## mayankdevnani 3 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

1. mayankdevnani

@mathslover @UnkleRhaukus @Callisto @abs4

2. mayankdevnani

@amistre64 @.Sam.

3. mayankdevnani

@ganeshie8 @hba

4. mayankdevnani

@LoveYou*69 @RaphaelFilgueiras

5. mayankdevnani

somebody plzzz.. help me

6. mayankdevnani

@TranceNova

7. mathslover

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

ok!! then

9. mathslover

*s_1 =200 m s_2 =300 m

10. hba

Its a simple objective question

11. mathslover

draw a rough diagram of the situation first

12. mayankdevnani

@hba whats the asnwer

13. mayankdevnani

|dw:1348059879133:dw| is it be??

14. mathslover

*very rough* :P wait

15. mayankdevnani

ok

16. RaphaelFilgueiras

|dw:1348059879801:dw|

17. RaphaelFilgueiras

50s

18. mathslover

easy ; $\large{\frac{s_{total}}{u+v} = \frac{200+300}{20+30}=\frac{500}{50} =10?}$

19. mayankdevnani

@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

@mathslover it's not the right answer......

21. RaphaelFilgueiras

look they have a relative velocity of v=30-20 m/s rigth?

22. mayankdevnani

right...

23. mayankdevnani

i.e is 10 m/s

24. mathslover

$\large{\frac{s_1+s_2}{v_1 - v_2}=\frac{300+200}{30-20}=\frac{500}{10}=50 \textbf{ms^{-1}}}$

25. mathslover

gotcha now..

26. mathslover

v_1 - v_2 because of Relative velocities ;)

27. mayankdevnani

ok!! then.

28. mayankdevnani

but why you divide v1-v2 @mathslover

29. RaphaelFilgueiras

imagine now in the moment the front of the train reach the other it must travel 300 m with this velocity

30. mayankdevnani

ok

31. RaphaelFilgueiras

now all train must pass throught other with the same velocity(10m/2),so the back part must travel 200 m

32. mayankdevnani

ok

33. RaphaelFilgueiras

now what is the total time?

34. mayankdevnani

what??

35. RaphaelFilgueiras

|dw:1348060501508:dw|

36. mayankdevnani

ok

37. mayankdevnani

neat and clean diagram?? wow! awesome

38. RaphaelFilgueiras

the back part of the train must run 500m with 10m/s what is the time to do it?

39. mayankdevnani

50m/s.. thnx..

40. RaphaelFilgueiras

yw 50s

41. mayankdevnani

ok!!

42. mayankdevnani

@RaphaelFilgueiras i have a problem??

43. woleraymond

@RaphaelFilgueiras is right

44. vannayen

|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