The motion of a particle along a straight line is described by the function `x=(2t -3)^2,` where x is in metres and t is in seconds. Find
(a) the position, velocity and acceleration at `t=2 s.`
(b) the velocity of the particle at origin.

Explanation

(i) Posiotion of the particle, `x = (2t-3)^(2)`
At `t=2s, x=(2xx2-3)^(2)=1.0 m`
Velocity, `v=(dx)/(dt)=4(2t-3)ms^(-1)`
At `t = 2 s, v= 4(2xx2-3)4ms^(-1)`
and acceleration of the particle, `a=(dv)/(dt)=8ms^(-2)`
(ii) At origin, x = 0 or `(2t-3)^(2)=0`
Velocity of the particle at origin, `v=4xx0=0`