1. Nov 25, 2003

### ero

Yeah, yet another nimrod with absolutely no clue on how to solve problems. I seriously think I need to stick with literature... something I actually get. >.<;

Anyhow, I have this problem I need to solve, but have no idea how. It's AP Physics B, so we're still based on Algebra, not Calculus. :)

Here's my problem:

Q: How deep was the Scorpion when its torpedo exploded?

Given/Background: In 1968, the attack submarine U.S.S. Scorpion mysteriously sank in the North Atlantic Ocean. On the night of October 26, 1993, CBS news aired a report on the sinking of the Scorpion with the loss of over 90 seamen. The report interviewed a United State's Navy Admiral who said the Scorpion sank itself when one of its torpedoes accidentally armed itself and exploded. The Admiral also stated that the wreck of the Scorpion was found in 11,500 of water. The evidence from the wreckage indicated that the sub. hit the ocean floor at 30 mph and took only 90 seconds to reach the ocean floor after the explosion. The U.S.S. Scorpion was 30 meters in length, 9 meters in width, and weighed (mass) 3.6 million kg. Water weighs about 1 kg per liter.

So, yeah... all I know is this: *sigh*

• The formula for finding acceleration of a body through a fluid:
a= W-R/M
• a= acce. of sub. through water
• W= weight of sub.
• R= resistance of water
• m= mass of falling object through water

So far, for W, I have 352800000 Newtons (w= m x g -> w= 3.6 x $$10^6$$ x 9.8) and 3,600,000 kg for m.

... And that's all. -_-;
To tell the truth, I don't see how solving for a helps much. That and I have no idea how to solve for R (resistance of water). I know I have to take in count the time when the sub. reaches terminal velocity (so that within 90s, it's going at constant velocity - 30mph.)I have to get the answers in feet, so I know I have to go from meters to feet and that something has to be cubed.

So, please help. I was able to get the equation and decipher what each letter represents, but I have no idea what to do, or how to proceed next!

I would really, really appreciate the help! :)

2. Nov 25, 2003

### chroot

Staff Emeritus
Where did you get this formula a = W-R/M? The drag force is actually dependent upon the square of the velocity.

- Warren

3. Nov 25, 2003

### ero

Hm? But isn't the velocity changing? How can it be squared when a number is constantly changing? Wouldn't it have to be at constant velocity?

Do I know anything? No, not really. >.<;

Oh, and that was our only given equation. We're to use that for finding the acceleration of the sub. though the water (keeping in mind water resistance, I guess.).

4. Nov 26, 2003

### chroot

Staff Emeritus
If the velocity is changing, and the drag depends upon the velocity, then the drag is also changing.
What do you plug in for this 'R' when you solve the problem? Your teacher is probably just approximating the drag force as a constant to make the problem easier to solve. Truth be told, however, the drag force is not at all constant, but depends upon the square of the velocity.

In any case, you need to solve for the acceleration of the sub. If R is constant, so is a. You could then use the equation

$$v(t) = v_0 + a t$$

to find the velocity as a function of time. That's why your teacher is approximating the drag force as constant -- if it weren't constant, you'd have to use calculus to solve the problem. You're more interested in the position, though, so you can use the equation

$$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$$

to find the position as a function of time.

Do these equations make sense to you? Let me know if they don't, and we can talk about where they come from.

- Warren

5. Nov 26, 2003

### ero

Well, one of the biggest reasons for my confusion is the fact that we have to solve for R, and I've no idea how. No one (in class) does, actually. Somehow, R, water resistance, has to be solved and left in Newtons.

Does the this: $$H_2O$$ = 1g = 1$$cm^3$$ play into the problem at all?

I believe it does, since something has to be cubed, and the only cube I see is in that "formula". But then, would dimensions come into play, to get the cube, I mean? So far, there's only two... but what about the third?

The first equation does make sense, but that's only because I've seen that before. But yet, how to get the velocity to plug in?
The second, I'm afraid, I've never seen before. I guess i'm taking dummy-physics or something... and even then... Oh, jeez.

Oh, and thanks for the help thus far. I'm so grateful! :)