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 70 m/s.minimum time in which train can complete a journey of 1000m ending at rest is 347/2@ sec. Where @ is an integer.find @

Answer 1

Since we are given the acceleration and deceleration and the maximum speed, we can get the distance the train moved during acceleration and deceleration.

Acceleration :

t = 70 / 10 = 7s

Deceleration :

70 / 5= 14 s

This two motions form a triangle with H = 70 and base = 7 + 14 = 21s

Distance = Area under the graph.

0.5 × 21 × 70 = 735 m

The total distance is equal to 100m meaning the remaining distance is travelled at a constant speed of 70m/s

This equals to :

1000 - 735 = 265m

Time = 265 / 70s

The total time taken for the entire motion is :

21 + 265 / 70 = 1777 / 72 s

This equals to :

347/2@ = 1777 / 72

24984@ = 3554

@ = 0.14225

Answer 2

Answer:

Explanation:

Since we are given the acceleration and deceleration and the maximum speed, we can get the distance the train moved during acceleration and deceleration.

Acceleration :

t = 70 / 10 = 7s

Deceleration :

70 / 5= 14 s

This two motions form a triangle with H = 70 and base = 7 + 14 = 21s

Distance = Area under the graph.

0.5 × 21 × 70 = 735 m

The total distance is equal to 100m meaning the remaining distance is travelled at a constant speed of 70m/s

This equals to :

1000 - 735 = 265m

Time = 265 / 70s

The total time taken for the entire motion is :

21 + 265 / 70 = 1777 / 72 s

This equals to :

347/2@ = 1777 / 72

24984@ = 3554

@ = 0.14225