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Trolley motion problem

  1. Jun 5, 2005 #1
    "A double-track trolley line runs between two stations 6 miles apart. From each station cars set out for the other station every 7.5 minutes and proceed at uniform speed. A pedastrian leaves one station just as one car is arriving and another is leaving. He walks beside the right-of-way at constant speed until he reaches the other station just as one car is arriving and another is leaving. Including these 4, he saw 62 cars on his way, 19 going in the same direction and 43 in the opposite direction. What was his speed, and what was the speed of the cars?"

    I figured that 18 cars caught up with the pedastrian (since the 19th left with him), and so that's 18x7.5=135 minutes to walk the 6 miles. That's 6 miles divided by 2.25 hours = 2+2/3 miles/hour = his rate.

    From there, I'm a little lost. Help, please?

    Thanks in advance,

    - Alisa
     
    Last edited: Jun 5, 2005
  2. jcsd
  3. Jun 6, 2005 #2
    Hi Alisa,

    You are taking the time to be only those cars that have passed the pedestrian in one direction, but have not counted the cars that are still on the tracks inbetween stations (that have left, but not yet caught up to the person). You can find this number of cars using information given about both directions (solving simultaneously for the unknown), and then the times will follow.

    Just be careful you count the right number of cars. I think by your previous reasoning you should've had 19, not 18.

    Hope that helps.
     
  4. Jun 7, 2005 #3
    Eh I still can't get it. = / The answer is supposed to be the cars move at 10 m/h and the man walks at 4 m/h. But how..? heh

    - Alisa = /
     
  5. Jun 7, 2005 #4

    OlderDan

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    What are you getting for an answer? I get speeds with a ratio of 10:4, but not those speeds. Your initial calculation was not correct, but because you are walking away from the trains that are going your way it is actually going to take longer than the time you calculated between trains. That means your total walking time will be greater, and your walking speed will be slower than what you calculated.

    Do you have all the numbers posted correctly. Are you sure the answers are 10m/h and 4m/h?
     
  6. Jun 8, 2005 #5
    My teacher said 10 and 4 something. Perhaps you are correct. How did you get that, OlderDan?

    Thanks for your help,
    Alisa
     
  7. Jun 8, 2005 #6

    OlderDan

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    |------------------|------------|------------------------------|
    |------------------S-----------D
    |<-------------L------------->|<------- -----L----- ------->|

    The mark in the middle, D, represents the destination station. The mark to the left of that, S, represents the starting point of the pedestrian. The marks at the far right and far left represent the starting positions of the trains that arrive at the destination station at the same time as the pedestrian. They are each at initial distance L from the destination. Each dash represents the separation distance, d, between trolleys. At the starting time, there are trolleys on each track between every pair of dashes. There are 42 dashes to the right of S, and 18 dashes to the left of S, correseponding to the 43 and 19 trolleys that will be seen by the pedestrian.

    From the number of trolley separations between S and D you can find d, and then you can find L. To a stationary observer, each trolley must move a distance d in the 7.5 minutes between trolleys in each direction. You can calculate the trolley speeds from this. You can then calculate the time it takes for those most distant trolleys to reach the destination, which is also the total walking time of the pedestrian. Use that to find the pedestrian's speed.

    If the time between trolleys had been 3 minutes instead of 7.5 minutes, the answers would be 10 m/h and 4 m/h.
     
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