Let the original speed and time is S and T
then distance = S x T
Now the speed changes to 2/3S and T is T+20
As the distance is same
S x T = 2/3Sx(T+20)
solving this we get t = 40 minutes
=40/60 = 2/3 hour
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.
Let the length of the stationary train Y be LY
Given that length of train X, LX = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 ---> ( 1 )
(300 + LY) / V = 60 ---> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + LY)/12 = 60
=> 300 + LY = 60 (12) = 720
=> LY = 720 - 300 = 420 m
Length of the stationary train = 420 m
Speed of train relative to jogger = 45 - 9 = 36 km/hr.
= 36 x 5/18 = 10 m/sec.
Distance to be covered = 260 + 140 = 400 m.
Time taken = 400/10 = 40 sec.
Relative speed = 60 + 90 = 150 km/hr.
= 150 x 5/18 = 125/3 m/sec.
Distance covered = 1.10 + 0.9 = 2 km = 2000 m.
Required time = 2000 x 3/125 = 48 sec.
Relative speed = 120 + 80 = 200 km/hr.
= 200 x 5/18 = 500/9 m/sec.
Let the length of the other train be L mts.
Then, (L + 260)/9 = 500/9 => L = 240 mts.
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