Question

The maximum possible acceleration of a train starting from the rest and m
train starting from the rest and moving on straight track is
10 m/s2 and maximum possible retardation is 5 m/s2. The maximum speed
retardation is 5 m/s2. The maximum speed that train can achieve is
ending at rest is
70 m/s. Minimum time in which the train can complete a jou
time in which the train can complete a journey of 1000m
347/2a sec. Where a is an integer. Find a.
​

Answer 1

Answer:

n = 28.84

Explanation:

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