A train is moving and crossing a man who is running on a  platform of 100 m at a speed of 8 kmph in the direction of the train, in 9 sec. If the speed of the train is 74 kmph. Find the length of the train ?

Let the speed of the slower train = p kmph

ATQ,

Speed of the faster train = (p + 6) kmph

Then,

(p + p + 6) x 5 = 160

10p + 30 = 160

10p = 130

p = 13 kmph

Then, speed of the faster train = p + 6 = 13 + 6 = 19 kmph.

We know that,  Speed = Distance/time

Let the speed of the train be = x kmph

           NORMAL SPEED                           SPEED 5 KMPH MORE

          Distance = 360 Km                         Distance =360
 
          speed    =  X Kmph                        Speed = (x+5)
 
            Time =360/X                              Time =360/(x+15)
          

For the same journey if the speed increased 10 kmph more it will take 1 hour less

Time with original speed - time with increased speed = 1

 => (360/x) - (360/x+10) =  1                               

 =>  (360(x+5)-360x)/x(x+5) =1

 =>   360X + 1800 -360X  = X(X+5)

  =>   X^2 + 5x - 1800 = 0

  X = 40 or -45

 X cannot be negetive value
                  

 X = 40 kmph.

Given speed of the first train = 60 km/hr = 60 x 5/18 = 50/3 m/s

Let the speed of the second train = x m/s

Then, the difference in the speed is given by

503 - x = 12018

=> x = 10 m/s

=> 10 x 18/5 = 36 km/hr

Let the length of the 1st train = L mts

Speed of 1st train = 48 kmph

Now the length of the 2nd train = L/2 mts

Speed of 2nd train = 42 kmph

Let the length of the bridge = D mts

Distance = L + L/2 = 3L/2

Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)

Time = 12 sec

=> 3L/2x25 = 12

=> L = 200 mts

Now it covers the bridge in 45 sec

=> distance = D + 200

Time = 45 sec

Speed = 48 x5/18 = 40/3 m/s

=> D + 200/(40/3) = 45

=> D = 600 - 200 = 400 mts

Hence, the length of the bridge = 400 mts.