Calculus

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.

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  1. 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.

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    bobpursley

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