A particle moves according to a law of motion s = f(t) = 0.01t4 - 0.04t3, t > 0, where t is measured in seconds and s in feet.

Question

A particle moves according to a law of motion s = f(t) = 0.01t4 - 0.04t3, t > 0, where t is measured in seconds and s in feet.

a. Find the velocity at time t

b. What is the velocity after 2s?

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 5s.

g. find the acceleration at time t and after 2s

i. when is the particle speeding up? When is it slowing down?

Answer 1

a) velocity = ds/dt = 0.04t³ - 0.12 t²

b) substitute t = 2 into ds/dt

c) find t when ds/dt = 0

d) moving in positive direction when ds/dt > 0. Graph ds/dt, and find t values when ds/dt > 0 from the graph

e) Find s, when t=0 and t=5. Difference will be the answer

g = dv/dt = 0.12 t² - 0.24 t , substitute t = 2 here

i speeding up when dv/dt > 0, slowing down when dv/dt < 0. Plot dv/dt and use your graph to find these

Answer 2

The best way to answer this question is to graph f(t).

You will see that the particle first moves in the negative direction, comes to a minimum, then moves in the positive direction comes to rest at t=4 and then continues in the positive direction.

Remember that velocity is simply f'(t) and acceleration is f"(t).

This information should enable you to answer all the questions.