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Grade: 12th pass
        
(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s2 and subway stations are located 806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, v, and a versust for the interval from one start-up to the next.
5 months ago

Answers : (1)

Arun
11598 Points
							
a) Distance available to accelerate = (806/2) = 403m. 
Velocity = sqrt. (2ad), = sqrt. (2.68 x 403), = 32.86m/sec. Speed max = 118.31 kph. 
b) Time to accelerate = (v/a), = 32.86/1.34, = 24.52 s. Doube it, = 49.04s. 
c) (49.04+ 20) = 69.04 s. 
In 69.04 s., it has travelled 806m. 
(806/69.04 ) = 11.67 m/sec average. Speed average = (11.67 x 3600/1000) = 42.012 kph.
5 months ago
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