calculus
posted by Anonymous .
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^224t6 [0,5]

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)(t1) = 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