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Having trouble with relative motion

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data

    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?


    2. Relevant equations

    x motion of ship: x = vt

    x motion of torpedo: x' - x[itex]_{}0[/itex]' = v[itex]_{}x[/itex]'t

    y motion of torpedo: y - y[itex]_{}0[/itex] = v[itex]_{}y[/itex]t

    3. 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[itex]_{}0[/itex] = v[itex]_{}y[/itex]t

    Then x = x'

    In which you get x[itex]_{}0[/itex] + v[itex]_{}x[/itex] (-y[itex]_{}0[/itex] / v[itex]_{}y[/itex]) = x[itex]_{}0[/itex]' + v[itex]_{}x[/itex]' (-y[itex]_{}0[/itex] / v[itex]_{}y[/itex])

    Do some algebra and you get x[itex]_{}0[/itex] - x[itex]_{}0[/itex]' = (v[itex]_{}x[/itex]' - v[itex]_{}x[/itex]) / v[itex]_{}y[/itex]

    But I can't solve for [itex]\Theta[/itex]
     
    Last edited: Jun 30, 2011
  2. jcsd
  3. Jun 30, 2011 #2
    The answer is 46.8º east of north...can someone help me how to handle this problem?
     
  4. Jun 30, 2011 #3

    SteamKing

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    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.
     
  5. Jun 30, 2011 #4
    Consider the battleship as the reference substance
     
  6. Jun 30, 2011 #5
    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.
     
  7. Jun 30, 2011 #6
    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)
     
  8. Jun 30, 2011 #7
    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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