Relative Motion problem driving me crazy

[Libohove90]

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

 

[ideasrule]

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.

 

[Libohove90]

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.

 

[gneill]

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.

In the figure, triangle ABC is a right triangle. You can calculate y_o 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.