Question

Acceleration of particle moving rectilinearly is $a=4-2x$ (where $x$ is position in meter and $a$ in ${ms}^{-2}$). It is at instantaneous rest at $x=0$. At what position $x$ (in meter) will the particle again come to instantaneous rest?

Answer 1

$a=4-2x$

$a=vdv/dx$

$vdv=(4-2x)dx$

Integrating from x = 0 to x = x1 where particle will come to instantaneous rest again,

$\left[v^2/2\right]=[4x-x^2]$

$0-0=4x_1-x_1^2$

$x_1=4m$ .