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A particle is moving in a straight line and its distance x from a fixed point on the line at any time t seconds is given by, x = t^4/12 – 2t^3/3 + 3t^2/2 + t + 15. At what time is the velocity minimum?(a) 1(b) 2(c) 3(d) 4I got this question in quiz.This interesting question is from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

Answer»

Right answer is (C) 3

To explain: Assume that the velocity of the PARTICLE at time t SECOND is vcm/sec.

Then, v = dx/dt = 4t^3/12 – 6t^2/3 + 6t/2 + 1

So, v = dx/dt = t^3/3 – 2t^2/ + 3t + 1

Thus, dv/dt = t^2 – 4t + 3

And d^2v/dt^2 = 2t – 4

For maximum and minimum value of v we have,

dv/dt = 0

Or t^2 – 4t + 3 = 0

Or (t – 1)(t – 3) = 0

Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3

Now, [d^2v/dt^2]t = 3 = 2*3 – 4 = 2 > 0

Thus, v is minimum at t = 3.


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