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mayankdevnani

  • 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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  1. mayankdevnani
    • 2 years ago
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    @mathslover @UnkleRhaukus @Callisto @abs4

  2. mayankdevnani
    • 2 years ago
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    @amistre64 @.Sam.

  3. mayankdevnani
    • 2 years ago
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    @ganeshie8 @hba

  4. mayankdevnani
    • 2 years ago
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    @LoveYou*69 @RaphaelFilgueiras

  5. mayankdevnani
    • 2 years ago
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    somebody plzzz.. help me

  6. mayankdevnani
    • 2 years ago
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    @TranceNova

  7. mathslover
    • 2 years ago
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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

  8. mayankdevnani
    • 2 years ago
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    ok!! then

  9. mathslover
    • 2 years ago
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    *s_1 =200 m s_2 =300 m

  10. hba
    • 2 years ago
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    Its a simple objective question

  11. mathslover
    • 2 years ago
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    draw a rough diagram of the situation first

  12. mayankdevnani
    • 2 years ago
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    @hba whats the asnwer

  13. mayankdevnani
    • 2 years ago
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    |dw:1348059879133:dw| is it be??

  14. mathslover
    • 2 years ago
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    *very rough* :P wait

  15. mayankdevnani
    • 2 years ago
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    ok

  16. RaphaelFilgueiras
    • 2 years ago
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    |dw:1348059879801:dw|

  17. RaphaelFilgueiras
    • 2 years ago
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    50s

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

  19. mayankdevnani
    • 2 years ago
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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

  20. mayankdevnani
    • 2 years ago
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    @mathslover it's not the right answer......

  21. RaphaelFilgueiras
    • 2 years ago
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    look they have a relative velocity of v=30-20 m/s rigth?

  22. mayankdevnani
    • 2 years ago
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    right...

  23. mayankdevnani
    • 2 years ago
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    i.e is 10 m/s

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

  25. mathslover
    • 2 years ago
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    gotcha now..

  26. mathslover
    • 2 years ago
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    v_1 - v_2 because of Relative velocities ;)

  27. mayankdevnani
    • 2 years ago
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    ok!! then.

  28. mayankdevnani
    • 2 years ago
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    but why you divide v1-v2 @mathslover

  29. RaphaelFilgueiras
    • 2 years ago
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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

  30. mayankdevnani
    • 2 years ago
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    ok

  31. RaphaelFilgueiras
    • 2 years ago
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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

  32. mayankdevnani
    • 2 years ago
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    ok

  33. RaphaelFilgueiras
    • 2 years ago
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    now what is the total time?

  34. mayankdevnani
    • 2 years ago
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    what??

  35. RaphaelFilgueiras
    • 2 years ago
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    |dw:1348060501508:dw|

  36. mayankdevnani
    • 2 years ago
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    ok

  37. mayankdevnani
    • 2 years ago
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    neat and clean diagram?? wow! awesome

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

  39. mayankdevnani
    • 2 years ago
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    50m/s.. thnx..

  40. RaphaelFilgueiras
    • 2 years ago
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    yw 50s

  41. mayankdevnani
    • 2 years ago
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    ok!!

  42. mayankdevnani
    • 2 years ago
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    @RaphaelFilgueiras i have a problem??

  43. woleraymond
    • 2 years ago
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    @RaphaelFilgueiras is right

  44. vannayen
    • 2 years ago
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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

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