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The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s^2 and maximum possible retardation is 5m/s^2. The maximum speed that train can achieve is 85m/s. Minimum time in which the train can complete a journey of 1000m ending at rest, is n*(sqrt2)/(sqrt3) sec, where n is integer. Find n?

The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s^2 and maximum possible retardation is 5m/s^2. The maximum speed that train can achieve is 85m/s. Minimum time in which the train can complete a journey of 1000m ending at rest, is n*(sqrt2)/(sqrt3) sec, where n is integer. Find n?

Grade:11

1 Answers

Piyush
23 Points
5 years ago
Starting from rest the train accelerates for about 8.5 second to achieve max speed of 85 m/ s. The distance covered will be S1 and let time be t1. Similarly, calculate time and distance for other two conditions( ie. For uniform veolcity and retardation) using kinematical equation of motion. The total time taken by the train would be approx 26 sec.

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