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Homework Help: I need help with this question about vectors

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  1. Dec 25, 2017 #21
    Q must be at a 45 degrees bearing to P
     
  2. Dec 25, 2017 #22

    LCKurtz

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    I posted: "So now you can show us how you calculate Q and P's positions at time t". Which you ignored.
     
  3. Dec 26, 2017 #23
    We use r=r0 + vt
     
  4. Dec 26, 2017 #24

    Mark44

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    LCKurtz's question was more specific than this. In other words, what were your calculations for Q and P at an arbitrary time t?
     
  5. Dec 26, 2017 #25
    Let's say at time t the position of P is p and the position of Q is q so in terms of p and q and t I can give the new position of P as
    r=r0+vt
    = (i+7j) + (5i-5j)t
    =i+7j + 5it-5jt
    =(1+5t)i + (7+5t)j

    And new position for Q as
    r=r0+vt
    =(3i-8j) + (6i+5j)t
    =3i-8j + 6it+5jt
    =(3+6t)i + (5t-8)j
     
  6. Dec 26, 2017 #26

    LCKurtz

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    Is there some reason you don't simplify those? Anyway, now you have them, what next? There was more in post #20. Tell us what you are thinking or why you are stuck.
     
  7. Dec 27, 2017 #27
    Well you can see how I have drawn the vector diagram I drew it true to the i j to x y vector coordinates. I just don't know how to go about the question. Some hints would be helpful
     
  8. Dec 27, 2017 #28

    LCKurtz

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    You have had some hints. In post #20 I asked you about Q - P. Have you even calculated that yet? Are you going to show us what you get? Have you thought about whether it points Northeast? Or how you would tell? Neither I nor anyone else on this board is going to work it for you.
     
  9. Dec 30, 2017 #29
    You don't need any diagrams. The key is to subtract the vector velocity and position of P from Q. This makes P the new origin and gives you a new location and new vector velocity for Q. You can now use those velocity components to figure out when the two position components will be equal which is when the position of Q is north-east of P.
     
  10. Jan 4, 2018 #30
    Thank you all for your help
    So the position of
    P = (i+7j) with v= (5i-5j)
    Q=(3i-8j) with v=(6i+5j)
    Already Q is on the east side of P and the rate at with Q moves in x axis is faster than P
    So therefore Q will always be on the east side of P
    When we look at the y axis it seems that P is moving down at the same rate as Q is moving up
    So the moment these two meet in y plan would be the time that Q is exactly on the east of P
    So using speed = distance /time
    Distance= 7-(-8)= 15
    To find mid position 15/2 =7.5
    Speed = 5
    Time=distance/speed
    =7.5/5=1.5hrs
    At this time the two ships will be is line in the
    With a gap in the x axis
    Anything more than 1.5 hours and Q will be in the NE section of P

    How's this?
    Ok?
     
  11. Jan 5, 2018 #31

    LCKurtz

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    How's that? Not good. You haven't even used ##t##.

    If you had actually read and responded to my posts you would have had this problem solved a week ago. You haven't simplified your answer in post #25 or responded to post #28. Until you do you will hear nothing more from me.
     
  12. Jan 14, 2018 #32
    I'm sorry about that
    Was away for some time and no internet there
    But I'll start working on this now.
    Ok so simplifying the new position of P
    (1+5t)i + (7+5t)j
    P= i+5ti + 7j+5tj
    New position of Q
    (3+6t)i + (5t-8)j
    Q= 3i+6ti + 5tj-8j

    Q-P
    (3i-8j)-(i+7j)
    =(2i-15j)
     
  13. Jan 14, 2018 #33

    LCKurtz

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    Make sure you are using the correct numbers as given in post #18.
    When you "simplify" a vector it should look like (...)i + (...)j, one i term and one j term.
    Your final answer for Q-P at time t must have a "t" in it.
    Then we can talk about whether it points NE.
     
  14. Jan 14, 2018 #34
    Post 18

    At noon two boats P and Q have a position vector (i+7j)km and (3i-8j)km respectively to the origin O. i and j are unit vectors in direction to East and north respectively P is moving south east at 5√2 km/h and Q is moving at a constant velocity of (6i+5j) km/h
     
  15. Jan 14, 2018 #35
    I m very confused
    In post #25 it seems that the vectors are in the format that you asked (....)i+(...)j and you asked me to simply them. So I expanded them and now your asking me to revert them back again.
     
  16. Jan 14, 2018 #36
    Are you asking me to subtract the new coordinates of P and Q or the initial coordinates given in the question?
     
  17. Jan 15, 2018 #37

    LCKurtz

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    OK, never mind about the i and j comments.
    This thread is getting very jumbled. There are multiple arithmetic errors and/or typos. In post #25 you have an arithmetic mistake in the new position of P. Then in post #32 you carried the same mistake through. Also in post #32 you dropped the ##t## in the calculation of Q-P, as I told you in post #33. Fix those. Once you finally get Q-P correct, with a ##t## in it, it is a simple one more step to finish your problem.
     
  18. Jan 15, 2018 #38
    That's what I was clarifying in post 36
    The initial P and Q position vectors given in the question has no t in it
    So which P and Q are you referring to?
    Is it the answers on post 25?
     
  19. Jan 15, 2018 #39

    LCKurtz

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    Think about the problem you are trying to solve. You are given original positions P and Q and their velocities. They are moving. Then you are asked at what time ##t## is Q northeast of P. So you need their positions at time ##t## don't you? So you need to know for what ##t## does Q-P point northeast. Fix your arithmetic and show Q-P as a function of ##t## and you are almost done.
     
  20. Jan 16, 2018 #40
    Ok
    Let's think here
    I'm just trying to clear my confusion please do tell if I'm thinking right

    Let's keep the vector situation and go to a simpler linear one
    Let's say P is on (1,1) on a grid
    And Q is on ( 5,-5) on the same grid
    P is moving on positive x axis (to the right) at let's say 1 unit/s
    And Q is moving in the positive y axis (upward) 1 unit/s
    Let's say the question is asking to find the time that P moves east of Q

    Is this the same or similar sort of thing?
     
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