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Relative Velocity Question

  1. Dec 6, 2003 #1
    I’m just having a bit of trouble with this relative velocity question:

    This is what ive got so far:

    Work so far.

    I think ive gotten a part a right, and ive made a start on a part b, but im pretty much lost on part b (i) and (ii).
    Any help would be appreciated.

    (v = velocity, Va = velocity of A, Vb = velocity of B, Vab = velocity of B, relative to A)
  2. jcsd
  3. Dec 6, 2003 #2


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    1. I am told I don't have "permission" to look at your website!
    2. Are both cyclists moving toward the intersection?
    3. Does the problem really have (i) and (ii) under (b) but a. and b. under (a)?
    4. Is the problem really asking for relative velocity rather than the rate at which they are closing?

    Assuming all this is true then:
    Set up a coordinate system with origin at P, the positive x-axis to the east, and positive y axis to the north, then A's velocity (relative to the ground) is (3,0) and B's is (0,4). B's velocity relative to A is
    (0,4)-(3,0)= (-3, 4).
    A's position after t seconds is (-80+ 3t,0) while B's position is (0,-40+4t). The distance between them (squared) is (3t-80)2+(4t-40)2. That is a minimum when its derivative 6(3t-80)+ 8(4t-40)= 50t- = 800 or t= 16 seconds. At that point the distance between them is √((-192)2+ (24)2)= √(37440)= 193.5 m.

    Now we assume that A decelerates at 0.1 m/s2. Then his velocity at time t is (3-0.1t, 0) and his position is (-80+ 3t- 0.05t2,0). B decelerates at q m/s2. His velocity at time t is (0,4-qt) and position (0,-40+ 4t- (q/2)t2).

    To find the velocity of B relative to A, subtract the velocity vectors relative to P as before: (0,4-qt)-(3- 0.1t)= (-3+0.1t,4-qt).

    To have them arrive at P at the same time, we must solve
    -80+ 3t- 0.05t2= 0 for t (to find when A arrives there), substitute that value into -40+ 4t- (q/2)t2= 0 to find the value of q that gets B there at the same time.
  4. Dec 7, 2003 #3
    1. Thislink should work.
    2. From my reading of the question, i would think that yes, they are both moving toward the intersection at junction P.
    3. Yes there is an A part a and b and a B part i and ii. If you dont get confused by the question then youll just get confused with the order, either way youre screwed.
    4. I would think that part A (a and b) are both looking for RV, whereas part B (i and ii) are looking for the rate at which theyre closing.

    Thanks again for the help, i hope this clears up the question for you a little bit.
  5. Dec 7, 2003 #4
    Ive just been looking through your answer Hallsofivy, and ive just got a few questions for you:

    What the question was asking was how far the two cyclists are from p when they are nearest together: from reading your reply it seemed to me like you were finding the closest distance they were from each other - like i was doing, but our answers are very different. You say
    i dont really follow this part very well, could you maybe explain a bit more?

    Where did the "-80+3t - 0.05t2" equation come from, particulary the 0.05t2 part?

    When i tried to solve your equation, i got a complex answer (iotas), so somewhere along the way theres been an error. Any idea? Ive spent a while looking through your solution, the bits i follow seem ok, i dont know if you or anybody else can maybe find some mistakes.
    Thanks again
  6. Dec 7, 2003 #5
    Hi, I get slightly different results to HallsOfIvy.


    I also get the time till there distance is a minimum = 16 seconds
    At this point, A is 32m from junction, B is 24m.
    (i used same method as HallsOfIvy)

    my equation was [tex]d^2=(-40+4t)^2+(-80+3t)^2[/tex]

    differentiating this gave a minimum value of t as 16.

    I got new relative veloctiy as


    I found a value for t in the quadratic as 17.01562

    This gave q=2.1157

    Do you have the correct answers?
  7. Dec 7, 2003 #6
    I dont have the answers as of yet, i should have them by the end of the week though. Im stil not too clear on how both of you got the equations to differentiate, so i guess im missing something fundamental to the understanding of the question. Thanks again for the help.
  8. Dec 8, 2003 #7
    The formula comes from finding their distances from (0,0) the origin after t seconds. If you draw the diagram with A at (-80,0) and B at (0,-40) and you consider their velocities. A is moving towards the junction therefore the distance is getting smaller. After 1 second A is at the point (-77,0) and after two seconds is at the point (-74,0).

    Therefore the distance of A from the junction after t seconds is (-80+3t) metres

    Similar reasoning is used for the position of B after t seconds.

    Then its just pythagorus with the distance between them being the hypotenuse of the right angled triangle, the right angle is the junction. Thats where the expression comes from
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