Physics Question
 

This is really more math but I can't figure it out:

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.75 seconds apart. How far away did the impact occur? Assume speed of sound in concrete is 3000 m/s and speed of sound in air is 343 m/s.

Clearly, the sound traveling through the concrete will be heard first. I tried to relate it to the distance=velocity*time equation but that did not yield satisfactory results. I built a spreadsheet and I think I have the right answer but would like to know how to do it using only pencil and paper.

Thanks.

Re: Physics Question
 

d = 343(t + 0.75)
d = 3000t

and solve? looks like 290m to me

Re: Physics Question
 

So you have two equations where the distance is the same and the time is different by .75 seconds. The slower speed will be farther away.
343 * x+.75 = d
3000 * x = d
3000 * x = 343 * x+.75
3000x = 343x + 257.25
2657x = 257.25
x = .1 second

d = 3000 * .1 = 300m
d = 343 * .85 = 292m

There are rounding errors all over, but that is the basic idea I think.

Re: Physics Question
 

It looks like algebra to me. We know that x (the distance) divided by t1 (the time it takes to travel through the concrete) is 3000. We know that x divided by t2 (the time it takes to travel through the air) is 343. We know that t1 is .75 seconds less than t2.

x/3000 = t1
x/343 = t2
t1 = t2-.75
x/3000 = (x/343)-.75

I'm solving this in a few different ways, but none of them are very elegant. I'm sure there's a simple trick I'm missing to make the algebra clean, but the answer is pretty consistent. I'm pretty sure it's x=(343*3000*.75)/(3000-343) or 771750/2657 or ~290.46.

(It looks like kerowo has a cleaner solution, maybe that's a better approach?)

Re: Physics Question
 

Much appreciated.

Thanks.

Re: Physics Question
 

I'm frustrated with how sloppy this all is. Someone please show me how to make it clean? I find it looks much neater if I do it in the abstract, so here's what I have...

a=3000
b=343
c=.75

x/a = x/b-c
x/a = (x-cb)/b
bx/a = x-cb
bx/a + cb = x
(bx+cba)/a = x
bx+cba = ax
cba = ax - bx
cba/a-b = x(a-b)/a-b
cba/a-b = x

Re: Physics Question
 

I'll take a stab at it:

a=3000
b=343
c=.75
x=time, what we're solving for

ax=b(x+c)
ax=bx+bc
ax-bx=bc
x(a-b)=bc
x=bc/(a-b)

substitute x into ax or b(x+c) and you've got your answer.

Re: Physics Question
 

You guys are killing me.

http://i27.photobucket.com/albums/c1...equation-1.png

Edit: Let me know if I have to explain that.

Re: Physics Question
 

Okay, that's the answer we got, right? I mean, vc is the same as a (3000), va is the same as b (343), and dt is the same as c (.75). So it's cba/(a-b), isn't it? ~290m?

I didn't know it was an "official" principle...

Re: Physics Question
 

May you all burn in UNH (Unitless Numerical Hell).

Not really.

Maybe a little.

My point being that these sorts of problems would just be SO much easier if people would just solve them algebraicly, rather than plugging in a bunch of numbers and leaping off a bridge with them.

http://i27.photobucket.com/albums/c1.../equation1.png