Nick G.

asked • 07/15/18

give solution

A train leaves statoin A at 8:00 hrs to reach station B at 09:30 hr ,travelling at 90 km/hr at 10 hr it leaves station B and travels at 110 km/hr to reach station C at 12hrs.what was the average speed of train in travelling from station A to C .

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Mark M. answered • 07/15/18

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Distance = (Rate)(Time)
 
Distance traveled from 8 AM to 9:30 AM: (90 km/hr)(1.5 hr) = 135 km
 
Distance traveled from 9:30 AM to 10 AM:  0 km
 
Distance traveled from 10 AM to 12:00 PM:  (110 km/hr)(2 hr) = 220 km
 
Average speed = (Total distance) / (Total time) = (355 km) / (4 hr) = 88.75 km / hr
 
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Arthur D.

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Hi Mark M., I don't see why you would consider the time that the train was idle in determining its average speed. What if the train sat in the station for 2 or 3 or 5 hours. Would you add the 2 or 3 or 5 hours to the 3.5 hours that it was travelling to get the average rate ? If the train was idle for 5 hours, the average rate would be 355/8.5=41.76 km/hr which doesn't make sense to me. There was a similar problem on this site a couple of years ago where a person sat and rested during a walk. There was a difference of opinion on that problem also.
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07/15/18

Mark M.

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I guess that the problem could be interpreted either way--counting only the time when the train is moving or counting the total time that it takes the train to get from its origin to its destination.  When I travel to NYC on the LIRR, the train schedule gives the departure time and the time of arrival at Penn Station in Manhattan.  The times that the train is stopped at various intermediate stations are included in the difference between the original station departure time and the final destination arrival time. 
 
The problem is ambiguous.  Either answer is acceptable, as far as I'm concerned.  
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07/15/18

Arthur D.

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Hi Mark, Thanks for your feedback. The problem definitely is ambiguous. I never thought about your example of the train to NYC. The layover times have to be included in the total time from departure to destination. Have a nice day, Mark.
Arthur D.
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07/16/18

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