Average Value and Work (using graphs of functions)

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soleilion

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1. In each part, the velocity versus time curve is given for a particle moving along a line. Use the curve to find the average velocity of the particle over the time interval t=[0,3]

http://img147.imageshack.us/img147/711/76ae7.png

These are two graphs. I draw these by word. Left is (a), right is (b). There are one black square in front of 1 means negative in each graph.
I hope helper can understant because I draw those by Word.

2. Suppose that f is a linear function. Using the graph of f, explain why the average value of f on [a,b] is f { (a+b)/2}

3. For the variable force F(x) in another exercise, consider the distance d for which the work done by the force on the particle when the particle moves from x=0to x=d is half of the work done when the particle moves from x=0to x=5. Find the exact value of d.

anther exercise: a variable force F(x) in the positive x-direction is graphed in the accompanying figure, Find the work done by the force on a particle that moves from x=0 to x=5

http://img337.imageshack.us/img337/5042/77op2.png

Here is the graph


All of these three, I have no idea how to solve, Please help me . Thank you!!!
 
Re: Average Value and Work

1.
\(\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}\)

You can find the net displacement by finding the net area under the curve.

2. Mmm, the graph would be nice since you are asked specifically to use it

3. Given a force vs. distance graph, the work done is equal to the area under the curve. Once you find that, it's just simple algebra to find d.

Show us your work and we might be able to help you point something out.
 
After your prompt

1. I did and I got correct answer
2. I misunderstood what you mean
3. W=A=40*2+1/2 *40*3=140 half work is 70
d= W/F=???? the correct answer is 7/4 so F should be 40
from the graph, we can see Force is not constant number, why shoule F be 40???
 
2. Your question says: "Using the graph of f ..." What graph of f?

3. Look at your graph. The total area is 140 J as you pointed out. If you want half the work done, the area under the graph must total 70 J. You know from x = 0 to x = 2, the total work done is 80 J. So, you know that the 70 J of work is done before the applied force starts to vary.
 
2. there is no graph with the question on my book, I hope we have to graph by ourselves

3. I understood, thank you
 
2. Oh. Well in any case, you know that linear functions take the form of: f(x) = mx + b. Now, for convenience I'll use p and q instead of a and b. So for this particular linear function on the interval [p, q]:

\(\displaystyle f(p) = mp + b \quad \mbox{and} \quad f(q) = mq + b\)

The middle value of this interval is (p + q) / 2, so:

\(\displaystyle f\left(\frac{p+q}{2}\right) = m\left(\frac{p+q}{2}\right) + b\)

Now, the average value of f would be [ f(p) + f(q)] / 2. So all you have to do is prove that:

\(\displaystyle \frac{f(p) + f(q)}{2} = f\left(\frac{p+q}{2}\right)\)
 
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