Between two stations a train accelerates uniformly at first then moves with constant velocity and then retards uniformly. If the ratio of the time taken is1:8:1,and the maximum speed attained by the train is 60 km per hour ,then what is the average speed of the whole journey?
Between two stations a train accelerates uniformly at first then moves with constant velocity and then retards uniformly. If the ratio of the time taken is1:8:1,and the maximum speed attained by the train is 60 km per hour ,then what is the average speed of the whole journey?
Let it accelerates with say ‘a’ in time ‘x’ then it reaches a velocity ‘v’ and continues motion for time ‘8x’ abd finally then retards with ‘b’ in time ‘x’.
v = 0 + ax
=> v = ax
=> 60 = ax
=> a = 60/x
And,
0 = v – bx
=> b = 60/x
Distance covered in time of acceleration,
S1 = 0 + ½ at2
=> S1 = ½ ax2 = ½ × 60/x × x2 = 30x
Distance covered while moving with constant velocity is,
S2 = vt = 60 × 8x = 480x
Distance covered in time of retardation,
02 = v2 - 2bS3
=> S3 = 602/(2 × 60/x)
=> S3 = 30x
So, total distance = 30x + 480x + 30x = 540x
Total time = x + 8x + x = 10x
So, average velocity = 540x/10x = 54 km/h
Let it accelerates with say ‘a’ in time ‘x’ then it reaches a velocity ‘v’ and continues motion for time ‘8x’ abd finally then retards with ‘b’ in time ‘x’.
v = 0 + ax
=> v = ax
=> 60 = ax
=> a = 60/x
And,
0 = v – bx
=> b = 60/x
Distance covered in time of acceleration,
S1 = 0 + ½ at2
=> S1 = ½ ax2 = ½ × 60/x × x2 = 30x
Distance covered while moving with constant velocity is,
S2 = vt = 60 × 8x = 480x
Distance covered in time of retardation,
02 = v2 - 2bS3
=> S3 = 602/(2 × 60/x)
=> S3 = 30x
So, total distance = 30x + 480x + 30x = 540x
Total time = x + 8x + x = 10x
So, average velocity = 540x/10x = 54 km/h
