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Acceleration and deceleration

  1. Oct 2, 2004 #1
    The question reads:
    A commuter train travels between two downtown stations. Because the stations are only 1.00km apart, the train never reaches its maxium possible cruising speed. The engineer minimizes the time t between the two stations by accelerationg at a rate a1=0.100m/s^2 for a time t1 and then by breaking with acceleartion a2=-0.500m/s^2 for a time t2. Find the minimum time of travel t and the time t1.

    hmmm, not sure where to begin. I ws thinking about finding the distance first, then relate it to time. I tried to set the distance of t1 to the distance of t2, something on the lines of:

    Vi^2=2ad-Vf^2...Vi would be 0 since the train starts at zero m/s
    Vf^2=2ad-Vi^2...Vf would be 0 since the train will make a complete stop

    so...2ad-Vf^2=2ad-Vi^2

    however, that wouldn't work because the distances are not the same :frown:

    Any suggestions?
     
  2. jcsd
  3. Oct 2, 2004 #2

    Pyrrhus

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    Well divide this in two parts, you know that after Time t1, the train will start from the final speed of the first to accelerate at -0.500 m/s^2 and then has a final speed of 0, so there's your link variable.
     
  4. Oct 2, 2004 #3
    you're saing that v1=v2...as in v1=the velocity after ti and v2=the inital velocity of t2.

    so...
    vf=vi+at
    vi+at=at-vf
    0+(-.5)(t)=(.1)(t)-0
    ????
     
  5. Oct 2, 2004 #4
    I tried to link the variables you talked about in my first statemet:

    2ad-Vf^2=2ad-Vi^2
    but the distance are not the same for t1 and t2

    not quite sure what you're getting at
     
  6. Oct 2, 2004 #5

    Pyrrhus

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    Divide this in two parts.

    1) Starts from Rest (Vo = 0) with an Acceleration a, covers a distance x in a time t1, and does a final speed of (V)

    2) Starts where the first part achieved with an initial speed of (V), with an Acceleration b, covers a distance y in a time t2, and does a final speed of 0

    Some extra stuff you know

    t = t1 + t2

    Total Distance = x + y

    if the Train covers a distance x, that distance will be equal to the total distance - y (distance covered in the second part).
     
    Last edited: Oct 2, 2004
  7. Oct 2, 2004 #6
    Okay, so I should first find the distance x or the time t1? Thoes are both unknown. I dont know how I should set up a problem with two unknown variables.
     
  8. Oct 2, 2004 #7

    Pyrrhus

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    Try to understand what i say, put all your values of info of the two parts on a side, and see all the kinematic equations with uniform acceleration that can help you, if its needed, go do something else, then come back to it.
     
  9. Oct 2, 2004 #8
    Here's what I got do far:

    For ti:
    vi=om/s
    d=D-y=(y/.2) <----shown below
    vf=V
    a=.1m/s^2
    t1=T-t2=(t2/.2)<---shown below

    for t2:
    vi=V
    d=y
    vf=0m/s
    a=-0.5m/s^2
    t=t2

    I tried using the equation vf=vi+at...which would combine to vi+at=at-vf. The vi and vf would cancel out since they are zero. so...
    (.1m/s^s)(T-t1)=(-.5m/s)(t2)
    -.2(T-t2)=t2
    so...T-t2=(t2/.2)

    Then I used vf^2=vi^2+2ad...which when set to each other...
    2ad=2ad
    2(.1)(D-y)=2(.5)y
    .2(D-y)=1y
    (D-y)=y/.2

    am I heading in the right direction?
     
  10. Oct 2, 2004 #9

    Pyrrhus

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    Just an advice, Find Y then find X, you can get the times then easily (thinking about it.. not so easily, but keep going), also If you're going to plug in numbers, remember their signs. Now that i stop to think.. consider their link variable to calculate time :smile:
     
    Last edited: Oct 2, 2004
  11. Oct 2, 2004 #10
    Sorry i didn't get a chance to respond I will work on it tomorrow If I can find some time... i ended up having someone close the aim window... feeel free though to contact me again sometime (its just I am incrediably busy at the moment with college application essays)
     
  12. Oct 2, 2004 #11
  13. Oct 2, 2004 #12

    Pyrrhus

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    Explain in your second displacement for the 2nd part, you have

    (0.100 t1)t, ,<--- ????

    Note: You learnt to integrate :smile:

    vt = initial position, i guess.
     
    Last edited: Oct 2, 2004
  14. Oct 2, 2004 #13
    sorry the writing is so light it is very difficult to read
     
  15. Oct 2, 2004 #14

    Pyrrhus

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    Your answers are correct. I finally read it. Well done.
     
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