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deff fn [24]
3 years ago
11

The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21

6 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 0.050g, what is the smallest radius of curvature for the track that can be tolerated? 7347 m (b) If there is a curve with a 0.90 km radius, to what speed must the train be slowed to keep the acceleration below the limit?
Physics
1 answer:
andrezito [222]3 years ago
8 0

Answer:

Explanation:

Given

average speed of train(v_{avg})=216 kmph\approx 60 m/s

Maximum acceleration=0.05g

Now centripetal acceleration is

a_c=\frac{v^2}{r}

0.05\times 9.8=\frac{60^2}{r}

r=7346.93 m

(b)Radius of curvature=900 m

therefore a_c=\frac{v^2}{r}

v=\sqrt{a_cr}

v=\sqrt{0.05\times 9.8\times 900}

v=\sqrt{441}=21 m/s

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