Relative Motion problem driving me crazy

AI Thread Summary
A battleship is moving east at 24 km/h while a submarine fires a torpedo at 50 km/h from a distance of 4 km, with the ship observed at 20º east of north. To solve for the direction to fire the torpedo, a coordinate system should be established, placing the origin at the submarine's position. By setting the coordinates of the ship and torpedo equal after a time t, two equations can be formed to solve for the unknowns, t and the firing angle theta. The discussion emphasizes using trigonometric relationships and the Pythagorean theorem to simplify the problem. Proper manipulation of the equations will lead to the solution for theta, allowing the torpedo to hit the moving battleship.
Libohove90
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Homework Statement



I figured I'll try this again.

A battleship steams due east at 24 km/h. A submarine 4.0 km away fires a torpedo that has a speed of 50 km/h. If the ship is seen 20º east of north, in what direction should the torpedo be fired to hit the ship?


Homework Equations



Battleships motion: x = x_{}0 + v_{}xt

Torpedo's motion: x' = v cos \phi t
y' = y_{}0' + v sin \phi t



The Attempt at a Solution



In my earlier thread, someone suggested that this is a related rate problem. I am not sure how to do that.

I tried vector addition, but two are velocity vectors (ship and torpedo's velocity) and one is a position vector (the distance between them). This does not seem like a difficult problem, but I need some guidance real bad here, it's halting my progress to doing other problems because I am too stubborn. I appreciate any useful help...thank you
 
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First, to make positions easier to reason about, set up a coordinate system. I think it's easiest to place the origin at the submarine's initial position, but you can choose whatever origin you prefer. Now, assume that the torpedo heads off in some direction theta.

Suppose the torpedo and ship collide after t seconds. What will the ship's coordinates be, after t seconds? What will the torpedo's coordinates be? Setting these coordinates equal to one another will give you 2 equations, one for x and one for y. You now have 2 unknowns--t and theta--and can solve for theta.
 
ideasrule said:
First, to make positions easier to reason about, set up a coordinate system. I think it's easiest to place the origin at the submarine's initial position, but you can choose whatever origin you prefer. Now, assume that the torpedo heads off in some direction theta.

Suppose the torpedo and ship collide after t seconds. What will the ship's coordinates be, after t seconds? What will the torpedo's coordinates be? Setting these coordinates equal to one another will give you 2 equations, one for x and one for y. You now have 2 unknowns--t and theta--and can solve for theta.

The equation governing the motion of the ship is x = xo + vt.
The equations governing the motion of the torpedo are x' = vx' t and y' = vy' t.

I need to set x = x', but before I do that I must solve for t in the third equation and that is t = y' / vy'. I plug this equation for t in the x = x' set up and I get xo + vx ( y'/vy' ) = vx' ( y'/vy' ).

Am I on the right track? If so, I am still not able to solve for theta after doing algebraic manipulations.
 
Libohove90 said:
The equation governing the motion of the ship is x = xo + vt.
The equations governing the motion of the torpedo are x' = vx' t and y' = vy' t.

I need to set x = x', but before I do that I must solve for t in the third equation and that is t = y' / vy'. I plug this equation for t in the x = x' set up and I get xo + vx ( y'/vy' ) = vx' ( y'/vy' ).

Am I on the right track? If so, I am still not able to solve for theta after doing algebraic manipulations.

You can go that route, substituting vx' = Vsin(θ) and vy' = Vcos(θ), where V is the torpedo speed and θ is the firing angle (from the vertical). The algebra may get a bit hairy, since you'll have to deal with the sin and cos terms in "unfriendly" places. You'll probably have to square things to use sin2 + cos2 = 1 to change one of the sines or cosines.

If I might make a small suggestion? Take advantage of the right-triangle formed by the sub, the north-south axis, and the ship+torpedo intersection point. The side lengths must be related by Pythagoras, and you have formulae for the side lengths with respect to time.
attachment.php?attachmentid=36898&stc=1&d=1309622516.gif

In the figure, triangle ABC is a right triangle. You can calculate yo easily enough. Side lengths BC and AC depend only upon time (you don't need to figure x or y components!). Solve for time. With time you have all the side lengths, so the firing angle becomes trivial to find.
 

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