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mayankdevnani
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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
 2 years ago
 2 years ago
mayankdevnani Group Title
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
 2 years ago
 2 years ago

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mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@mathslover @UnkleRhaukus @Callisto @abs4
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 @.Sam.
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@ganeshie8 @hba
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@LoveYou*69 @RaphaelFilgueiras
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
somebody plzzz.. help me
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@TranceNova
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
ok!! then
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
*s_1 =200 m s_2 =300 m
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Its a simple objective question
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
draw a rough diagram of the situation first
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@hba whats the asnwer
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
dw:1348059879133:dw is it be??
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
*very rough* :P wait
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
dw:1348059879801:dw
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
50s
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
easy ; \[\large{\frac{s_{total}}{u+v} = \frac{200+300}{20+30}=\frac{500}{50} =10?}\]
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@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
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@mathslover it's not the right answer......
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
look they have a relative velocity of v=3020 m/s rigth?
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right...
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
i.e is 10 m/s
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
\[\large{\frac{s_1+s_2}{v_1  v_2}=\frac{300+200}{3020}=\frac{500}{10}=50 \textbf{ms^{1}}}\]
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
gotcha now..
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
v_1  v_2 because of Relative velocities ;)
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
ok!! then.
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
but why you divide v1v2 @mathslover
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
imagine now in the moment the front of the train reach the other it must travel 300 m with this velocity
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
now all train must pass throught other with the same velocity(10m/2),so the back part must travel 200 m
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
now what is the total time?
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
what??
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
dw:1348060501508:dw
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
neat and clean diagram?? wow! awesome
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
the back part of the train must run 500m with 10m/s what is the time to do it?
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
50m/s.. thnx..
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.3
yw 50s
 2 years ago

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
@RaphaelFilgueiras i have a problem??
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@RaphaelFilgueiras is right
 2 years ago

vannayen Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago
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