Having trouble with relative motion

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Homework Help Overview

The problem involves relative motion, specifically the scenario of a battleship moving east and a submarine firing a torpedo at it. The challenge is to determine the direction in which the torpedo should be fired to successfully hit the moving battleship.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to account for the battleship's movement when aiming the torpedo, with some suggesting the use of related rates to find the solution. Others express uncertainty about the variables involved and how to apply calculus to the problem.

Discussion Status

Some participants have offered hints regarding the need to aim ahead of the battleship due to its motion. There is an ongoing exploration of different methods, including geometric approaches and related rates, but no consensus has been reached on how to proceed effectively.

Contextual Notes

Participants mention the relative positions and speeds of the battleship and torpedo, as well as the initial distance from the submarine. There is a noted lack of clarity regarding certain variables and their meanings in the context of the problem.

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Homework Statement



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



x motion of ship: x = vt

x motion of torpedo: x' - x_{}0' = v_{}x't

y motion of torpedo: y - y_{}0 = v_{}yt

The Attempt at a Solution



The torpedo will hit the ship when its at y = 0, thus solve for t in the y motion equation of the torpedo.

0 - y_{}0 = v_{}yt

Then x = x'

In which you get x_{}0 + v_{}x (-y_{}0 / v_{}y) = x_{}0' + v_{}x' (-y_{}0 / v_{}y)

Do some algebra and you get x_{}0 - x_{}0' = (v_{}x' - v_{}x) / v_{}y

But I can't solve for \Theta
 
Last edited:
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The answer is 46.8º east of north...can someone help me how to handle this problem?
 
Hint: You must aim the torpedo ahead of the battleship in order for it to arrive where the battleship will be after the torpedo has made its run. You know the speeds of the battleship and the torpedo, as well as the relative bearing of the target from the sub at the moment of firing. You'll have to solve a related rate problem in order to obtain the complete firing solution.
 
Consider the battleship as the reference substance
 
azureth said:
Consider the battleship as the reference substance

So that means that the submarine is moving westward at 24 km/h and when the hypotenuse is 4 km, the torpedo is fired.

That means we use x^2 + y^2 = c^2.

d/dt (x^2 + y^2) = d/dt (c^2) which gives us 2x dx/dt + 2y dy/dt = 2c dc/dt...and since there is no change in y, the equation reduces to 2x dx/dt = 2c dc/dt.

x = 4 cos 70º
dx/dt = 24 km/h
I am not sure what c and dc/dt stand for...I am not sure what to solve. Sorry, I haven't done a related rate problem n a while.
 
I do not know how to solve this problem by using calculus.In fact you can draw the vectors and solve it geometrically.(I cannot post a picture for you now.If you still cannot work it out,I will reply to you a few days later)
 
SteamKing said:
Hint: You must aim the torpedo ahead of the battleship in order for it to arrive where the battleship will be after the torpedo has made its run. You know the speeds of the battleship and the torpedo, as well as the relative bearing of the target from the sub at the moment of firing. You'll have to solve a related rate problem in order to obtain the complete firing solution.

I tried the related-rate way a few posts above but I am not sure if I am doing it right.
 

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