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. What is the minimum velocity?

(a) 1 cm/sec

(b) 2 cm/sec

(c) 3 cm/sec

(d) 4 cm/sec

I had been asked this question in my homework.

I need to ask this question from Calculus Application in portion Application of Calculus of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

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The correct choice is (a) 1 cm/sec

The best explanation: 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.

Putting t = 3 in (1) we get,

3^3/3 – 2(3)^2/ + 3(3) + 1

= 1 cm/sec.