A particle moves according to a law of motion below, 0 t < 6, where t is measured in seconds and s in feet. s = f(t) = cos((pi t)/3)

Question

A particle moves according to a law of motion below, 0  t < 6, where t is measured in seconds and s in feet.

(b) What is the velocity after 2 s?

(c) When is the particle at rest?

(d) When is the particle moving in the positive direction?

(e) Find the total distance traveled during the first 6 s.

(g) Find the acceleration at time t and after 2 s

When is the particle speeding up? When is it slowing down?

Answer 1

velocity = ds/dt so take the derivative and plug in t = 2. The chain rule is required so you should get

s' = -(π/3)sin((π/3)t)

The particle is at rest when velocity = zero so solve s' = 0. remember sin(x) = 0 when x = 0, π, 2π, 3π, etc.

The particle is moving in the positive direction whenever velocity is positive. You can graph the derivative function (aka velocity function) to see where it is positive if you need help.

To find the total distance traveled, you'll need to add up all the distances in the intervals where v is positive and where v is negative (since it goes back and forth).

To find the acceleration, take the second derivative.

The particle is speeding up when s'' is positive and it is slowing down when s'' is negative.

Let me know if you need more help than this.