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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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  1. mayankdevnani Group Title
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    @mathslover @UnkleRhaukus @Callisto @abs4

    • one year ago
  2. mayankdevnani Group Title
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    @amistre64 @.Sam.

    • one year ago
  3. mayankdevnani Group Title
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    @ganeshie8 @hba

    • one year ago
  4. mayankdevnani Group Title
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    @LoveYou*69 @RaphaelFilgueiras

    • one year ago
  5. mayankdevnani Group Title
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    somebody plzzz.. help me

    • one year ago
  6. mayankdevnani Group Title
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    @TranceNova

    • one year ago
  7. mathslover Group Title
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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

    • one year ago
  8. mayankdevnani Group Title
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    ok!! then

    • one year ago
  9. mathslover Group Title
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    *s_1 =200 m s_2 =300 m

    • one year ago
  10. hba Group Title
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    Its a simple objective question

    • one year ago
  11. mathslover Group Title
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    draw a rough diagram of the situation first

    • one year ago
  12. mayankdevnani Group Title
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    @hba whats the asnwer

    • one year ago
  13. mayankdevnani Group Title
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    |dw:1348059879133:dw| is it be??

    • one year ago
  14. mathslover Group Title
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    *very rough* :P wait

    • one year ago
  15. mayankdevnani Group Title
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    ok

    • one year ago
  16. RaphaelFilgueiras Group Title
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    |dw:1348059879801:dw|

    • one year ago
  17. RaphaelFilgueiras Group Title
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    50s

    • one year ago
  18. mathslover Group Title
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    easy ; \[\large{\frac{s_{total}}{u+v} = \frac{200+300}{20+30}=\frac{500}{50} =10?}\]

    • one year ago
  19. mayankdevnani Group Title
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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

    • one year ago
  20. mayankdevnani Group Title
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    @mathslover it's not the right answer......

    • one year ago
  21. RaphaelFilgueiras Group Title
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    look they have a relative velocity of v=30-20 m/s rigth?

    • one year ago
  22. mayankdevnani Group Title
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    right...

    • one year ago
  23. mayankdevnani Group Title
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    i.e is 10 m/s

    • one year ago
  24. mathslover Group Title
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    \[\large{\frac{s_1+s_2}{v_1 - v_2}=\frac{300+200}{30-20}=\frac{500}{10}=50 \textbf{ms^{-1}}}\]

    • one year ago
  25. mathslover Group Title
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    gotcha now..

    • one year ago
  26. mathslover Group Title
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    v_1 - v_2 because of Relative velocities ;)

    • one year ago
  27. mayankdevnani Group Title
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    ok!! then.

    • one year ago
  28. mayankdevnani Group Title
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    but why you divide v1-v2 @mathslover

    • one year ago
  29. RaphaelFilgueiras Group Title
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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

    • one year ago
  30. mayankdevnani Group Title
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    ok

    • one year ago
  31. RaphaelFilgueiras Group Title
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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

    • one year ago
  32. mayankdevnani Group Title
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    ok

    • one year ago
  33. RaphaelFilgueiras Group Title
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    now what is the total time?

    • one year ago
  34. mayankdevnani Group Title
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    what??

    • one year ago
  35. RaphaelFilgueiras Group Title
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    |dw:1348060501508:dw|

    • one year ago
  36. mayankdevnani Group Title
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    ok

    • one year ago
  37. mayankdevnani Group Title
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    neat and clean diagram?? wow! awesome

    • one year ago
  38. RaphaelFilgueiras Group Title
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    the back part of the train must run 500m with 10m/s what is the time to do it?

    • one year ago
  39. mayankdevnani Group Title
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    50m/s.. thnx..

    • one year ago
  40. RaphaelFilgueiras Group Title
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    yw 50s

    • one year ago
  41. mayankdevnani Group Title
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    ok!!

    • one year ago
  42. mayankdevnani Group Title
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    @RaphaelFilgueiras i have a problem??

    • one year ago
  43. woleraymond Group Title
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    @RaphaelFilgueiras is right

    • one year ago
  44. vannayen Group Title
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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

    • one year ago
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