The distance s of a particle moving along a straight line from a fixed-pointO on the line at time t seconds after start is given by x = (t – 1)^2(t – 2)^2. What will be the distance of the particle from O when its velocity is zero?

(a) 4/27 units

(b) 4/23 units

(c) 4/25 units

(d) 4/35 units

The question was posed to me in semester exam.

This is a very interesting question from Calculus Application topic in division Application of Calculus of Mathematics – Class 12

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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.