Asked by Anonymous
A train takes 3.3 hours to travel the first leg of a trip at 17 m/s. How fast must the train move over the second leg of 200 km to have a total average speed of 30 m/s?
Answers
Answered by
drwls
The distance traveled on the first leg is 3.3*3600*17 = 201,960 m = 201.96 km. Call it 202 km. The total trip distance is 402 km. Require that the total trip time is
402,000 m = Vaverage* T
T = 13,400 s = 3.72 hours
That leaves only 0.42 hours for the second leg.
The train needs a very high speed on the second leg to cover 200 km in 0.42 hours. It will have to be a "bullet", TGV or maglev train. There are trains that go that fast, but not in the USA.
402,000 m = Vaverage* T
T = 13,400 s = 3.72 hours
That leaves only 0.42 hours for the second leg.
The train needs a very high speed on the second leg to cover 200 km in 0.42 hours. It will have to be a "bullet", TGV or maglev train. There are trains that go that fast, but not in the USA.
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