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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
 one year ago
 one year 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
 one year ago
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
right...
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

mayankdevnani Group TitleBest ResponseYou've already chosen the best response.0
ok!!
 one year ago

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

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