Find the average velocity of the object from points A to B, B to C , and A to C .

- anonymous

Find the average velocity of the object from points A to B, B to C , and A to C .

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- schrodinger

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- anonymous

|dw:1328486272548:dw|

- anonymous

the points are a(0)=0 B(3)=25 c(6)=0

- anonymous

My answer was 8.33,7, and zero

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## More answers

- anonymous

i mean -7

- anonymous

Average velocity is given by [f(b)-f(a)]/b-a.
So, for the first part, you'd want to do [f(3)-f(0)]/3-0

- anonymous

i did that

- anonymous

Then what is your question?

- anonymous

my answer is wrong

- anonymous

can you solve for it and tell me if our answers match up

- anonymous

Yes

- anonymous

Okay. What I get is this:
[25-0]/[3] = 8.333 That should be correct.
Next, [f(6)-f(3)]/(6-3) = (0-25)/3 = -8.333
And lastly: [f(6)-f(0)]/(6-0) = (0-0)/6 = 0

- anonymous

why is the second one 8.33

- anonymous

ohh i apologize you are right thank you

- anonymous

Well, a good way to look at this visually is to look at the graph:
I'm assuming that that parabola is symmetrical. The curve comes up at the average velocity (slope) of 8.333. Then it comes back down with a similar slope, but negative. Hence the -8.333

- anonymous

Oh, good! I'm glad!

- anonymous

can you also help me with this For the intervals in above would the average speed be less than, equal to, or greater than the values you found in that Part?

- anonymous

Yes, let me think for a moment.

- anonymous

Well, speed is different from velocity in the fact that velocity has direction. For example, the answer we got from A to B and B to C were the same except for their sign, right? That's because velocity is specific to the direction. A to B was going up, B to C was going down. Speed doesn't have that distinction. If you were to take the average speed from A to B, it'd be the same, but from B to C, speed doesn't care about direction; it would be 8.333 without the negative. Does that help?

- anonymous

so speed and velocity are the same magnitude but speed doesnt have direction . so speed tells you how fast you are going but not where you going

- anonymous

Yes! So do you think the average speed from B to C would be greater than, equal to, or less than the average velocity from B to C?

- anonymous

i think it would be greater

- anonymous

I would agree :)

- anonymous

how about from ac

- anonymous

As far as I know, there would be no change from average velocity to average speed.

- anonymous

the answer choices has one of the them stay equal and the other 2 be greater

- anonymous

Could you list the available choices? Thanks.

- anonymous

yah
equal for ab, greater for bc, ac
equal for bc, greater for ab ac
equal for ac greater for ab ac

- anonymous

Huh. One moment.

- anonymous

yah i know

- anonymous

Oh. Okay. I think I know. So, we said that the velocity from A to B was 8.333, right? And we said that the velocity from B to C was -8.333. If you wanted to find the average for the entire thing, you would do (8.333+[-8.333])/2. Make sense? You just add them (because they cover the entire graph and divide it to find average. Now, let's look at speed. We decided that from A to B stayed the same: 8.333, but we also decided that B to C changed to 8.333. Let's take the average:

- anonymous

(8.333+8.333)/2. This is the average from A to C! And it's greater with speed than it is velocity! That answers our question, I think. Same for AB, but greater for BC and AC.

- anonymous

Did you follow that?

- anonymous

kinda of im a little confused to be honest

- anonymous

Okay. Let's see if I can clear this up.

- anonymous

The average velocity of the entire curve can be obtained by adding together the avg velocity of one half of the curve and the other half and then dividing it by 2. Does that make sense?

- anonymous

yah

- anonymous

Okay, and the average velocity for the entire curve is the same thing as average velocity from A to C, yes?

- anonymous

yah bevause the entire curve goes from a to c

- anonymous

Our velocity: \[[8.333+(-8.333)]\div2\]
Our speed: [8.333+8.333]div2
Do you see and understand the difference?

- anonymous

Oops.\[[8.333+8.333]\div2 \]

- anonymous

yah our velocity is zero and our speed is somethings else

- anonymous

so it shows that it changed?

- anonymous

Right! And did it become greater than or less than? What do you think?

- anonymous

it became greater

- anonymous

Yes. So, in conclusion, we agreed that AB is the same, BC is greater and just now, AC is also greater. Is that an option?

- anonymous

yah you are right it is correct. so when we have tofind speed we have to always find the average

- anonymous

Well, at this point, if you do not have a function for speed, you can only 'guess'. In Calculus it is possible to find the instantaneous speed, but otherwise you can only take averages.

- anonymous

can you explain quickly how you find the average of anything

- anonymous

Well, you add up all of the terms (whatever they may be) and then divide it by the number of terms. Good?

- anonymous

but didnt we have 3 terms

- anonymous

No, we had 1 half of the curve and the other half of the curve, so we divided by two, right?

- anonymous

okay . sohow do we know if we have a function for speed

- anonymous

Well, in the example you were just given a position function; the graph showed the position of the object over time. If the question had included a v(t) function, i.e. velocity over time, you could use it to find the velocity. Or, if the question had included a s(t) function, speed over time, you could use it to find the speed. Neither of these were included so we don't worry about it.

- anonymous

so just as a refresher what is a function

- anonymous

In general (and by no means is this an all inclusive definition) it is an equation that you put in an input (typically x) and receive and output (typically y). It has to pass the 'vertical line test' meaning that there are no two x values that equal the same y value.

- anonymous

okay thank you for all your help and patience.

- anonymous

You're welcome :) Good luck!

- anonymous

thanks.

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