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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 -

    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 -

    That is wrong^ av=11-2/3= 9/3=3

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