Sound traveling through air/concrete

[gmmstr827]

Homework Statement

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 s apart. How far away did the impact occur?

The velocity of sound through the indicated mediums are as follows:
v_air(20°C) = 343 m/s
v_concrete = 3000 m/s

Homework Equations

t = d/v >>> d = tv

The Attempt at a Solution

d = tv
d = (0.75 s)(343 m/s) = 257.25 m

t = d/v
t = (257.25 m)/(3000 m/s) = 8.575 s

d = (8.575 s)(3000 m/s) = 25725 m

So, the rock was 25725 meters away?

I feel like that is a huge distance, and that perhaps I should have stopped at 257.25 m and used that as my answer instead of plugging the value back into the equation. How do I find the time that passed before he heard the rock through the concrete?
If I use 257.25 m as the answer, it would be assuming that the time it took for him to hear it through the concrete was almost instantaneous.

 

[verty]

I want to be helpful, so I'll say, think about all the important moments in time and what happens at those moments. Then draw a sketch and think about what you need to calculate.

 

[cupid.callin]

d = tv
d = (0.75 s)(343 m/s) = 257.25 m

what are you doing?
first you take time to find distance ... then you use that distance to find time again, then again you find distance from it

"t" is not time taken by sound in air but time difference.

 

[gmmstr827]

what are you doing?
first you take time to find distance ... then you use that distance to find time again, then again you find distance from it

"t" is not time taken by sound in air but time difference.

So the part you quoted is the correct answer?

 

[cupid.callin]

No.

You need to find the distance. let it be d

what are the time taken by sound in air and concrete?

 

[gmmstr827]

No.

You need to find the distance. let it be d

what are the time taken by sound in air and concrete?

You quoted the distance formula
d = tv
and what I used to solve it
d = (0.75 s)(343 m/s) = 257.25 m

I stated the velocity at which sound travels through those mediums in the original post:
v_air(20°C) = 343 m/s
v_concrete = 3000 m/s

So what's wrong with what I did?

 

[cupid.callin]

why are you thinking that .75 is the time taken by sound to travel total distance in air?

 

[gmmstr827]

The time it took for the sound to travel through the concrete is not given. Due to this, and a lack of a way to find it, I'm assuming the large velocity at which time travels through concrete will make that time near instantaneous, giving a total of ABOUT .75 s.

 

[cupid.callin]

just take time as t for concrete and ? for air

distance is same for both ... just use d=vt for both

 

[gmmstr827]

I can't do that if 0 s is when it is heard through the concrete. At t = 0s he hears it through the concrete, at t = .75 s he hears it in the air, it occurs at a negative time to this scale, which is not given. Can you elaborate if you understood that before?

 

[cupid.callin]

I can't do that if 0 s is when it is heard through the concrete. At t = 0s he hears it through the concrete, at t = .75 s he hears it in the air, it occurs at a negative time to this scale, which is not given. Can you elaborate if you understood that before?

Why is it negative time?
remember that speed is more in concrete than in air
so time in air is (less or more ?) than concrete?

 

[gmmstr827]

Why is it negative time?
remember that speed is more in concrete than in air
so time in air is (less or more ?) than concrete?

Did you read the problem? The object hit the ground, then he heard it in the concrete, THEN 0.75 s AFTER he heard it in the concrete, he heard it in the air, so on the scale I just gave, the object hit the ground at a negative time in respect to that scale (x seconds before he heard it in the concrete).

Sound obviously takes longer to travel through air than concrete, which is why I said I assumed the time it took to travel through the concrete to be nearly instantaneous.

If you know the answer, don't keep it a secret, because you aren't really telling me anything that I don't already know.

 

[cupid.callin]

i can't tell you the answer ... PF rules!

just solve these eqn's:
d = v_air x (t + .75)
d = v_concrete x (t)

 

[gmmstr827]

Ok, so then:

d = v_air x (t + .75)
d = (343 m/s)(t+.75)
d = 343t + 257.25

d = v_concrete x (t)
d = (3000 m/s)(t)
d = 3000t
343t + 257.25 = 3000t
2657t = -257.25
t = -0.097 s when the object hit the concrete?

so d = 343(-0.097) + 257.25 = 224.04 m?

I think the value should be larger than the one I found myself, so I'll try with time as a positive value.

d = 343(0.097) + 257.25 = 290.52 = 291 m
or d = (3000)(0.097) = 291 m

Well, that looks much better. Am I on target this time?

 

[cupid.callin]

yes its correct

 

[gmmstr827]

Okay, thank you!

 

[verty]

Consider this advice for the future. One can learn a lot more a lot faster by thinking about what people say and why they say it. If it sounds like you already know it, it often means you haven't thought enough about it.

People are here to help, after all.