. Two trains A and B of length 200 m each are moving on two parallel tracks with a
uniform speed of 10 1 ms−
in the same direction, with the train A ahead of B. The
driver of train B decides to overtake train A and accelerates by 1 2 ms−
. If after 50
s, the guard of train B brushes past the driver of train A, what was the original
distance between the two trains?
Answers
Answer:
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Explanation:
Given that,
u
A
=u
B
=72kmh
−1
=72×
18
5
=20ms
−1
Using the relations, s=ut+
2
1
at
2
, we get
S
B
=u
B
t+
2
1
at
2
=20×50+
2
1
×1×(50)
2
S
B
=1000+1250=2250m
Also, let S
A
be the distance covered by the train A, then S
A
=u
A
×t
=20×50=1000m
Original distance between the two trains
= S
B
−S
A
=2250−1000=1250m
this is a same type of question that u want.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h
−1
in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s
−2
. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
this is the same question.
for train A
u = 101 m/s
t = 50 sec
a = 0 m/s²
by II equation of motion ,we get
s = ut + ½at
s = 101 × 50 + ½ × 0 × 40²
s = 5050 m
for train B
u = 101 m/s
t = 50 sec
a' = 12 m/s²
by II equation of motion , we get
s' = ut + ½a't
s' = 101 × 50 +½ × 12 × 50²
s' = 5050 + 6 × 2500
s' = 5050 + 15000
s' = 20050 m
ditance between the guard of the train B and the
driver of train A = 20050 - 5050 = 15000 m
HINT : the position of guard of train at the last of train and the position of driver in the engine of train which is attached to the train