Physics, asked by veerpal076kaur, 9 months ago

Two train a and b of of length 400m each are moving on two parallel tracks with uniform speed of 72 km h -1 in the same direction with a ahead of B the driver of B decide to overtake A accelerates by the 1m s-2 if after 50 s the guard of b just passes the driver of a what was the original distance between?

Answers

Answered by Anonymous
13

ANSWER :

For train A,

(u)A = 72kmph = 72×5/18 = 20mps

(a)A = 0

t = 50s

✏ (S)A = (u)A×t = (20)(50)

(S)A = 1000m

For train B,

(u)B = 72kmph = 72×5/18 = 20mps

(a)B = 1m/s^2

t = 50s

✏ (S)B = (u)B×t + 1/2(a)B×t^2

✏ (S)B = (50×20) + 1/2×1×(50)^2

(S)B = 2250m

☞ Original distance b/w A and B

☞ (S)B - (S)A = 2250 - 1000

ΔS = 1250m

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