Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1:8:1 and the maximum speed attained be 60km/hr, then what is the average speed over the whole journey?
a) 48km/h
b) 52km/h
c) 54km/h
d) 56Km/h
Between two stations, a train accelerates from rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is 1:8:1 and the maximum speed attained be 60km/hr, then what is the average speed over the whole journey?
a) 48km/h
b) 52km/h
c) 54km/h
d) 56Km/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
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
