Two trains, each 80 m long, pass each other on parallel lines. If they are going in the same direction, the faster one takes one minute to |pass the other completely. If they are going |in opposite directions, they completely pass each other in 3 seconds. Find the speed of each train in metre per seconds:
Answers
Answer:
speed of faster train 28 m/second
and other 76/3 m/second.
explanation:
let the speeds of trains
be a and b respectively.
we know
speed=distance/time.
According to question,
first case:
If they are going in the same direction,
the faster one takes one minute
to pass the other completely.
distance=80+80=160
time=1 minute=60 seconds
- travel in same direction
relative speed [a-b]
a-b=160/60..............................(1)
second case:
If they are going in opposite directions
they completely pass each other
in 3 seconds
distance=80+80=160
time=3 seconds
- travel in opposite direction
relative speed [a+b]
a+b=160/3...........................(2)
adding equations (1) +(2)
a-b+a-b=160/60+160/3
2a=(160+3200)/60
2a=3360/60
2a=56
a=28 m/sec
substitute a value in equation (1)
28-b=160/60
28-b=8/3
-b=8/3-28
-b=(8-84)/3
-b=-76/3
b=76/3 m/sec
Given,
Length of train = 80 metre
Time taken to pass in the same direction = 60 seconds
Time taken to pass in opposite direction = 3 seconds.
To Find,
The speed of the trains.
Solution,
Let the speeds of trains be x and y respectively.
Speed = Distance/Time.
First case (x - y):
Distance = 80 + 80 = 160
Time = 60 seconds
Speed = 160/60..............................(1)
Second case (x + y):
Distance = 80 + 80 = 160
Time = 3 seconds
Speed = 160/3...........................(2)
Now, adding equations (1) & (2)
x - y + x + y = 160/60 + 160/3
⇒2x = (160+3200)/60
⇒2x = 3360/60
⇒2x = 56
⇒x = 28 m/sec
Now, substituting the value of x in equation (1). We get,
28 - y = 160/60
⇒28 - y = 8/3
⇒-y = 8/3-28
⇒-y = (8-84)/3
⇒-y = -76/3
⇒y = 76/3 m/sec
⇒y = 25.33 m/sec
∴ The faster train (x) was moving at 28 m/sec speed and the slower train (y) was moving approximately at 25 m/sec speed.
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