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A train of length 200 m travelling at 30 m/s overtakes another train of length 300 m travelling at 20 m/s. The time taken by the first train to pass the second is

Physics
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@amistre64 @.Sam.

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Other answers:

somebody plzzz.. help me
A train of length 200 m travelling at 30 m/s overtakes another train of length 300 m travelling at 20 m/s. The time taken by the first train to pass the second is s = 200 m v _1= 30 m/s v_ 2 = 20 m/s
ok!! then
*s_1 =200 m s_2 =300 m
  • hba
Its a simple objective question
draw a rough diagram of the situation first
@hba whats the asnwer
|dw:1348059879133:dw| is it be??
*very rough* :P wait
ok
|dw:1348059879801:dw|
50s
easy ; \[\large{\frac{s_{total}}{u+v} = \frac{200+300}{20+30}=\frac{500}{50} =10?}\]
@RaphaelFilgueiras i don't know your process of taking your answer....but it is correct....plz do it by step by step ...don't draw
@mathslover it's not the right answer......
look they have a relative velocity of v=30-20 m/s rigth?
right...
i.e is 10 m/s
\[\large{\frac{s_1+s_2}{v_1 - v_2}=\frac{300+200}{30-20}=\frac{500}{10}=50 \textbf{ms^{-1}}}\]
gotcha now..
v_1 - v_2 because of Relative velocities ;)
ok!! then.
but why you divide v1-v2 @mathslover
imagine now in the moment the front of the train reach the other it must travel 300 m with this velocity
ok
now all train must pass throught other with the same velocity(10m/2),so the back part must travel 200 m
ok
now what is the total time?
what??
|dw:1348060501508:dw|
ok
neat and clean diagram?? wow! awesome
the back part of the train must run 500m with 10m/s what is the time to do it?
50m/s.. thnx..
yw 50s
ok!!
@RaphaelFilgueiras i have a problem??
@RaphaelFilgueiras is right
|dw:1348140084541:dw| x=v1t +x1,x1=0 x=v2t +x2 , x2=500m first train over take second train unless x=x v1t=v2t +500 t=50s

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