Question

If the relation between acceleration and time for an object is given by $a=2t+4t^2$.
Calculate the position of object from the origin at $t=4s$ (At $t=0,v=0,x=0$)

Answer 1

It is given that $a=2t+4t^2$

$\because a=\frac{dv}{dt}$

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

$\int dv=\int (2t+4t^2)dt$

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

$\because v=\frac{dx}{dt}$

$dx=(t^2+\frac{4}{3}{t}^3)dt$

$\int dx=\underset{0}{\overset{4}{\int }}(t^2+\frac{4}{3}{t}^3)dt$

$x={{\frac{t^3}{3}+\frac{t^4}{3}|}_0}^4$

$x=\frac{1}{3}[4^3+4^4]$

$x=\frac{1}{3}(64+256)$

$x=106.66m$