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Magnitude & Direction!

  1. Oct 8, 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 80.7 km/h to 58.9 km/h in the 12.4 s that it takes to round the bend. The radius of the curve is 165 m. Compute the acceleration at the moment the train speed reaches 58.9 km/h. Assume that it continues to slow down at this time at the same rate.

    Magnitude?
    Direction? ____° backward (behind the radial line pointing inward)

    2. Relevant equations

    a = change in V/ change in T

    a = v^2/r


    3. The attempt at a solution

    I've attempted the equations above and got 42.37 and 0.489 for magnitude, which are both wrong. Can someone please help?
     
  2. jcsd
  3. Oct 8, 2008 #2

    LowlyPion

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    This might be a better way to determine a:

    [tex] v^2 = v_0^2 + 2 a \Delta x [/tex]

    Careful with your units. You have km/h.
     
  4. Oct 8, 2008 #3
    when I plugged it into that equation, I got ..

    16.36^2 = 22.42^2 +2a (165)
    a = -.712

    is that correct?
     
  5. Oct 8, 2008 #4

    LowlyPion

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    Assuming your math is correct, now what is the centripetal acceleration?

    Since these are both vectors, you can add them to determine total magnitude and direction.
     
  6. Oct 8, 2008 #5
    so to find V, I used the eq. V = (2 pi r)/t and got V = 83.61

    then plugged the V into V^2 / r and got centripetal acceleration of 42.37 m/s/s

    then from what you said, I added the two vectors to determine magnitude

    -.712^2 + 42.37^2 = 1795.72^2 so ultimately the magnitude would equal 42.38

    right?

    would the direction be inverse tan (42.37/.712) = 89 degrees?
     
    Last edited: Oct 8, 2008
  7. Oct 8, 2008 #6

    LowlyPion

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    Whoa. Wait a minute. The tangential v is given as 58.9km/h or 16.36 m/s

    v = 16.36 m/s
    a = v2/R = 1.622 directed radially
     
  8. Oct 8, 2008 #7
    hm, how did you get v = 16.36 m/s? I'm not getting that number ..
    would you mind showing me your steps please?
     
  9. Oct 8, 2008 #8

    LowlyPion

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    That's the problem statement.
     
  10. Oct 8, 2008 #9
    So then do you add 1.622 and -.712 together to get the magnitude? How do I calculate the direction? inverse tan of the numbers?
     
  11. Oct 8, 2008 #10

    LowlyPion

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    Yes and yes.

    Tangential acceleration/Radial acceleration

    Remember it is trailing the radius.
     
  12. Oct 8, 2008 #11
    Okay, just to double check..

    so it's 1.622 + -.712 = .91
    and inverse tan (1.622/-.712) = -66.3 degrees? or do I add 180 degrees to that?

    or do I have to do 1.62^2 + -.712^2 = c^2?
     
    Last edited: Oct 8, 2008
  13. Oct 8, 2008 #12

    LowlyPion

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    No. Pythagoras is your guide.

    And no. Tangential acceleration/Radial acceleration
     
  14. Oct 8, 2008 #13
    thank you so much for your help LP !
     
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