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Quick question about kinematics problem?

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A commuter train can minimize the time t between two stations by accelerating (a1 = 0.1 m/s^2) for a time t1 then undergoing a negative acceleration (a2 = -0.5 m/s^2) by using his brakes for a time t2. Since the stations are only 1 km apart, the train never reaches its maximum velocity. Find the minimum time of travel t, and the time t1.


    2. Relevant equations



    3. The attempt at a solution
    basically I listed
    a1 = 0.1 m/s^2
    a2 = - 0.5 m/s^2
    Then I integrated the equations into a piecewise equation for velocity and position.
    from 0 < t <= t1 v(t) = t/10
    from t1 <= t < t2 v(t) = -t/2 + v(t1)

    from 0 < t <= t1 x(t) = t^2 / 20
    from t1 <= t < t2 x(t) = -t^2 / 4 + (t1)(t) / 10 + t1^2 / 20

    Then I set v(t2) = 0
    getting t1 = 5t2.
    And I set x(t2) = 1000m using the second equation to solve for t1 and t2.
    I eventaully got t1 = 129 s. and t = 154.8 s.

    I'm not sure about this answer because from the piecewise function for x(t).
    The first piece of the function doesn't equal the second piece at t = t1 so I just want to know what I did wrong. I'm pretty sure I integrated correctly.

    thanks.
     
  2. jcsd
  3. Sep 21, 2012 #2

    SammyS

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    Well, it is a bit hard to follow, but t1 should be 5 times t2, which is your result.

    The train goes 1000 meters in 154.8 seconds, which is an average velocity of about 6.46 m/s.

    So the maximum velocity was twice that. This does give (approximately) the correct accelerations.

    Looks good.
     
  4. Sep 21, 2012 #3
    For the one you changed, I think it was supposed to be like that. I just substituted v1 into the previous equation making it t1 / 10. Then integrating it I got t1 t /10
     
  5. Sep 21, 2012 #4

    SammyS

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    The only (intentional) change I made was putting the v in to make it v(t1) and then placing parentheses around that.

    You had x(t) = -t^2 / 4 + (t1)(t) / 10 + t1^2 / 20 .

    I changed the quantity in red to v(t1), which is what I thought you intended.
     
  6. Sep 21, 2012 #5
    yeah that's what I meant lol. v(t1) is the same thing as t1 / 10.
    btw, how come for the x(t) piecewise function, the first piece isn't equal to the second at t1? isn't it supposed to be a continuous function.
     
  7. Sep 21, 2012 #6

    SammyS

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    What do you mean "first piece" ?
     
  8. Sep 21, 2012 #7
  9. Sep 21, 2012 #8
    A commuter train can minimize the time t between two stations by accelerating (a1 = 0.1 m/s^2) for a time t1 then undergoing a negative acceleration (a2 = -0.5 m/s^2) by using his brakes for a time t2. Since the stations are only 1 km apart, the train never reaches its maximum velocity. Find the minimum time of travel t, and the time t1.
    ---------------------------------

    I think there is no minimum time or maximum time of travel.
    There is only one maximum velocity thus one unique time.
    For other values of time, either it runs short of destination or overshoot.

    a1=0.1m/s2
    a2=-0.5m/s2
    s=1000m
    v0=0
    v2=0

    Unknown, t1,t2,v1

    (1/2)v1(t1+t2)=1000 ...(1)
    v1=a1t1 .....(2)
    v1=-a2t2.......(3)
     
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