Question

The acceleration of a particle varies with time $t$ seconds according to the relation $a=6t+6m{s}^{-2}$. Find velocity and position as functions of time. It is given that the particle starts from origin at $t=0$ with velocity $2m{s}^{-1}$.

Answer 1

We know that $\int dv=\int adt\Rightarrow {\int }_2^{v}dv={\int }_0^t(6t+6)dt$

$\Rightarrow v-2={[\frac{6t^2}{2}+6t]}_0^t\Rightarrow v=3t^2+6t+2$

Now $\int dx=\int vdt\Rightarrow {\int }_0^{x}dx={\int }_0^t(3t^2+6t+2)dt$

$\Rightarrow x=t^3+3t^2+2t$