A particle moves according to a law of motion s=f(t), t≥0, where t is measured in seconds and s in feet. f(t)= t^3-6t^2+9t

Question

I have the answers for a, b, and c but I have a problem with the questions d and e.

(a) Find the velocity at time t.

v(t)= 3t^2-12t+9

(b) What is the velocity after 2 s?

v(2)=-3 ft/s

(c) When is the particle at rest? (Enter your answer as a comma-separated list.)

t= 3,1

(d) When is the particle moving in the positive direction? (Enter your answer in interval notation.)

(e) Find the total distance traveled during the first 4 s.

Answer 1

When the velocity is positive, that's when it's moving in the positive direction. From graphing the parabola, it's (0,1)U(3,inf)

As for the distance, you want to integrate the velocity function, but there's a catch; if you graph the function and it's under the x-axis, you're going to take the absolute value. Abs(int(3t^2-12t+9)|₀³)+(int(3t^2-12t+9)|₃⁴)=4+4=8

Note had it been displacement, you wouldn't have to break it up.