Question

The maximum possible acceleration of a train starting from rest and moving on straight track is 10m/s² and maximum possible retardation is 5m/s².the maximum speed that train can achieve is 85 m/s.minimum time in which train can complete a journey of 1000m ending at rest is n√2/3 sec. Where n is an integer.find n

Answer 1

Time for acceleration :

85 / 10 = 8.5 s

Time for deceleration :

85 / 5= 17s

Total time for retardation and acceleration :

8.5 + 17 = 25.5 s

This motion forms a triangle of height 85 and length 25.5

Distance covered in retardation and acceleration is :

0.5 × 25.5 × 85 = 1083 m

The distance moved here is 1000m

We therefore look for the value of the that gives 1000

0.5 × t × 85 = 1000

t = 1000 / 42.5 = 23.53 s

This is equal to the total time :

23.53 = n√2/3

23.53 = 0.816n

n = 23.53 / 0.816 = 28.84

n = 28.84

Answer 2

Answer: it will come 29.53 approx 30

Explanation: