posted by Alex .
A particle moves on a vertical line. Its position, s, in metres at t seconds is given by s(t) = t^3 - 9t^2 + 24t, t>0/
I found the velocity and acceleration functions.
s'(t) = 3t^2 - 18t + 24
s''(t) = 6t-18
b) When is the particle moving up? down?
c) Find the distance the particle travels between t=0 and t=6.
Right on the first.
Now when it is going up> When is s'positive? When is it negative? The sign of s'is direction. Hint: factor the quadratic and test domains to see when it is +-
The distance it travels? Integrate velocity over time. The area under a velocity time graph is distance.