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`        (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.`
8 months ago

Arun
13863 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.
```
8 months ago
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## COUPON CODE: SELF10

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
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• Discussion Forum
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions