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mathslover

  • 3 years ago

A 2m wide car is moving uniformly with a speed of 8m/s along the edge of a straight horizontal road. A pedestriun starts to cross a road with a speed "v" when the car is 12 m away from him. The minimum value of v for the pedestrian to cross the road safety is ?

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  1. ash2326
    • 3 years ago
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    There can be two cases, either the man crosses the road before the car reach him or he crosses it after the car has crossed

  2. mathslover
    • 3 years ago
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    well if we take the man to reach more safely there is that is the first case

  3. ash2326
    • 3 years ago
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    see the diagram|dw:1361020408653:dw| Time taken for car to reach the man, \[12 m\div 8 m/s= 1.5 seconds\] In this time the man should cover the 2 meter distance, fairly simple now Can you find v?

  4. DLS
    • 3 years ago
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    |dw:1361020539994:dw| @ash2326 just trying..i dont think there will be 2 cases,he should move making an angle theta with the axis not straight..

  5. mathslover
    • 3 years ago
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    yeah @DLS has a point\

  6. ash2326
    • 3 years ago
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    But we need to find the minimum V for that, he has to walk straight

  7. mathslover
    • 3 years ago
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    the man will make an angle theta first we have to find the componenets ( x and y ) and then solve for theta

  8. ash2326
    • 3 years ago
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    |dw:1361020702285:dw| We aren't given D, we can't find angle without it

  9. mathslover
    • 3 years ago
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    |dw:1361020857966:dw|

  10. DLS
    • 3 years ago
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    \[v=\frac{d}{t} \] \[BC=2cosec \theta\] \[t_{bus}=t_{boy} \] \[\frac{4+2\cot \theta}{8}=\frac{8}{v}=2\sin \theta+\cos \theta \] \[v=\frac{8}{2\sin \theta+\cos \theta}\] for v to be minimummmmmmmmmmm \[\frac{dv}{d \theta}=0\] \[\frac{d8(2 \sin \theta+\cos \theta)^-1}{d \theta}=0\]

  11. DLS
    • 3 years ago
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    sorry i took the distance 4m correct that value..

  12. DLS
    • 3 years ago
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    |dw:1361020933104:dw|

  13. DLS
    • 3 years ago
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    if my solution is close enough

  14. mathslover
    • 3 years ago
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    how r those angles : 2 cot theta and 2 cosec theta ?

  15. DLS
    • 3 years ago
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    \[\LARGE \frac{12+2\cot \theta}{8}=\frac{8}{v}=\frac{6+\cot \theta}{4}\]

  16. DLS
    • 3 years ago
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    final thing

  17. ash2326
    • 3 years ago
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    How do you have 8/v? man isn't travelling 8 meters

  18. DLS
    • 3 years ago
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    @ash2326 I equated the times.. Tbus=Tboy

  19. DLS
    • 3 years ago
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    V=s/t

  20. ash2326
    • 3 years ago
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    I know that but man has to travel \[\sqrt{2^2+(2\cot \theta)^2}\] so it should be \[\frac{\sqrt{2^2+(2\cot \theta)^2}}{v}\]

  21. ash2326
    • 3 years ago
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    |dw:1361021378137:dw|

  22. DLS
    • 3 years ago
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    yeah i got ur query,hold on

  23. DLS
    • 3 years ago
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    i took that 12+2cot theta that is the total distance he has to travel i guess distance is scalar :o so why are u adding it vectorally? im getting confused too :S

  24. ash2326
    • 3 years ago
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    I solved this, there is no global minima for v. We need to revisit the problem

  25. mathslover
    • 3 years ago
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    got to go now,,,... sorry but please post your opinions

  26. DLS
    • 3 years ago
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    sure

  27. ash2326
    • 3 years ago
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    @DLS the man is travelling with an angle, his speed is v I just found the hypotenuse distance

  28. DLS
    • 3 years ago
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    i thought that was 2 cosec theta :O

  29. ash2326
    • 3 years ago
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    Yes, if you solve the square root, you will get 2 cosec theta

  30. DLS
    • 3 years ago
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    :D

  31. ash2326
    • 3 years ago
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    I'm not convinced about the car travelling an extra distance 2 cot theta

  32. ash2326
    • 3 years ago
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    @mathslover do you have answer for this?

  33. DLS
    • 3 years ago
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    offline~

  34. DLS
    • 3 years ago
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    Just do the whole thing for him :P From the beginning :D

  35. mathslover
    • 3 years ago
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    yep --- 4/3 --- ans..

  36. Mashy
    • 3 years ago
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    bahhhh!!!!..

  37. Mashy
    • 3 years ago
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    doing all that other crap!! :P.. good thing you put up the answer!!

  38. mathslover
    • 3 years ago
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    @Mashy you cn put up your soln... I posted it becz ash wnted to check the ans... Please post your soln

  39. Mashy
    • 3 years ago
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    4/3 is the simple method.. of going straight :O..

  40. DLBlast
    • 3 years ago
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    before i can start i need to know the in physics what is the start point for all formula is = to ? or 0.00 or what the name of all start point plz? start point =

  41. ash2326
    • 3 years ago
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    @mathslover Use my first post, I had told you for minimum it has to be straight

  42. mathslover
    • 3 years ago
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    DLS can you tell me how you got 2 cot theta there ?

  43. DLS
    • 3 years ago
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    hold on

  44. mathslover
    • 3 years ago
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    |dw:1361184323999:dw| ok :)

  45. DLS
    • 3 years ago
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    isnt the sol clear? they are taking for it to be going straight

  46. mathslover
    • 3 years ago
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    But I am trying to do for the angle theta... please tell how you got 2 cot theta there

  47. mathslover
    • 3 years ago
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    ??

  48. DLS
    • 3 years ago
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    wait I got confused :P sec

  49. DLS
    • 3 years ago
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    |dw:1361185375230:dw| AC=2 cot theta

  50. DLS
    • 3 years ago
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    \[\LARGE \cot \theta=\frac{1}{\tan \theta}\]

  51. DLS
    • 3 years ago
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    idk why i got confused !

  52. DLS
    • 3 years ago
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    \[\LARGE \cot \theta=\frac{AC}{2}\]

  53. mathslover
    • 3 years ago
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    OH got it thanks a lot

  54. DLS
    • 3 years ago
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    :P

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