Question

If the relation between acceleration and time of an object is given by $a=2t+4t^2$. Calculate the position of the object from the origin at $t=4s$.

Answer 1

$a=2t+4t^2$

$\frac{dv}{dt}=2t+4t^2$

${\int }_0^{v}dv={\int }_0^t(2t+4t^2)dt$

$v=t^2+\frac{4t^3}{3}$

or, $\frac{ds}{dt}=t^2+\frac{4t^3}{3}$

${\int }_0^{s}ds={\int }_0^4(t^2+\frac{4t^3}{3})dt$

$s=[\frac{t^3}{3}+\frac{t^4}{3}{]}_0^4$

$=\left(\frac{64+256}{3}\right)=106.6m$