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Uniform Circular Motion of a train

  1. Feb 4, 2007 #1
    I was wondering if someone could help me figure out what formulas to use solving this problem. My professor gave me the answer (as a way to check our answers) : 1.48 m/s^2 inward and 29.9 degrees backward

    The problem is: A train slows down as it rounds a sharp horizontal turn slowing from 90.0km/h to 50.0km/h in the 15.0s that it takes to round the bend. The radius of the curve is 150m. Compute the acceleration at the moment the train reaches 50.0km/her. Assume it continues to slow down at this rate.

    Thank you in advance for helping me!
  2. jcsd
  3. Feb 4, 2007 #2
    There are 2 components of acceleration. Find 1 with a kinematic and the other with the equation for centripetal acceleration. Then resolve the vectors to find magnitude and acceleration
  4. Feb 4, 2007 #3
    Ok. So do I use the kinematic equation: r(final) = r(initial) + Vi(t) +.5a(t^2)? When I do I get -2m/s^2.

    I know centripital acceleration is (v^2)/r. But do I use the 13.89m/s (50km/h converted)?
  5. Feb 4, 2007 #4
    I am still very lost! Can someone please further help me!
  6. Feb 4, 2007 #5
    I figured it out!!!! You use the formula Vf= Vi + at to find the acceleration, then use the formula a(centripital) = (v^2)/r. Then you find the acceleration by the formula: a = [(a^2) + (a(centripital))^2]^1/2
  7. Feb 4, 2007 #6


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    The kinematic equation applies to the tangential deceleration along the curved bend. You are not given the distance traveled along that bend. But you are given the time and change in speed. Can you find another equation of kinematics that will give you the tangential deceleration without knowing the tangential displacement?
  8. Feb 4, 2007 #7
    Thank you for all of your help!!!
  9. Feb 4, 2007 #8


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    yes! That gives you the magnitude of the acceleration. What about its direction? BTW, this is non-uniform circular motion.
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