Average Velocity and Average Speed of Integral

[Ocasta]

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

<br /> \frac{1}{b-a} \int_a^b f&#039;(x) dx <br />

avg value of a function

The Attempt at a Solution

<br /> \frac{1}{2-0} \int_0^2 [-t^2 + 1] dt<br />

<br /> \frac{1}{2} [- \frac{t^3}{3} + t]_0^2<br />

<br /> \frac{1}{2} [- \frac{8}{3} + \frac{6}{3}]<br />

<br /> \frac{1}{2} [- \frac{2}{3}]<br />

<br /> [- \frac{1}{3}]<br />

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/" ) Instructed me to evaluate the integral without the averaging \frac{1}{b-a}, but I ended up with:

<br /> - \frac{2}{3}<br />

But according to the solutions manual, the answer is 1

 

[process91]

The Average Speed is
\frac{\text{total distance traveled}}{\text{time}},

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.

 

[Ocasta]

The Average Speed is
\frac{\text{total distance traveled}}{\text{time}}

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