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Acceleration and time

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    How much time does it take for this train to travel 9.0 km?

    The stations at which the trains must stop are 1.8 km apart (a total of 6 stations, including those at the ends).
    Assume that at each station the train accelerates at a rate of 1.1 m/s^2 until it reaches 94 km/h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at -2.1 m/s^2. Assume it stops at each intermediate station for 25 s.

    2. Relevant equations

    i thought i would need the equation d=vi(t) + 1/2(a)(t^2)

    3. The attempt at a solution

    I know it only needs to travel to 5 stations to be 9.0 km but, i do not even really know how to start this one.

    I converted all the units to m and m/s so....
    94 km/h = 26.1 m/s
    1.8 km = 1800 m
    9 km = 9000 m

    HELP PLEASE :)
     
    Last edited: Jan 24, 2010
  2. jcsd
  3. Jan 24, 2010 #2
    I tried working this out and came up with 385.8 seconds or 6.43 minutes but this is wrong.

    i think its wrong because i didnt do anything with the deceleration rate because i do not know where to include it or how to really go about this one :(
     
  4. Jan 24, 2010 #3
    Hi mybrohshi5

    Let's consider when the train travels from station 1 to station 2 (1.8 km) :

    1.
    The train starts from rest with a = 1.1 ms-2 until it reaches 94 km/h, so you can find the time needed and distance traveled for this case.

    2.
    To stop, the train decelerates at -2.1 ms-2 from 94 km/h, so you can find the time needed distance traveled for this case.

    3.
    Now let's take a look when the train travels at constant speed. You can find the distance traveled using the above results then time needed.

    Finally, you have total time needed to travel 1.8 km :smile:
     
  5. Jan 24, 2010 #4
    Thank you for the reply and your response makes total sense, i am just having trouble with your first part 1.

    How would i find the time needed and distance traveled with only knowing those two points of information: 1.1 m/s^2 and 94 km/h????

    is there an equation i need to be using that i am forgetting besides the equation d=vi(t) + 1/2(a)(t^2)
     
  6. Jan 24, 2010 #5
    Yes, you need another kinematics formula. Can you find it in your book?

    The information is not just the acceleration and final velocity, but also including initial velocity. Do you know the initial velocity?

    :smile:
     
  7. Jan 24, 2010 #6
    initial velocity would be 0.

    i believe the equation im looking for is vf = vi+at

    26.1m/s = 0 +(1.1m/s^2)t
    t=23.7s
    d = 0 + .5(1.1)(23.7^2)
    d = 308.93 m

    0 m/s = 26.1 + (-2.1)t
    t=12.43s
    d=26.1(12.43) + .5(-2.1)(12.43^2)
    d=162.19m

    can you confirm that these are correct?

    if these are right how can i find the distance traveled using the above results then time needed as stated in part 3????
     
  8. Jan 24, 2010 #7
    well u know that total distance traveled between each station is 1.8km so [tex]d_{tot}=d_{1}+d_{2}+d_{3}[/tex]
    and you already told us the speed it will be going during [tex]d_{2}[/tex] (94km/h) so do u have any equations that u have the change of displacement and the constant velocity, and it will give u the time it took to travel that distance?
     
  9. Jan 25, 2010 #8
    i used d=vt

    so 1800-(308.9+162.2)=(26.1)t

    t=50.9 seconds

    So now i know the total time it took to accel (d1), stay at a constant velocity (d2) and decel (d3) for a distance of 1.8 km.

    so the total time to travel 1.8 km i calculated to be 87.05 seconds.

    i multiplied this by 5 (because it takes this amount of time to travel between the 6 stations) then added 25 multiplied by 4 (because of the 4 intermediate stations between the starting and ending stations) so....

    .... i get an answer of 535.228 seconds or 8.92 mins

    can anyone let me know if this answer is correct?

    thank you
     
    Last edited: Jan 25, 2010
  10. Jan 25, 2010 #9
    Yay its right!

    thank you everyone for all the help!
     
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