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Help needed please

  1. May 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A train accelerates uniformly from rest to a velocity of 60km/h in 6 minutes after which the velocity is kept constant calculate the time taken to travel 10km



    2. Relevant equations
    S=u×t
    V=u+at



    3. The attempt at a solution
    I have worked out the 60km/h to meters per seconds I get 16.67m/s and do not know how to contintue
     
  2. jcsd
  3. May 20, 2008 #2
    Thank you for the help this has sent me on me way
     
  4. May 20, 2008 #3
    Sorry, let me repost after making a complete hash of the last attempt to answer you!

    Still seperate the journey into two legs;

    The accelerating leg.
    The constant speed leg.


    Start by working out the acceleration of the train in the first leg of the journey:

    [tex]v=u+at[/tex]

    then using this you can find the distance travelled in the first leg using the equation:

    [tex]s = 0.5at^2[/tex]

    so now we're onto the second part of the journey, we know the speed and the total distance we want to travel, but we have already done some that distance in accelerating, so the distance we still need to travel is total distance - distance done in first leg.

    Now simply use [tex]s = ut[/tex] and solve for t.

    Adding the 6 minutes we are told it takes for the train to accelerate you will get your answer.
     
    Last edited: May 20, 2008
  5. May 20, 2008 #4
    where does the 0.5 come from
     
  6. May 20, 2008 #5
    It's another one of the equations of motion formulas for constant acceleration.

    http://en.wikipedia.org/wiki/Equations_of_motion

    Written in its entirety it's:

    [tex]s = ut+0.5at^2[/tex]

    but in your case, the train accelerates uniformly from rest so that u = 0, thus:

    [tex]s = 0.5at^2[/tex]
     
  7. May 20, 2008 #6
    Thank you
     
  8. May 20, 2008 #7
    No problem.
     
  9. May 22, 2008 #8
    to work out the initial velocity in the first equation

    V=u+at

    do I convet 60 km/h in to 17.8819 m/s
     
  10. May 22, 2008 #9
    Yes, it's always wise to convert units you're given into SI units.
     
  11. May 22, 2008 #10
    so to work out the acceleration I will use the formula

    accerleration (m/s2) = change in speed m/s / time taken for change

    for my case it will be

    accerleration (m/s2) = 17.8819 m/s divided by 360 = 20.132
    changing the 6 minutes in to seconds which is 360

    Is this correct
     
    Last edited: May 22, 2008
  12. May 22, 2008 #11
    60km/h = 60'000/3600m/s ~= 16.67m/s - you actually stated this in your initial post :)

    Then divide through by the 360 to get the acceleration.
     
  13. May 22, 2008 #12
    do i divide the 360 / 16.67 = 21.595
     
  14. May 22, 2008 #13

    Compare these two.

    It should be 16.67 / 360.
     
  15. May 22, 2008 #14
    which is 0.0463

    V=u+at

    17.8819 + 0.0463 * 360 = 34.5499
     
    Last edited: May 22, 2008
  16. May 22, 2008 #15
    Correct, now use this in the equation:

    [tex]s = 0.5at^2[/tex]

    You now know a (0.046) and t (360).

    This will give you the distance travelled in the accelerating leg of the journey.
     
  17. May 22, 2008 #16
    s= 0.5 * 0.0463 * 360 2 = 3000.24
     
  18. May 22, 2008 #17
    correct, now the question asks for how long it takes for the train to travel 10km (10000m)

    We know it takes 6 minutes to accelerate travelling 3000.24m in the process.

    So all that's left is to travel ~7000m at a constant speed.

    We use:

    distance / speed = time

    we know the distance (7000) and the speed (16.67m/s) so we can work out the time required to do this leg of the journey (in seconds).

    If we add this time onto the 6 minutes required to accelerate, you'll get your answer.
     
  19. May 22, 2008 #18
    time = distance / time = 7000 / 16.67 = 419.91 + 6 = 425.91

    does the 6 mintues have to be converted in to seconds
     
  20. May 22, 2008 #19
    Yes,

    The 419.91 figure is in seconds, and the 6 is in minutes. These need to be the same unit.
     
  21. May 22, 2008 #20
    419.91 + 360 = 779.91 seconds
     
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