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.
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.
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.
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
Speed of train relative to man = 67 - 7 = 60 km/hr.
= 60 x 5/18 = 50/3 m/sec.
Time taken to pass the man = 600 x 3/50 = 36 sec.
Given speed of jogger = 9 kmph
Speed of train = 45 kmph
As they are in same direction,
Relative speed = 45 - 9 = 36 kmph = 36x5/18 = 10 m/s
Distance = 240 + 120 = 360 mts
Time = distance/speed = 360/10 = 36 sec.
Let length of train A be ?L? m and speed be ?V? m/s
ATQ ?
V =
Ans.
Given that length of first train + length of second train = 660
Speed of first train : Speed of second train = 5:8 =5x : 8x
Time taken to cross the poll by two trains = 4:3 =4y:3y
Now ,
(5x*4y) + (8x*3y) = 660
44xy = 660
xy = 15
L1 - L2 = 24xy -20xy = 4xy = 60 m
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.