Question

A particle starts from the origin and moves along the $x-$axis such that its velocity $v$ varies with time $t$ as $v=4t^3-2t$, where $v$ is in $m/s$ and $t$ is in $second$. What is the distance of the particle from the origin when its acceleration is $46$ $m/s^2$?

Answer 1

$v=4t^3-2t$

$\frac{dv}{dt}=12t^2-2=a$

if $a=46$

$\Rightarrow 12t^2-2=46$

$\Rightarrow t^2=\frac{48}{12}$

$\Rightarrow t=2$ $sec$

as, $v=4t^3-2t$

$\frac{dx}{dt}=4t^3-2t$

$\Rightarrow x=\int (4t^3-2t)dt$

$\Rightarrow x=t^4-t^2$

$\Rightarrow x$ at $t=2=16-4=12$ $m$