Posted by **Anonymous** on Saturday, July 7, 2012 at 10:51pm.

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]

- calculus -
**Damon**, Saturday, July 7, 2012 at 11:41pm
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

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