A particle moves along X-axis obeying the equation x = t (t – 1) (t – 2) where x is in meter & t in second.

(a) Find the initial velocity of particle

(b)initial acceleration of the particle

(c) find 't' time when its displacement of particle is zero

(d) find the acceleration of the particle when its velocity is zero.

a)  x = t (t –1) (t – 2)                     . . . (i)

      ⇒  x = t3 – 3t2 + 2t

⇒ v = dx/dt  = 3t2 – 6t + 2               . . . (ii)

⇒ At time t = 0 , v = 2 m/sec.  &

a = dv/dt = 6t – 6 = 6(t –1)        . . . (iii)

⇒ at time t = 0,  a = -6m/sec2  .

(c) Displacement of the particle is zero at time t given by

x = t (t – 1) (t−2) = 0

⇒    t = 0 , t = 1 & t = 2

putting the values of t in eq. (ii) & (iii) we obtain,

velocity v = +2 , -1 & +2 m/sec.  respectively.

Acceleration a = −6 , 0 & +6 m/sec2 respectively.

(d) velocity v = 0

⇒ v = 3t2 – 6t + 2 = 0

⇒ t = 1 + (1/√3)    & t = 1 − (1/√3)

By putting these values of t in eq. (i) we obtain,

x = (−2/3√3) & (2/3√3) m respectively & the corresponding  acceleration can be obtained

by putting the values of t in  equation (iii) given by

a = 2√3 m/sec2  & – 2√3 m/s2 respectively