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How fast must the women run?

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    A woman is on a train leaving the station at3.5 m/s , while a friend waving goodbye runs alongside the car she's in. Call the train's direction of motion the +x direction.
    Once the train has reached a speed of 10 m/s, how fast must the woman walk, and in which direction, to keep up with her friend?

    2. Relevant equations
    vxw=vxt+vtw

    3. The attempt at a solution
    v= 10-3.5
     
  2. jcsd
  3. Oct 9, 2016 #2

    phinds

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    This seems to me like a VERY poorly worded question. I see various interpretations, but yes, your answer is correct using one of them, although even with that, I don't think the "keep up with her friend" is anything like standard English using since it is the friend who is "keeping up". It would have been more correct to say "stay even with her friend"

    Also, you did not specify the direction of the passenger's motion relative to the train. You have v, which normally stands for "velocity", a vector, shown as a scalar.
     
  4. Oct 9, 2016 #3
    its not the correct answer though. mastering physics says it wrong
     
  5. Oct 9, 2016 #4

    phinds

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    Well, then perhaps we are both misinterpreting the problem. As I said, I think it is poorly worded. Not clear, to me at least, exactly what they are asking for.
     
  6. Oct 9, 2016 #5
    i still need help finding the answer
     
  7. Oct 9, 2016 #6

    phinds

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    Well, good luck with that.
     
  8. Oct 9, 2016 #7
    do you have any suggestions ?
     
  9. Oct 9, 2016 #8

    I like Serena

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    Did you try -6.5?
    Negative since she moves to the back of the train, while the train is moving in the positive x direction.
    To be honest, it doesn't really make sense, since the problem says she's "walking", while 6.5 m/s is "sprinting".
    Then again, seeing how poorly worded the problem is, perhaps it's still the correct answer.
     
    Last edited: Oct 9, 2016
  10. Oct 9, 2016 #9
    Yea I tried negative and still wrong
     
  11. Oct 10, 2016 #10
    It is assumed that both activities occur at the same time interval...they move in same direction since one must approach the other from assumed reference start or origin.however,the question lacks good construct....The v=8.25m/s.... one way is to linearise it with equations of motion....using v^2= u^2 + 2ax, for the train ,v= 10, u= 6.5, both in m/s...so,we obtain ,ax=28.875m/s ..(1),also considering the time interval for covering x,we use v=u+ at, so,at= 3.5m/s ...(2),therefore, using both equations,we can divide 1 by 2 ,velocity with which the woman must move to catch up with the friend or thingy is given by v= 28.875/3.5 m/s. = 8.25m/s....
     
  12. Oct 10, 2016 #11
    Please, I want people to check this.....
     
  13. Oct 10, 2016 #12
    that anwser was incorrect too
     
  14. Oct 10, 2016 #13

    SammyS

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    Is that the question as given to you, word for word ?
     
  15. Oct 10, 2016 #14
    exact whole question:
    A woman is on a train leaving the station at3.5 m/s , while a friend waving goodbye runs alongside the car she's in. Call the train's direction of motion the +x direction.
    Part A)If the friend is running at 7.0 m/s and moving in the same direction as the train, how fast must the woman walk, and in which direction, to keep up with him? answer: 3.5 m/s
    Part B) Once the train has reached a speed of 10 m/s, how fast must the woman walk, and in which direction, to keep up with her friend?
    Answer:???
     
  16. Oct 10, 2016 #15

    SammyS

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    Now let's see your attempt(s) .
     
  17. Oct 10, 2016 #16
    10-3.5= 6.5 m/s
     
  18. Oct 10, 2016 #17

    SammyS

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    In part B), What is it that moves at 3.5 m/s ?
     
  19. Oct 10, 2016 #18
    should it be 10-7?
     
  20. Oct 10, 2016 #19
    since the bus is moving at 10 and the friend is running at 7
     
  21. Oct 10, 2016 #20

    SammyS

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    Try it .
     
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