Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector , Cosine Rule Problem

  1. Oct 16, 2006 #1
    Question :
    Two ships A and B leave port P at the same time . Ship A travels due north at a steady speed of 15km/h and ship B travels N 60 degree E at a steady speed of 10km/h. what is the distance and direction from A to B after 1 hour ? what is the velocity of B relative to A ?

    Solution :
    I couldnt get the velocity part right .....

    I resolved the two velocities along North and East , since 15 km/hr is due north it doesnt have a component along the East ...the second velocity has components along both the directions so .....

    v = [ (-15 + 10 cos60)^2 + (10 sin60)^2 ]^0.5 = 13.22 km/hr

    1).what i cant get is why dont i get this answer when i use the cosine rule .... is it because im using the relative velocity of A which BTW seems to be negative ???

    2). I get the rigt answer if i changed the '-' part in the cosine rule to a '+'
  2. jcsd
  3. Oct 16, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    I don't understand how your last statement can be true. If the statement of the problem means that ship B is heading 60 degrees North of East, then you have the sine and the cosine reversed. The northward component is the sine term and the eastward component is the cosine term. But then you should have gotten the first part wrong. If the problem statement means 60 degrees East of North, then you have done it correctly.

    If your first part is indeed correct, then the magnitude of relative velocity is just the distance between the ships after 1 hour divided by 1 hour. What distance did you get for the first part? The direction of the relative velocity is the same as the direction from the first part.
  4. Oct 16, 2006 #3
    thats 60 degrees east of north and the displacement was the same as the velocity . I got it right but I cant understand why I get a wrong answer when i use the cosine rule ... is it because im using a negative velocity ???
  5. Oct 16, 2006 #4
    the 60 degrees I think is negative because it is east of north. So you are going clockwise.
  6. Oct 16, 2006 #5
    im pretty sure that 60 degree east of north is positive .... its not about that ...

    what im confused about is how do you find the resultant of a vector subtraction operation with the cosine rule ???? Does the formula have to be changed or something for negative sides of traingles ??
  7. Oct 16, 2006 #6


    User Avatar
    Science Advisor
    Homework Helper

    I don't understand what you are saying.. What cosine rule? The equation you posted and the answer are correct for 60 degrees east of north. The - sign on the 15 is because you are calculating relative velocity of B to A velocity- velocity of B minus velocity of A
  8. Oct 16, 2006 #7
    I usually solve vector problems with cosine rule , the method above was done just so that i could know the correct answer and i understand why it is -15.

    Im just wondering why im not getting the same answer when i use the cosine law ....... is it because of the -15 ???
  9. Oct 16, 2006 #8


    User Avatar
    Science Advisor
    Homework Helper

    Again I ask you, what cosine rule? Are you talking about the Law of Cosines?

    c^2 = a^2 + b^2 - 2abcosC.

    If so, then all the distances have to be positive. This is a formula relating positive lengths to positive angles in a triangle. It will give the correct lengths and angle magnitudes if applied correctly, but then you have to interpret the results to state a direction relative to compass directions.
  10. Oct 16, 2006 #9
    Thank you for your response ..... so what options do i have if i am to use the 'law of cosines' to solve this problem ??
  11. Oct 16, 2006 #10


    User Avatar
    Science Advisor
    Homework Helper

    The expression you used for v was based on the Pythagorean theorm. Since you squared the vector components you had the sum of positive numbers. When you took the square root, you got the length of the hypotenuse of a right triangle. It does not matter if you put in negatives for one or both of the legs of the triangle in this case.

    The Pythgorean theorem is a special case of the more general Law of Cosines (LoC) for theta = 90 degrees, but the LoC only works for positive (or a and b both negative) lengths. The product ab must be a positive number.

    When you said earlier that "it worked" when you changed the - to +, what you really meant was that it worked when you changed the sign of both the terms in the first square. That is to say, you computed the net North component which turned out to be negative, and changed the sign of the result to apply the LoC. That is exactly what you have to do if you are going to use it, but it is not the same thing as changing the -15 to +15. I misinterpreted what you were saying about changing the sign because I did not know that what you meant by the "cosine rule" was the LoC. I thought you were just changing the -15 to +15 and that definitely will not do.
    Last edited: Oct 16, 2006
  12. Oct 16, 2006 #11
    thank you ! :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook