Displacement-time equation of a particle moving along x-axis is `x=20+t^3-12t` (SI units)
(a) Find, position and velocity of particle at time t=0.
(b) State whether the motion is uniformly accelerated or not.
(c) Find position of particle when velocity of particle is zero.

Options

A. `12m//s`
B. `22m//s`
C. `-22m//s`
D. `-12m//s`

Answer

D

Explanation

(a) `x=20+t^3-12t` …(i)
At t=0, `x=20+0-0=20m`
Velocity of particle at time t can be obtained by differentiating Eq. (i) w.r.t. time i.e.
`v=dx/dt=3t^2-12` …(ii)
At t=0, `v=0-12 =-12m//s`
(b) Differentiating Eq. (ii) w.r.t. time t, we get the acceleration `a=(dv)/(dt) =6t`
As acceleration is a function of time, the motion is non-uniformly accelerated.
(c) Substituting v=0 in Eq. (ii), we have `0=3t^2-12`
Positive value of t comes out to be 2s from this equation. Substituting `t=2s` in
Eq. (i), we have `x=20+(2)^3-12(2)` or `x=4m`