Two trains A and B of length 300m.Each are moving on two parallel tracks with a uniform speed of 54km/h in the same direction with the train A ahead of .The driver of train B decides to overtake A and accelerates by 2m/s .If after 25s the guard of train B just brushes part the drives of A . What was the original distance between them
Answers
Answer:
Train A:-
Speed = 72km/h = 72×5/18 = 20 m/s
Time = 50 seconds
Distance = Speed × Time
= 20 × 50
= 1000 m
Train B:-
u = 72 km/h = 20 m/s
t = 50 seconds
a = 1 m/s²
s = ut + 1/2at²
= 20×50 + 1/2×1×(50)²
= 1000+ 1/2×2500
= 1000+1250
= 2250 m
Distance travelled more by train B = 2250-1000
= 1250 m
Length of 1 train = 400 m
Distance between them = 1250-400
= 850 m
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solution:
u=54kmph
u=15m/s
Distance covered in 25s is X1=ut
=15×25
=375
time t=25s, acceleration a=2m/s
the distance covered in 25s is
X2=ut+1/2at^2
=15×25+1/2×2×(25)^2
=375+625 X2 =1000m
original distance b/w the two trains is
X= X2-X1
=1000-375
=625