Breaking down a velocity

1. Oct 24, 2012

Physicist1231

I have a scenario that is driving me crazy and I am hoping someone can help.

My setup:

Person A is at coords (0u,0u,0u)
Person B is at coords (10u,0u,0u)

Person A has a bell and at the same time he rings it Person B moves with a velocity of (1u,1u,1u)
At this point Person B has not "heard" the bell.

For the sake of simplicity lets say the sound of the bell travels at an overall speed of 5u/s.

Notice that Person B velocity is broken down into 3 coords whereas we just know the overall speed of the sound.

My questions are:

a) How long does it take for Person B to hear the bell?
b) What is the breakdown of the bells velocity into 3d coords?

I know it can be done... Just dont know how. Let me know if any more information is needed

2. Oct 24, 2012

Staff: Mentor

This may belong in the homework forum... But if it does, it will get moved in due time, so no big deal.

And to get you started: Think of the sound wave as a sphere expanding out from the origin; you can write its distance from the origin as a function of time. And you should be able to write person B's distance from the origin as a function of time as well.

3. Oct 24, 2012

Physicist1231

Yeah, I got that part about a sphere. My problem is I have two (at least) unknowns . How do I solve for that?

The time it takes to intersect and the breakdown of the sound wave into its three coords so I can add the velocities...

Any thoughts? I have been trying to rip this apart and feel that an extra brain or two can crack this one.

Btw this is not homework as I could not even solve this with my college teacher... course she was not that interested and said I would fail. Rather this is a personal endeavor for something more grand...

4. Oct 24, 2012

Physicist1231

Given the example I can start with:

(radius)/speed of sound. = (distance walked)/speed of person b

(Initial distance + distance walked)/SOS = (distance walked)/ SOP

(Di + (SOP * T)/SOS = (SOP * T) /SOP

This just looks like its going to just cancel out or something...

5. Oct 25, 2012

JustinRyan

Try this,,

Initial distance + (Velocity of person x T) = Speed of sound x T

solve for T

You will need to consider the velocity as a vector wrt the origin. Not just speed.

6. Oct 25, 2012

Physicist1231

Ok doing that I get:

Di / (Sos - Vop) = T

What next?

7. Oct 25, 2012

nasu

This equation suffers from inconsistency. The right hand side is a scalar whereas the RHS has at least one vector quantity (velocity x time).
You could write it correctly by using components, for example.
(xo+vx*t)^2+(yo+vy*t)^2=(ct)^2
where xo,yo are the coordinates of the initial position of the moving object,
vx and vy are the components of its velocity and c is the speed of sound.

8. Oct 25, 2012

Physicist1231

Granted.

Eventually you will need to compare C (speed of sound) with V (velocity of the person). To do this you will need to break down both into the three components. V is already done with (1,1,1).

How do you properly breakdown C into its three components without knowing how much time it took to intersect

--or--

How do you know how much time it took if you cannot break down C into its three parts properly???

9. Oct 25, 2012

nasu

You don't need to "break" c into any components. Solve the equation that I showed in my previous post and get t (you know everything else).
Your point (b) does not make sense as formulated.
You can resolve the velocity of sound into components at every point in the space around the source, but the components will change from point to point. So there is no unique way of "breaking it".
Is there another problem you want to solve? For the one formulated here, you don't even need it. You can find the time at which he hears the sound for the first time as well as and the position where this happens.

10. Oct 25, 2012

Physicist1231

Please let me know how I can make it clearer. I was trying to say that the source of sound is at (0,0,0) and person b was 10 units away from that point on the X axis.

You can make a formula to get the C components but you still cannot solve for the correct set without knowing the total time it traveled to intersect with person B.

Maybe i am just not following you. You are telling me i have enough information to solve the equasion (and i agree with you). I just do not see HOW to get there...

Can you solve this and show me the steps you took?

