Answer to Question #313345 in Calculus for Jhonny

Question

The velocity of a particle moving on the x-axis is given by v(t) = t^(3) − 6t^(2) for the time interval 0 ≤ t ≤ 10.

a) When is the particle farthest to the left?

b) When is the velocity of the particle increasing the fastest?

Answer 1

a) The farthest left point corresponds to the change of velocity from negative to positive value.

$v(t) = t^3 - 6t^2 = t^2(t-6).$ We see two values, namely 0 and 6, which make v(t)=0.

When t=6 v(t) turns from negative to positive, so when t=6 the particle is farthest to the left.

b) The fastest increase of the velocity corresponds to the greatest derivative of v(t).

$v'(t) = 3t^2 - 12t$ . Maximum derivative of v(t) corresponds to the zero value of v''(t) or takes place at the boundaries:

$v''(t) = 6t-12 = 0$ when t=2, but this value corresponds to the change of the sign from negative to positive, so the derivative of v(t) here is minimal. For t > 2 v''(t) is positive, so v'(t) increases, therefore max v'(t) on 0 ≤ t ≤ 10 corresponds to t=10.