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Homework Help: Tangential and radial train acceleration

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A train slows down as it rounds a sharp horizontal turn slowing from 90.0 km/h to 50.0km/h in the 15 s that it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50km/h. Assume it continues to slow down at this time at the same rate.


    2. Relevant equations



    3. The attempt at a solution

    I think this problem is asking to find the instantaneous velocity at t = 15 sec, which when the speed of the train is 15 sec. From the above information give, we can find the acceleration of the train during from 90-50, which is 11.55 using the formula Vf = Vot + 1/2at^2. Then I am stuck here
     
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  3. Sep 29, 2008 #2

    tiny-tim

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    Hi -EquinoX-! :smile:
    eugh! :yuck:

    Vf = Vot + 1/2at2 (and the similar formulas) is only for uniform (constant in magnitude and direction) acceleration.

    Hint: the clue's in the title

    tangential and radial acceleration! :wink:
     
  4. Sep 29, 2008 #3
    I know the formula of radial acceleration is ar = V^2/r so is it just 50^2/150?
     
  5. Sep 29, 2008 #4

    tiny-tim

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    That's right! :smile:

    (except, of course, you'll have to convert the 50km/h into m/s first :wink:)
     
  6. Sep 29, 2008 #5
    so it's 192.901/150 = 1.28,

    why does the book gives me the information about time?
     
  7. Sep 29, 2008 #6

    tiny-tim

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    (I make it nearer 1.29)

    You'll need the time for the tangential acceleration. :wink:
     
  8. Sep 29, 2008 #7
    and how is that related? between tangential and acceleration? as far as I know it's dv/dt, but I don't have an equation here to derive...
     
    Last edited: Sep 29, 2008
  9. Sep 29, 2008 #8

    tiny-tim

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    Oh come on, -EquinoX-!

    You titled this thread "tangential and radial acceleration" …

    so you tell us

    what's the formula for tangential acceleration? :smile:
     
  10. Sep 29, 2008 #9
    ok, my basic instinct says that the 90 can be utilized for something, do we find the acceleration first by (90-50)/15? and yes I know it's in km
     
  11. Sep 29, 2008 #10

    tiny-tim

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    Yup, that should do it! :smile:

    i hope you're not going to mention instinct in the exams :biggrin:
     
  12. Sep 29, 2008 #11
    and does that results in the total acceleration?

    and then we can find tangential acceleration from the formula a = sqrt(at^2+ac^2)
     
  13. Sep 29, 2008 #12

    tiny-tim

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    No, it's the other way round … the tangential acceleration is (90-50)/15 (I thought that's what you meant in your previous post).

    Then the total acceleration (if they want it, which they probably don't) is a = sqrt(at^2+ac^2)
     
  14. Sep 29, 2008 #13
    thanks tiny_tim :)
     
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