Problem : A particle moves along X-axis obeying the equation x = t (t – 1) (t – 2) where x is in metre & t in second. Find the (a) initial velocity and acceleration of the particle (b) velocity & acceleration of the particle when its displacement is zero (c) displacement & acceleration of the particle when its velocity is zero.

Solution :

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

* ⇒ * x = t^{3} – 3t^{2} + 2t

⇒ v = dx/dt = 3t^{2} – 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/sec^{2} .

**(b)** 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/sec^{2} respectively.

**(c)** velocity v = 0

⇒ v = 3t^{2} – 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/sec^{2 }& – 2√3 m/s^{2} respectively