (C).