Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two trains headed toward each other

  1. Jun 9, 2008 #1
    The problem statement, all variables and given/known data

    As two trains move along a track, their conductors suddenly notice that they are headed toward each other. Figure 2-27
    http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c02/pict_2_27.gif

    gives their velocities v as functions of time t as the conductors slow the trains. The figure's vertical scaling is set by vs = 43.0 m/s. The slowing processes begin when the trains are 204 m apart. What is their separation when both trains have stopped?

    The attempt at a solution

    I tried this problem and got it wrong. I was able to come up with d = 107.5 m for Train 1 but I think I am missing something when I try to calculate d for Train 2.

    The part of this question that confuses me the most if when it says "The figure's vertical scaling is set by vs = 43.0 m/s." I think I don't really understand how to read the graph correctly.

    Help is appreciated
     
    Last edited: Jun 9, 2008
  2. jcsd
  3. Jun 9, 2008 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Ok well, if 4 blocks up the side is 43m/s then what are three blocks going to represent for the velocity?
     
  4. Jun 9, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You found the distance traveled by Train 1 before it stops. Do the same thing for Train 2.
    I take that to mean that V_s = 43 m/s--which means that 4 vertical boxes = 43 m/s. So what's the initial velocity of Train 2?
     
  5. Jun 9, 2008 #4
    Well then is this right?


    if velocity equals -33 m/s then....

    Train 2

    Data:
    v = -33 m/s
    v2 = 0 m/s
    t = 4s
    a = ?
    ...then I used this equation....
    v2 = v + a2t
    a2 = (v2 - v) / t

    = (0m/s - (-33 m/s)) / 4s

    a = 8.25 m/s^2

    v2^2 = v^2 + 2ad
    d = (v2^2 - v^2) / 2a

    = ((0m/s)2 - (-33 m/s)2) / (2(8.25m/s^2)

    = -66m <--distance traveled by train 2
     
  6. Jun 9, 2008 #5

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It doesn't equal - 33m/s exactly. There are a couple of easier ways to calculate the distance traveled. Firstly one can calculate the area under the curve to the x-axis, or you can use the average speed formula which amounts to the same thing. There's nothing wrong with the method you've used, just a bit long.
     
  7. Jun 9, 2008 #6
    I don't quite get what you are saying by I can use the average speed formula? Isn't that formula S(avg)=(total distance traveled)/(time elapsed)? How does that relate?
     
  8. Jun 9, 2008 #7

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No you're right, ignore that its a load of rubbish I was thinking of something else. Any way my main point was 33m/s is not quite the right velocity.
     
  9. Jun 9, 2008 #8
    Ok so if 33 m/s isn't quite right how much is it off by and how else would I do this? I tried getting to the final answer using 33 m/s and my final answer was wrong so I am assuming that 33 m/s is off enough to cause me to get the problem incorrect.
     
  10. Jun 9, 2008 #9

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What have you done to get 33m/s as the speed? Remember that 4 divisions on the y-axis is equivalent to 43m/s. What does that make 1 division equivalent to? So what velocity does 3 divisions correspond to?
     
  11. Jun 9, 2008 #10
    Oh man......now i think i got it! 32.25 m/s ?
     
  12. Jun 9, 2008 #11

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sounds good.
     
  13. Jun 9, 2008 #12
    If the answer is 32 m

    try using areas method as mentioned before..


    edit: seems like you are doing something else wrong too.. because there isn't much different between 33 and 32.5 (or maybe ...)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook