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Mechanics please help

  1. Nov 11, 2007 #1
    Two ships P and Q are travelling at night with constant velocities. At midnight,P is at point with position vector (20i + 10j) km relative to fixed origin 0. At the same time, q is at the point with position vectro (14i - 6J)Km. Three hours later, P is at point with position vector (29i + 34j) Km . the ship Q travels with velocity 12jkm h^-1 . At time t hours after midnight, the position vector of P and Q are p km and qKm respectively. find

    a) the velocity of P, in terms of i and j.

    i got

    t=3= (20i + 10j)Km + 3( i + J)

    t = 3 = (20i + 10j)Km + 3( i + J) = (29 i + 34j)

    = 20 + 3i = 29

    i = 3

    then 10 + 3b = 34j

    so b = 8j

    Then velocity will be = 3i + 8j

    (ok now i need someone to see if this part is right please.)

    B) Then i got to find an expression for P and Q in terms of t, i and j

    I got

    P= (20i + 10j)+ ( 3i + 8J)t

    and For Q = (14i - 6j) + (12j)t

    then the questions goes :

    At time t hours after midnight, the disatnce between p and Q is d Km.

    c) by finding an expression For PQ show that

    d^2 = 25t^2 - 92t + 292.

    Now for this one i know that i have to take away Q from P but does it mean i take away the following :

    Q = (14i - 6j) + (12j)t - P= (20i + 10j)+ ( 3i + 8J)t

    Because i get the worng answer, so what i want from someone is to kindly, see if i have done part A and B right and to show me how to do C becuase i ahve been goign in circles, and no hope, so can someone please help thank you.
  2. jcsd
  3. Nov 11, 2007 #2


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    Science Advisor
    Homework Helper

    Just subtract Q from P.

    "I got

    P= (20i + 10j)+ ( 3i + 8J)t

    and For Q = (14i - 6j) + (12j)t"

    I.e. [(20i + 10j)+ ( 3i + 8J)t]-[(14i - 6j) + (12j)t].
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