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Average Velocity and Average Speed of Integral

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    An object moves with velocity v(t) = −t2 +1 feet per second between t = 0 and t = 2. Find the average velocity and the average speed of the object between t = 0 and t = 2

    2. Relevant equations

    [itex]
    \frac{1}{b-a} \int_a^b f'(x) dx
    [/itex]

    avg value of a function

    3. The attempt at a solution
    [itex]
    \frac{1}{2-0} \int_0^2 [-t^2 + 1] dt
    [/itex]

    [itex]
    \frac{1}{2} [- \frac{t^3}{3} + t]_0^2
    [/itex]

    [itex]
    \frac{1}{2} [- \frac{8}{3} + \frac{6}{3}]
    [/itex]

    [itex]
    \frac{1}{2} [- \frac{2}{3}]
    [/itex]

    [itex]
    [- \frac{1}{3}]
    [/itex]

    So I've got the average velocity down, but I don't see how they want me to come up with the average speed. I know that speed and velocity are similar, but speed has no direction.

    The book (http://www.whitman.edu/mathematics/multivariable/" [Broken]) Instructed me to evaluate the integral without the averaging [itex]\frac{1}{b-a}[/itex], but I ended up with:

    [itex]
    - \frac{2}{3}
    [/itex]

    But according to the solutions manual, the answer is 1
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 26, 2011 #2
    The Average Speed is
    [tex]\frac{\text{total distance traveled}}{\text{time}}[/tex],

    see this thread for a thorough discussion:
    https://www.physicsforums.com/showthread.php?t=133408

    What happens here is within the first second the object is moving with a positive velocity, but slowing down. Then, at t=1, it stops, and proceeds to move in reverse. Because you have a simple function, you should see an easy way to get the total distance here.
     
  4. Sep 26, 2011 #3
    th_smiley-bangheadonwall.gif

    I should have had that on the top of my head! Duh!
     
    Last edited by a moderator: Apr 14, 2017
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