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.

Explanation

(i) Given, displacement-time equation of a particle P
`x=20+t^(3)-12t` ….(i)
At `t =0, x = 20+0-0=20 m`
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 =-12 m//s`
(ii) 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.
(iii) Substituting v=0 in Eq. (ii), we have
`0=3t^(2)-12`
Positive value of t comes out to be 2 s from this equation. Substituting t = 2 s in Eq. (i), we have,
`x=20+(2)^(3)-12(2)` or `x = 4 m`