Posted by **Millie** on Sunday, November 13, 2011 at 9:38pm.

At time t >or= to 0, the position of a particle moving along the x-axis is given by x(t)= (t^3/3)+2t+2. For what value of t in the interval [0,3] will the instantaneous velocity of the particle equal the average velocity of the particle from time t=0 to time t=3?

- Calculus -
**Steve**, Monday, November 14, 2011 at 9:07pm
The average velocity for t in [0,3] is the distance traveled divided by the time spent:

av = [x(3)-x(0)]/(3-0) = 17/3

So, we want t where x' = t^2 + 2 = 17/3

t^2 = 11/3, so t=1.9 seconds

- Calculus -
**Erin**, Thursday, February 4, 2016 at 10:51pm
That is wrong^ av=11-2/3= 9/3=3

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