a) 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
The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h.
(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?
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2 answers