2 trains length 300m,400m have speeds 10m/s, 15m/s.They start at 2 ends of bridge(1200m) & move towards each other.Find following:
1)After how much time they Meet each other?
2)After how much time they Cross each other?
PLease explain the difference, methods.
THanks.
2007-01-02 00:52:23 · 5 answers · asked by ʞzɹәႨnɹ 2 in Science & Mathematics ➔ Mathematics
1)
They approach each other at 10 + 15 = 25m/s so they meet in 1200/25 = 48 seconds
48 seconds
2)
Imagine one train (A) is stationary and the other is running at 25m/s (B)
The front of A meets the front of B at time x. At time x + (lengthA/25) the front of B is at the back of A. For the back of B to be at the back of A, the time is x + (lengthA/speedB) + (lengthB/speedB) = x + (sum of lengths/speed) = 48 + (300 + 400)/25 = 48 + 700/25 = 48 + 28 = 76 seconds
76 seconds
2007-01-02 01:00:03 · answer #1 · answered by Tom :: Athier than Thou 6 · 0⤊ 0⤋
let trains be a and b
for a
s = ut =10t
for b
s = ut = 15 t
time to meet each other is t
therefore
1200 = 10t + 15 t
1200 = 25 t
t = 48 s
time to cross each other be x after they meet
for a
s = ux =10x
for b
s = ux = 15x
therefore
300 + 400 = 10x +15x= 25x
x = 700/25 =28s
total time taken to cross each other is 28 +48 = 76
good luck
2007-01-02 01:46:31 · answer #2 · answered by kainesh p 2 · 0⤊ 0⤋
The two trains are moving toward each other at 25m/s.
Therefore, they will meet at 1200m / 25m/s = 48 seconds.
I assume "cross" refers to when the last car in each train will be together. Then add the length of the trains to the problem.
(1200m + 300m + 400m) / 25m/s = 76 seconds.
2007-01-02 00:58:48 · answer #3 · answered by Steve A 7 · 0⤊ 0⤋
two trains travel 25 metres in 1 sec
bridge is 1200 metres long
1200 divided by 25 = 48 seconds
Part two:
1200 + 300 = 400 = 1900
1900 divided by 25 = 76 seconds
2007-01-02 04:46:40 · answer #4 · answered by David C 2 · 0⤊ 1⤋
Too long to type in./ Go on line and check it out
2007-01-02 00:59:46 · answer #5 · answered by Anonymous · 0⤊ 0⤋