# Finding Distance with given 2 speeds and a time

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1. Jun 11, 2016

### Kushal Gurung

1. Example (bomb): There is a ship and you need to calculate the distance between the bomb and the ship. A bomb explodes underwater, the underwater sound wave travels at a speed of 1400 m/s and in the air at a speed of 330 m/s (0°C). The sound wave underwater is 5 seconds faster than in the air. Calculate the distance between the bomb and the ship.

My attempt: ( to be honest I've tried my best, I've even talked to myself explaining things that doesn't make sense. I'm going mentally ill here since I can't solve the question. I stress out too much when I can't figure out and that basically makes it worse for me.)

2. Jun 11, 2016

### SteamKing

Staff Emeritus
Well, why don't you write out and post your best calculation about the distance. That's what PF expects from people seeking help with HW.

If you don't do this, how can we know what you understand about this problem and what you don't understand?

3. Jun 11, 2016

### Kushal Gurung

That's why I explained about my attempt. I don't know how to start it. I've used the formulas that we were given but I don't know where to start and where to end. Example of my attempt.

Since we have speed and time we are looking for the distance.

Distance = Speed x Time

Distance (underwater): 1400 m/s x time

Since we know that 5 seconds faster means it isn't the actual time so we go ahead and look for the real time.

Distance (in the air): 330 m/s x time

Well that's it. I don't know if I started correctly or I'm just outer space.

4. Jun 11, 2016

### SteamKing

Staff Emeritus
You're overlooking one important fact: it takes the sound 5 seconds longer to travel thru the air than to travel the same distance thru the water to the ship.

Do you think you could write an equation which expresses this fact?

5. Jun 11, 2016

### Kushal Gurung

Distance (underwater): 1400 m/s x 5s = 7000 m
Distance (in the air): 330 m/s x 10s = 3300 m

Distance (total) = 10.300 m

Distance = 10.300 m / 5s = 2030 m

Ok I'm lost again.

6. Jun 11, 2016

### SteamKing

Staff Emeritus
You're just making stuff up now. There is no evidence that the sound travels for 10 sec. in air. What if it takes sound 15 sec. to reach the observer from the ship thru the air?

Remember, the distance from the observer to the ship is the same. The sound takes two different times to travel this distance, but you only know that these times are 5 seconds apart.

Call the distance to the ship D. How would you calculate the time it takes the sound to travel the distance D in water, and then in air?

How would you express the difference in the two times such that the difference is 5 seconds?