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

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

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

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

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0A 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

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

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

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

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

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

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

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

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

mayankdevnani
 2 years ago
Best 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

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

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

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

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

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

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

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

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

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

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

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

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

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

vannayen
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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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