mayankdevnani
  • mayankdevnani
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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SOLVED
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chestercat
  • chestercat
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mayankdevnani
  • mayankdevnani
@mathslover @UnkleRhaukus @Callisto @abs4
mayankdevnani
  • mayankdevnani
@amistre64 @.Sam.
mayankdevnani
  • mayankdevnani
@ganeshie8 @hba

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More answers

mayankdevnani
  • mayankdevnani
@LoveYou*69 @RaphaelFilgueiras
mayankdevnani
  • mayankdevnani
somebody plzzz.. help me
mayankdevnani
  • mayankdevnani
@TranceNova
mathslover
  • mathslover
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
mayankdevnani
  • mayankdevnani
ok!! then
mathslover
  • mathslover
*s_1 =200 m s_2 =300 m
hba
  • hba
Its a simple objective question
mathslover
  • mathslover
draw a rough diagram of the situation first
mayankdevnani
  • mayankdevnani
@hba whats the asnwer
mayankdevnani
  • mayankdevnani
|dw:1348059879133:dw| is it be??
mathslover
  • mathslover
*very rough* :P wait
mayankdevnani
  • mayankdevnani
ok
anonymous
  • anonymous
|dw:1348059879801:dw|
anonymous
  • anonymous
50s
mathslover
  • mathslover
easy ; \[\large{\frac{s_{total}}{u+v} = \frac{200+300}{20+30}=\frac{500}{50} =10?}\]
mayankdevnani
  • mayankdevnani
@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
mayankdevnani
  • mayankdevnani
@mathslover it's not the right answer......
anonymous
  • anonymous
look they have a relative velocity of v=30-20 m/s rigth?
mayankdevnani
  • mayankdevnani
right...
mayankdevnani
  • mayankdevnani
i.e is 10 m/s
mathslover
  • mathslover
\[\large{\frac{s_1+s_2}{v_1 - v_2}=\frac{300+200}{30-20}=\frac{500}{10}=50 \textbf{ms^{-1}}}\]
mathslover
  • mathslover
gotcha now..
mathslover
  • mathslover
v_1 - v_2 because of Relative velocities ;)
mayankdevnani
  • mayankdevnani
ok!! then.
mayankdevnani
  • mayankdevnani
but why you divide v1-v2 @mathslover
anonymous
  • anonymous
imagine now in the moment the front of the train reach the other it must travel 300 m with this velocity
mayankdevnani
  • mayankdevnani
ok
anonymous
  • anonymous
now all train must pass throught other with the same velocity(10m/2),so the back part must travel 200 m
mayankdevnani
  • mayankdevnani
ok
anonymous
  • anonymous
now what is the total time?
mayankdevnani
  • mayankdevnani
what??
anonymous
  • anonymous
|dw:1348060501508:dw|
mayankdevnani
  • mayankdevnani
ok
mayankdevnani
  • mayankdevnani
neat and clean diagram?? wow! awesome
anonymous
  • anonymous
the back part of the train must run 500m with 10m/s what is the time to do it?
mayankdevnani
  • mayankdevnani
50m/s.. thnx..
anonymous
  • anonymous
yw 50s
mayankdevnani
  • mayankdevnani
ok!!
mayankdevnani
  • mayankdevnani
@RaphaelFilgueiras i have a problem??
anonymous
  • anonymous
@RaphaelFilgueiras is right
anonymous
  • anonymous
|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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