Average Velocity and Average Speed of Integral

  • Thread starter Ocasta
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  • #1
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Homework Statement


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

Homework Equations



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

avg value of a function

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
 
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Answers and Replies

  • #2
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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.
 
  • #3
40
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The Average Speed is
[tex]\frac{\text{total distance traveled}}{\text{time}}[/tex]

th_smiley-bangheadonwall.gif


I should have had that on the top of my head! Duh!
 
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