v = ds/dt = -6 t^2 + 30 t -24
a = dv/dt = -12 t + 30
at t = 0 the point is at -6
at t = 1 it is at -17
at t = 2 it is at -10
etc until
at t = 5 it is at -1
graph that
the velocity changed sign in there, look for where v = 0
-6 t^2 +30 t -24 = 0
-t^2 + 5 t - 4 = 0
t^2 -5 t + 4 = 0
(t -4)(t-1) = 0
at t = 1 and at t = 4 the direction reverses
at t = 0 the velocity is -24 so it starts out at -6 moving south fast
at t = 1 it reverses course and starts up from -17 etc
at t = 4 it is at 10 and starts down toward reaching 1 at t = -5
A point moving on a coordinate line has position function s. Find the velocity and acceleration at the t, and describe the motion of the point during the indicated time interval
s(t)=-2t^3+15t^2-24t-6 [0,5]
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