Calculating distance using the speed of sound through two different mediums

[Joe Schmoe]

Homework Statement

David is swimming when he hears a beaver sap the water with its tail, first through the water (v= 1400 meters/second) and then 0.95 seconds later (after the sound reaches him through the water) through the air (v= 340 meters/second).

Homework Equations

Find the distance from David to the beaver.

The Attempt at a Solution

This one stumped me. I thought maybe I 'd have to compare the two velocities and do some cross multiplication but I really don't know.

 

[gneill]

Homework Statement

David is swimming when he hears a beaver sap the water with its tail, first through the water (v= 1400 meters/second) and then 0.95 seconds later (after the sound reaches him through the water) through the air (v= 340 meters/second).

Homework Equations

Find the distance from David to the beaver.

The Attempt at a Solution

This one stumped me. I thought maybe I 'd have to compare the two velocities and do some cross multiplication but I really don't know.

What equations do you think apply here? If you happened to know the distance and velocity, how would you find the time? Write the expressions for the time of travel for both modes of transport (air, water) using a variable to represent the unknown distance.

 

[Joe Schmoe]

Well, I obviously know that d=t×v. This would mean that in the air the sound travels 323 m (d= 0.95×340). The problem is that isn't the total distance, there remains an additional distance that I can't figure out. If I sub in the time value into an equation with the sound velocity in water (d=0.95×1400) then the sound will have traveled 1330 m. But I am sure that neither of these two values are the actual distance. There remains another step but I am unsure what to do.

 

[MrWarlock616]

The distance traveled in both mediums is a constant. 0.95 seconds is a relation between the time taken by the sound wave in both mediums to cover this distance.

 

[gneill]

Well, I obviously know that d=t×v. This would mean that in the air the sound travels 323 m (d= 0.95×340).

The problem is that isn't the total distance, there remains an additional distance that I can't figure out. If I sub in the time value into an equation with the sound velocity in water (d=0.95×1400) then the sound will have traveled 1330 m. But I am sure that neither of these two values are the actual distance. There remains another step but I am unsure what to do.

No, the 0.95 seconds is the difference in time between the arrival of the sound by the two paths.

There is only one unknown distance, the distance between the beaver and the person. Call that distance d. Write the expressions for the two arrival times of the sound assuming distance is d; so t1 = ?, t2 = ?.

 

[Joe Schmoe]

OK, so t1 (water)= d/1400
t2 (air)= d/340

 

[gneill]

OK, so t1 (water)= d/1400
t2 (air)= d/340

Okay. Now, what is the relationship between t1 and t2? (This is where the .95 seconds comes into play).

 

[Joe Schmoe]

Then, t2=t1+0.95
so: d/1400=d/340+0.95

 

[MrWarlock616]

[STRIKE]Yes. Now write another equation using speed=distance/time.[/STRIKE]

 

[Joe Schmoe]

no, sorry, I mean: d/340=d/1400+0.95

 

[MrWarlock616]

no, sorry, I mean: d/340=d/1400+0.95

You can edit your posts.

 

[Joe Schmoe]

You mean like this: 340=d/(d/1400)+0.95

 

[MrWarlock616]

You mean like this: 340=d/(d/1400)+0.95

Using your equation d/340=d/1400+0.95, can you not solve for d?

 

[Joe Schmoe]

So would the answer be 690m?

 

[MrWarlock616]

I don't think so.

 

[Joe Schmoe]

OK, my bad. I am just unsure at how to rearrange d/340=d/1400+0.95 to isolate d. Any ideas?