between two stations a train accelerates uniformly at first then moves with constant velocity and finally retards uniformly if the ratio of the time taken by 1 ratio 8 ratio 1 and the max speed attained be 60 kmph then what is the average speed over the whole journey
Answers
Answered by
235
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
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
Answered by
55
Answer:
54 Km/h if we want is SI Unit The Answer will be 15 m/s
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