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Expressing tan θ in a triangle

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data


    The original question is from AP French's Classical Mechanics. It's about a boat trying to catch up with a ship. I've interpreted the question as follows: which is to express tan θ as a function of u and v.

    a0xfh1.png


    I tried working with vectors, where v is actually -v in vector notation.

    where tan θ = (-v . (u-v))/(-v x (u-v)

    but that only complicates things as we don't know much about the angles.

    So in this question, θ is fixed, u is fixed and v is fixed.

    Tried using the cosine rule, but that only gives the unknown length...
     
  2. jcsd
  3. Oct 21, 2012 #2

    PeterO

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    Your diagram does not give any indication of the direction of vectors u and v .
    Also, as you said, this was the way you interpreted the problem. We have no idea whether that is correct either.
     
  4. Oct 21, 2012 #3
    Okay, here's the original question:


    A ship is streaming parallel to a straight coastline, distance D offshore, at speed V. A coastguard cutter, whose speed is u (u<V) sets out from a port to intercept the sip.

    23vbwd0.png

    (A) Show that the cutter must start out before the ship passes a point a distance
    D√(V2 - u2)/u back along the coast.

    (B) If the cutter starts out at the latest possible moment, where and when does it reach the ship?


    My idea is: The fastest way for the cutter to approach the ship is to (in frame of the ship) move directly at a straight line to it.

    Then letting the distance be x, we have a right angled triangle with tan θ = D/x.

    The relative velocity diagram is drawn too, with the cutter moving at an angle θ.
     
  5. Oct 21, 2012 #4
    Doesn't the coastguard guy always go in the direction of the ship? So he would be actually moving in a curvy "trajectory". I mean yeah, he can go to the ship by..predicting the ship's position and going in a straight line to where the ship supposedly would be once he got there, but he doesn't know for sure. My guess is that the coastguard guy always is heading toward the ship, since that is the fastest way to get to the ship without taking chances.
     
  6. Oct 21, 2012 #5

    PeterO

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    A couple of things I notice:

    Your θ is in a different place this time, in the relative velocity diagram.

    While that may represent the fastest way for the cutter to reach the ship, the question was about the position of the ship at the latest time the cutter could leave port and still intercept. Perhaps I am misunderstanding your thinking.
     
  7. Oct 21, 2012 #6
    Let's try a different approach:

    1) Shortest time is achieved when the relative velocity (in the frame of the boat) is when the cutter moves directly towards the boat.


    Writing an expression for time:

    t = x/(u cosθ + v)

    t = D/(u sin θ)


    To find the shortest time with respect to x, we have to find dt/dx = 0...


    I tried to substitute sin θ = √ 1 - cos2θ but that made the expression very complicated..
     
    Last edited: Oct 21, 2012
  8. Oct 22, 2012 #7
    would appreciate any input..
     
  9. Oct 27, 2012 #8
    help, anyone??
     
  10. Oct 27, 2012 #9
    I'm heading out, but some very quick observations:

    I assume the port is located a horizontal distance x from the D vector.

    The cutter ship must travel the perpendicular distance D from shore at minimum to have any chance to catch the boat. Any direction at an angle to the shore will cover more distance and take longer. You should be able to draw a conclusion from that.

    The second thing that crossed my mind is at time t, the cutter can be on any point of a circle centred at the port with radius Ut. The streaming ship will be located at (Vt,D). The intersection of the line and circle (that changes with time) will show the course the cutter should take.

    I don't know if the latter is a dead end but a little more thought may lead to an analytical solution.
     
  11. Oct 27, 2012 #10

    PeterO

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    OK, let's assume the best strategy is for the cutter to sail at right angles to the shore.
    The time taken for the cutter to reach the point of interception is thus D/u.

    Steaming at speed V, the ship will cover distance V. D/u which I choose to write as (D/u).V
    So that would represent how far the point is back along the coast.

    The answer they want you to show can be re-written as (D/u).√(V2 - u2)

    Clearly different - and the √ part looks like part of result of a Pythagorean calculation?
     
  12. Oct 28, 2012 #11
    1) The latest possible time the cutter can sail out must be at right angles, cause at any other angles it will take a longer time. Consider a very small angle pointing to the left - D has to be very big in this case. If the angle is greater than 90 degrees, the time taken would clearly be longer as you are moving away from the ship, compared to moving towards the ship

    2) Time taken = D/u. Time in ship's frame = Time in cutter's frame = Time in laboratory frame

    Whether in the frame of the ship, cutter or the laboratory, this time is unchanged (classically). So in the ship's frame, the velocity of approach can be taken as

    V' = √V2 + U2

    Horizontal distance x = V' cosθ * t = vD/u

    I still get the same answer....

    Besides, if the distances were equal,

    (D/u).√(V2 - u2) = (D/u).V

    This would imply:
    u = 0

    which doesn't make sense..
     
  13. Oct 28, 2012 #12

    PeterO

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    Without getting my head around the actual calculations - because that is your job - I draw your attention to the following.

    If the cutter is to go perpendicular to the shore, it will take time D/u, and must leave when the ship is distance V.(D/u) back along the coast.

    If the cutter was to aim 1 km left, it would take slightly longer to reach the intercept line. That would mean it left when the ship was slightly further back from the intercept point - so clearly further than V.(D/u) back to the left.

    HOWEVER: If the cutter aimed that same 1 km right, the ship - while being slightly further back from the intercept point - so more than V.(D/u) back to the left of the intercept point - may have been less than V.(D/u) back from the cutter port.
     
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