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August 31, 2014

August 31, 2014

Posted by **Sam** on Saturday, July 10, 2010 at 5:58pm.

(a) If the train goes around a curve at that speed and the acceleration experienced by the passengers is to be limited to 0.060g, what is the smallest radius of curvature for the track that can be tolerated?

(b) At what speed must the train go around a curve with a 0.50 km radius to be at the acceleration limit?

- Physics -
**Sam**, Saturday, July 10, 2010 at 6:26pma) Use the formula a=(v^2)/r

a=.060g

g=9.8 m/(s^2)

so a=.588 m/s^2

v=216km/hr * 1hr/3600s * 1000m/1km = 60m/s

.588=(60^2)/r

solve for r, r = 6122.44898

b)Use the formula a=(v^2)/r

a=.060g

g=9.8 m/(s^2)

so a=.588 m/s^2

r=.50km * 1000m/1km =500m

.588=(v^2)/500

solve for v, v= 17.146 m/s

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