Relative speed = (60+ 90) km/hr
=
150 x
5
m/sec
18
=
125
m/sec.
3
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time =
2000 x
3
sec = 48 sec.
125
=
150 x
5
m/sec
18
=
125
m/sec.
3
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time =
2000 x
3
sec = 48 sec.
125
120
m/sec = 12 m/sec.
10
Speed of the second train =
120
m/sec = 8 m/sec.
15
Relative speed = (12 + 8) = 20 m/sec.
Required time =
(120 + 120)
sec = 12 sec.
20
Then, the length of the second train is
x
metres.
2
Relative speed = (48 + 42) kmph =
90 x
5
m/sec = 25 m/sec.
18
[x + (x/2)]
= 12 or
3x
= 300 or x = 200.
25
2
Length of first train = 200 m.
Let the length of platform be y metres.
Speed of the first train =
48 x
5
m/sec =
40
m/sec.
18
3
(200 + y) x
3
= 45
40
600 + 3y = 1800
y = 400 m.
240
m/sec = 10 m/sec.
24
Required time =
240 + 650
sec = 89 sec.
10
20 x
5
m/sec =
50
m/sec.
18
9
Length of faster train =
50
x 5
m =
250
m = 27
7
m.
9
9
9
So, 2x =
(120 + 120)
12
2x = 20
x = 10.
Speed of each train = 10 m/sec =
10 x
18
km/hr = 36 km/hr.
5
72 x
5
m/sec
= 20 m/sec.
18
Time = 26 sec.
Let the length of the train be x metres.
Then,
x + 250
= 20
26
x + 250 = 520
x = 270.
60 x
5
m/sec
=
50
m/sec.
18
3
Length of the train = (Speed x Time).
Length of the train =
50
x 9
m = 150 m.
3
Video Explanation: https://youtu.be/q6Xy5JXNp-k
=
200 x
5
m/sec
18
=
500
m/sec.
9
Let the length of the other train be x metres.
Then,
x + 270
=
500
9
9
x + 270 = 500
x = 230.
2 x
5
m/sec =
5
m/sec.
18
9
4 kmph =
4 x
5
m/sec =
10
m/sec.
18
9
Let the length of the train be x metres and its speed by y m/sec.
Then,
x
= 9 and
x
= 10.
y –
5
9
y –
10
9
9y – 5 = x and 10(9y – 10) = 9x
9y – x = 5 and 90y – 9x = 100.
On solving, we get: x = 50.
Length of the train is 50 m.