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
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mayankdevnani
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@mathslover @UnkleRhaukus @Callisto @abs4
mayankdevnani
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@amistre64 @.Sam.
mayankdevnani
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@ganeshie8 @hba
mayankdevnani
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@LoveYou*69 @RaphaelFilgueiras
mayankdevnani
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somebody plzzz.. help me
mayankdevnani
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@TranceNova
mathslover
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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
s = 200 m
v _1= 30 m/s
v_ 2 = 20 m/s
mayankdevnani
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ok!! then
mathslover
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*s_1 =200 m
s_2 =300 m
hba
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Its a simple objective question
mathslover
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draw a rough diagram of the situation first
mayankdevnani
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@hba whats the asnwer
mayankdevnani
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|dw:1348059879133:dw|
is it be??
mathslover
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*very rough* :P wait
mayankdevnani
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ok
RaphaelFilgueiras
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|dw:1348059879801:dw|
mathslover
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easy ;
\[\large{\frac{s_{total}}{u+v} = \frac{200+300}{20+30}=\frac{500}{50} =10?}\]
mayankdevnani
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@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
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@mathslover it's not the right answer......
RaphaelFilgueiras
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look they have a relative velocity of v=30-20 m/s rigth?
mayankdevnani
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right...
mayankdevnani
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i.e is 10 m/s
mathslover
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\[\large{\frac{s_1+s_2}{v_1 - v_2}=\frac{300+200}{30-20}=\frac{500}{10}=50 \textbf{ms^{-1}}}\]
mathslover
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gotcha now..
mathslover
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v_1 - v_2 because of Relative velocities ;)
mayankdevnani
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ok!! then.
mayankdevnani
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but why you divide v1-v2 @mathslover
RaphaelFilgueiras
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imagine now in the moment the front of the train reach the other it must travel 300 m with this velocity
mayankdevnani
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ok
RaphaelFilgueiras
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now all train must pass throught other with the same velocity(10m/2),so the back part must travel 200 m
mayankdevnani
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ok
RaphaelFilgueiras
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now what is the total time?
mayankdevnani
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what??
RaphaelFilgueiras
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|dw:1348060501508:dw|
mayankdevnani
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ok
mayankdevnani
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neat and clean diagram?? wow! awesome
RaphaelFilgueiras
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the back part of the train must run 500m with 10m/s what is the time to do it?
mayankdevnani
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50m/s.. thnx..
mayankdevnani
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ok!!
mayankdevnani
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@RaphaelFilgueiras i have a problem??
woleraymond
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@RaphaelFilgueiras is right
vannayen
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|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