Calculus/physics question

Suppose that an object has position function s(t) = 2 t2− 4 t+ 2. Find the average velocity over the interval [3, 6] and find the time at which the instantaneous velocity equals the average velocity.

So, I'm guessing that I use the Mean Value Theorem, but I'm not really sure what to do. Step-by-step would be great.

s' = velocity = 4t - 4
s' (6) = 20
s' (3) = 8

Average velocity = (20 + 8) / 2 = 14 m/s

4t - 4 = 14
4t = 18
t = 18/4 = 4.5s

Is that what you meant?

You don't need to use the mean value theorem. Remember that velocity = change in position / change in time

t = 3
t = 6
s(3) = 2 * 3^2 - 4 * 3 + 2 = 2 * 9 - 12 + 2 = 18 - 12 + 2 = 20 - 12 = 8
s(6) = 2 * 6^2 - 4 * 6 + 2 = 2 * 36 - 24 + 2 = 72 - 24 + 2 = 74 - 24 = 50

So we want:

(s(6) - s(3)) / (6 - 3) =>
(50 - 8) / (6 - 3) =>
42 / 3 =>
14

The average velocity equals 14

Now, take the derivative of s(t) and find when s'(t) = 14

s(t) = 2t^2 - 4t + 2
s'(t) = 4t - 4
s'(t) = 14
14 = 4t - 4
18 = 4t
18/4 = t
9/2 = t

There you go.

if s(t) is position, then s ' (t) is velocity and s''(t) is acceleration.

s ' (t) = 2*2*t-4=4t-4

Vavg = 1/(6-3) integral (t=3..6) of (4t-4)dt

=(1/3)* [4t^2/2-4t] on (t=3..6)
=(1/3) [ (2*36-24) - (2*9-4*3)]
=(1/3)[48 - 6]
=14

v(t) = s ' (t) = 4t-4 = 14

t=18/4
t=4.5sec