Question

Help Pleaseee!(:
A particle's position coordinates (x, y) are (3.0 m, 5.0 m) at t = 0; (6.0 m, 9.0 m) at t = 2.0 s; and (15.0 m, 15.0 m) at t = 5.0 s.
==>Find the magnitude of the average velocity from t = 0 to t = 5 s.

Answer 1

What's stopping you from using the distance formula? That should provide magnitude.

Distance = Rate * Time

When was the last time you pulled that formula out of your hat?

Answer 2

i tried it and i got the wrong answer. I don't know what I'm doing wrong. @tkhunny

Answer 3

Show me what you did and what happened.

Answer 4

\[vector:V(sub,m)= \frac{ r(sub,f) - r(sub,i) }{ \Delta t}\]
\[=\frac{ (15,15)-(3,5) }{ 2 } =(1.5,2.0)m/s\]
\[V(sub,m)=\sqrt{5^2+6^2} = 7.8 m/s\]
@tkhunny

Answer 5

I don't see a distance formula in there. Thus, your equations still have no idea how far we have traveled.

The notation is a little funny, but we should get \((15,15) - (3,5) = \sqrt{(15-3)^{2}+(15-5)^{2}}\). I can't really tell what your notation is trying to tell me.

Answer 6

Oh I did it wrong I guess, I tried to follow an example.

Answer 7

Too bad the example apparently wasn't clear enough. You understand the use of the distance formula here?

Answer 8

yes... could you guide me?

Answer 9

@tkhunny

Answer 10

The arithmetic is already spelled out up above. Please do the arithmetic indicated. What's the square root of 244?

Answer 11

15.62? but that answer was wrong as well.

Answer 12

@tkhunny

Answer 13

Please learn a little patience and stop looking at the answers. Let's solve the problem and THEN look at the answers.

Moving from (3,5) to (15,15) is a magnitude of \(\sqrt{(15-3)^{2} + (15-5)^{2}} = \sqrt{12^{2} + 10^{2}} = \sqrt{144 + 100} = \sqrt{244} = 2\sqrt{61} = 15.6\). This is the total magnitude of the change in position. Since this is not what the problem statement wants, we are not yet done.

Using the old, familiar forumla, Distance = Rate * Time, and solving for Rate, we get Rate = Distance / Time.

We then compute the average velocity like that: \(Average\;Velocity = \dfrac{Total\;Displacement}{Total\;Time}\). Once we have this value, we should be done.

Answer 14

so it would be 15.6/5?

Answer 15

Yup.

Answer 16

Thank you so much!!!(: @tkhunny

Answer 17

We got it? Excellent. Now I 'm really scratching my head about that example you tried to follow. Oh, well. Good luck with that. :-)

Answer 18

Yeah, it was an example my friend had... Thank you though! I really appreciate the help! @tkhunny