A train takes 4 min. to go between stations 2.25 km apart starting and finishing at rest. The acceleration is uniform for the first 40 sec and the deceleration is uniform for the last 20 sec. Assuming the velocity to be constant for the remaining time, then maximum speed of the train is ?
Answers
Maximum speed of the train = 10.71 m/s
Explanation:
Please see the attached picture for a velocity-time graph of the motion of the train.
Velocity is taken along the y-axis and time along the x-axis.
Let v be the maximum speed of the train.
x1 is the distance covered in first 40 seconds.
v/2 * 40 = x1
x1 = 20v
The total time is 4 minutes or 240 seconds.
Time corresponding to the AB of the graph = 240 - 40 - 20 = 80s.
Distance covered in AB = x2
So x2 = 180v
Distance covered in last 20 seconds = x3
So x3 = v/2 * 20 = 10v
Now x1 + x2 + x3 = 20v + 180v + 10v
2.25 * 1000 = 210v
2250 / 210 = v
v = 10.71 m/s
Acceleration = v/40 = 10.7 / 40 = 0.2675 m/s2
Retardation = v/20 = 10.7/20 = 0.535 m/s2
Speed = distance / time
S1 = x1/ time = 20v/ 20 = v = 10.71 m/s
Maximum speed of the train = 10.71 m/s
