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Constant Velocity Challenge Problem

  1. Jun 13, 2012 #1
    I need help solving this problem. To be clear, I AM NOT LOOKING FOR THE ANSWER....just need some hints as to HOW to go about solving. I've tried several paths/steps and I keep going in circles. Can someone please help me get started on a path?

    Person A left Town X at 10:18 am. He walked at a constant speed and arrived at town Z at 1:30 pm. On the same day, Person B left town Z at 9 am. Person B walked the same route in the opposite direction at a constant speed. Person B arrived at town X at 11:40 am. The road crosses a wide river. By coincidence, both arrived at the bridge on opposite sides of the river at the same instant. Person A left the bridge 1 minute later than Person B. At what time did they arrive at the bridge?
     
  2. jcsd
  3. Jun 13, 2012 #2
    Just to be clear...this is NOT a homework problem!! I'm actually a physics teacher...this problem was presented to a room of physics teachers as a problem to use with our students. I need to figure it out first...but I need help! :)
     
  4. Jun 13, 2012 #3
    What have you done so far? What kind of help do you need?
    Can you write the equations of motion for the two persons?
     
  5. Jun 14, 2012 #4
    Can you draw a diagram of the problem out? Drawing the diagram will help you greatly in visualizing the problem to use with the equations of motion.
     
  6. Jun 14, 2012 #5
    The information given allows one to calculate the ratio of the speeds.
    Doing this I calculate that B takes 160/32 minutes to cross the bridge.

    Is this enough of a start?
     
  7. Jun 14, 2012 #6
    Try converting the format hours:minutes into minutes since 12 am and solve the equations. Later, when you find the answer you can reconvert it into the original format.
    In this problem, it is impossible to find the length of the bridge and the distance between X and Z. The only thing you can find is the time at which the two people are at the opposite ends of the bridge.
    The ratio of speeds is not required.
     
    Last edited: Jun 14, 2012
  8. Jun 14, 2012 #7

    HallsofIvy

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    The most interesting part of the problem is "Person A left the bridge 1 minute later than Person B." So person A took one minute longer than person B to cross the bridge.
    Let y be the length of the bridge, vA the speed of person A, and vB the speed of person B. Then vAy= vBy+ 1. (The speeds are measured in distance per minute.)

    Further, let the distance from town X to the bridge be x and the distance from town Z to the bridge be y. Then the total distance from town X to town Z is x+ y+ z. The time, in minutes, it took A to walk from town x to town Z is (x+ y+ z)vA and we are told it took A 192 minutes. Similarly, it took B (x+ y+ z)vB minutes to walk and we are told that was 160 minutes.
     
  9. Jun 14, 2012 #8
    Hello HOI that was something like my start.

    Let the velocities be Va and Vb

    Let the distance between X and Z be D, The distance across the bridge be d.

    We are told the time of transit of D for A is 192 minutes and for B is 160 minutes.

    Let the time of transit of the bridge be tb for B and therefore it is tb+1 for A.

    Thus D = 192 Va = 160Vb

    leading to Va/Vb = 160/192.

    Also Va/Vb = tb/(tb+1)

    Substituting Va/Vb we find

    192tb = 160tb +160

    leading to tb = 160/32 = 5 minutes as I previously said.
     
  10. Jun 14, 2012 #9
    But I don't see how this helps to find what's required.
     
  11. Jun 14, 2012 #10
    I didn't want to post more so I posted a result not asked for edit but a necessary intermediate result.

    However you can carry on the method applying the known ratios of the velocities to each part of the journey allowing for the time differences to synchronise clocks.
     
    Last edited: Jun 14, 2012
  12. Jun 14, 2012 #11
    The actual value is not needed per se. You do need to express the length of the bridge as a function of velocities (and their ratio) and the value of the time to cross is a "byproduct" of this.
    It is interesting though that whereas the distance between cities and length of the bridge are not determined by the given data (they depend on the actual values of the velocities), the time to cross the bridge is always the same. As is the time before they reach the bridge.
     
  13. Jun 14, 2012 #12
    Yes this is a nice little problem and I raise my glass to those who thought it up.

    https://www.physicsforums.com/attachment.php?attachmentid=31861&d=1296667292


    For those who are interested in comparing their answer with mine, I could not make A and B 'meet on the bridge at midnight'.
    Instead I got them at opposite ends at 1100 hours.

    If anyone can tell me how to get the attachment to display directly I would be grateful.
     
    Last edited: Jun 14, 2012
  14. Jun 15, 2012 #13

    haruspex

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    But your equation implies minutes/distance, no?
     
  15. Jun 15, 2012 #14
    I got the same but it seems you did it differently.
     
  16. Jun 15, 2012 #15
    I am happy to display the rest of my solution when everybody interested is ready. In principle this is a repeat of the bridge transit time method with one extra simultaneous equation.

    It seems our new recruit who posted this has not been back. That is a pity.
     
  17. Jun 15, 2012 #16
    OMGoodness!! I have to say, this was my first visit to this site and I'm officially bookmarking!! I posted the other night out of pure frustration and then walked away. This is my first time back and I've since solved the problem...but it took a LOOONNNGGG time!!
    I did get a final answer of 11:00 am. My solution right now is a mess...on several pages. As soon as I make sense of what I did and put it in a clear, precise manner, I will post!
    Thank you all for replying!! I <3 Physics!!
     
  18. Sep 3, 2012 #17
    yeah, I'm in ap physics, and our teacher gave us a couple of days to work on this, but I still don't know how to exactly start it, so i was wondering if anyone on here could guide me. thank you.
     
  19. Oct 17, 2012 #18
    lol Ms. D! What a character...
     
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