Correct answer is (a) 4/27 units
Explanation: Let v be the velocity of the particle at time t seconds after start (that is at a distance s from O). Then,
v = ds/dt = d[(t – 1)^2(t – 2)^2]/dt
Or v = (t – 2)(3t – 4)
Clearly, v = 0, when (t – 2)(3t – 4) = 0
That is, when t = 2
Or 3t – 4 = 0 i.e., t = 4/3
Now, s = (t – 1)(t – 2)^2
Therefore, when t = 4/3, then s = (4/3 – 1)(4/3 – 2)^2 = 4/27
And when t = 2, then s =(2 – 1)(2 – 2)^2 = 0
Therefore, the velocity of the particle is zero, when its distance from O is 4/27 units and when it is at O.