11. Oct 25, 2012

nasu

Just draw a figure and use the Pythagorean theorem.
Put the point where it hears the sound, let say P. If the origin is O, then OP=ct.
On the other side, the coordinates of the point P are can be calculated in terms of the motion of the person. On the x axis, you have xo+vx*t and on the y axis yo+vy*t.
(In your specific case, yo=0).

12. Oct 25, 2012

JustinRyan

Sorry yes, what I meant to say was you need to convert the vector quantity velocity into a scalar -> distance from origin as a fn of time D.

D^2=(10+t)^2 + 2t^2

Set that equal to the distance of the sound wave front d=5t

I think you should find the time they meet t=sqrt(100/22)

That probably doesn't help much. One day I will figure out how to type the math functions properly.

Last edited: Oct 25, 2012
13. Oct 26, 2012

Physicist1231

I do see what you are saying. This would normally work if the paths made a right triangle. Then using pythagorean theorem that would make things easier.

14. Oct 26, 2012

nasu

If you draw a diagram you will see it even better. It's not the "paths" that make a right triangle but the x,y components of the position vector of the point P, where the sound catches with the person.

15. Oct 26, 2012

Physicist1231

After reading this whole thread through light bulbs are going off!!! this might be what I am missing. The sad thing is that it would have been so simple... but such is life and that it what an extra set of eyes are for!!!

16. Oct 29, 2012

Physicist1231

Well.... I ran in to another issue.
The following formula makes sense but i question it's usefulness or accuracy. Mainly because i can get two answers when there can only be one...

(CT)^2 = (Dx + (VxT))^2 + (Dy+ (VyT))^2 + (Dz + (VzT))^2

Standard P. Theorem for a 3d triangle.

Now we plug in the numbers:

(5T)^2 = (10 + (1T))^2 + (0+ (1T))^2 + (0 + (1T))^2

Then we get

25T^2 = 100 + 20T + 3T^2

this will transform into:

22T^2 - 20T - 100 = 0

using the Quadratic Formula equasion this comes out to:

T = 2.6345.....
T = -1.7254....

Obviously the Negative number is out but this still causes an issue with what I am trying to do.

Guessing that the first one to be accurate makes sense because if he had just stood still the sound should have reached him in 2 seconds. Walking (away) from the sound would increase that time naturally. Had he been walking in the opposite direction (toward) I would expect the absolute value of the second answer.

Of course when it comes to math guessing is not always an option. Any ideas? Please let me know if I am looking at this wrong... I am really counting on you guys!

17. Oct 29, 2012

JustinRyan

Sorry physicist1231 I may not have been clear what I did.

CT = Hypotenuse
Dx + VxT = Adjacent
Sqrt((Dy+VyT)^2 + (Dz + VzT)^2) = Opposite

I used pythagoras thoerem first to find the hypotenuse between Dy+VyT & Dz+VzT.
Then again with the result from the first used as the opposite side in the larger triangle to solve for T with CT as the hypotenuse.

Hope that makes more sense.

Last edited: Oct 29, 2012
18. Oct 30, 2012

nasu

You did not say what are you trying to do. The negative solution it is not an issue or problem by itself. It happens all the time when you have quadratic equations.

19. Oct 31, 2012

Physicist1231

It makes sense but Pothag theorem can be expanded in three dimentions by

D^2 = A^2 + B^2 + C^2

where D is the Hypotenuse. Your way works too!!

20. Oct 31, 2012

Physicist1231

Yeah and that was one of my fears when seeing the quadratic equasions because given my problem that I put in place I could only have one correct answer so I shunned the quadratic equasion.

I just forgot that since the Person B is traveling in a straight line that and that pusing the numbers through the equasion we loose the "direction" he is traveling (either forward or backward) on his path of travel. Thus we get two answers.

At that point we just need to remember which direction he is traveling (ultimately closer or further away) compared to the source of the noise.

You are also correct that I did not tell you what I was ultimately trying to do because I did not want you to call me crazy lol and drop the subject without giving it a second look.

You guys did great! Thanks!