Answer to Question #125779 in Physics for DEEPAK KUMAR

Question

Message: A body having a mass of M starts from rest at the point x=0. It accelerates by -t^2+4t.
The maximum distance that is covered by the body in the +ve x direction before moving backward in the -ve direction is

Answer 1

We have that

$a=-t^2+4t$ .

So,

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

and

$S=\int{vdt}=-\frac{t^4}{12}+2\frac{t^3}{3}$

$S=S_{max}$ at $v=0$

$v=-\frac{t^3}{3}+4\frac{t^2}{2}=0\to t=6s$ (if $t=0$ then $v=0$ )

$S=S_{max}=-\frac{t^4}{12}+2\frac{t^3}{3}=-\frac{6^4}{12}+2\frac{6^3}{3}=36m$