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 = t4/12 – 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity?

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 = t4/12 – 2t3/3 + 3t2/2 + t + 15. What is the minimum velocity? Correct Answer 1 cm/sec

Assume that the velocity of the particle at time t second is vcm/sec. Then, v = dx/dt = 4t3/12 – 6t2/3 + 6t/2 + 1 So, v = dx/dt = t3/3 – 2t2/ + 3t + 1 Thus, dv/dt = t2 – 4t + 3 And d2v/dt2 = 2t – 4 For maximum and minimum value of v we have, dv/dt = 0 Or t2 – 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, t = 3 = 2*3 – 4 = 2 > 0 Thus, v is minimum at t = 3. Putting t = 3 in (1) we get, 33/3 – 2(3)2/ + 3(3) + 1 = 1 cm/sec.