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
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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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