Math, asked by sharathreddy253, 1 year ago

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

Answered by Anonymous
43

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

Answered by AneesKakar
3

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.

#SPJ3

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