Question

s(t)=t^3 - 5/2 t^2 - 2t + 1 for all t>=0,
1. what was the displacement on 0<= t <= 4
2. at t =.5 is the function slowing down or speeding up?

Answer 1

The context of the (incomplete) question indicates that s(t) represents the displacement of a particle or an object with respect to time, t.

1.
Displacement is a vector, so the path does not count. Thus evaluate
displacement(t=0,t=4) = s(4)-s(0)
2. To find out if the function is increasing or decreasing, you would find the derivative (s'(t)) of s(t) with respect to t.
s'(t)=3t²-5t-2
Evaluate s'(t) at t=0.5.
If s'(t)>0, it is increasing (speeding up), if s'(t)<0, it is decreasing (slowing down).

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