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hba Group Title

A particle moves along x-axis.It starts from rest at origin O and its acceleration after t sec is given by a=3-t . Calculate the displacment from O when it is at rest.

  • one year ago
  • one year ago

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  1. hba Group Title
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    @saifoo.khan

    • one year ago
  2. hba Group Title
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    Can you do it ?

    • one year ago
  3. jennychan12 Group Title
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    find antiderivative twice for a = 3-t

    • one year ago
  4. saifoo.khan Group Title
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    integrate. a=3-t v = 3t - t^2/2 + C Now set it equal to zero to get time. Then integrate again, find the disp.

    • one year ago
  5. hba Group Title
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    Therefore, the answer is 18. Something which appeared in the (NUST) test.

    • one year ago
  6. hba Group Title
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    @saifoo.khan

    • one year ago
  7. saifoo.khan Group Title
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    REALLY? O_O

    • one year ago
  8. hba Group Title
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    Yeah

    • one year ago
  9. saifoo.khan Group Title
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    An O'levels student who have Add math can solve this too! D: D: D:

    • one year ago
  10. hba Group Title
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    Hmm,good :)

    • one year ago
  11. hba Group Title
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    What about hyperbolic functions ? You've done that ?

    • one year ago
  12. saifoo.khan Group Title
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    give me the problem, then i can tlel.

    • one year ago
  13. hba Group Title
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    \[\int\limits_{}^{}e^x[\sinh^{-1} x+\frac{ 1 }{ \sqrt{1+x^2} }]dx=\]

    • one year ago
  14. hba Group Title
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    @saifoo.khan Like this.

    • one year ago
  15. saifoo.khan Group Title
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    INtegration by parts?

    • one year ago
  16. saifoo.khan Group Title
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    idk how to deal with sinh^-1

    • one year ago
  17. hba Group Title
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    So you don't know hyperbolic functions ?

    • one year ago
  18. experimentX Group Title
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    use e^x(f(x) + f'(x)) rule differentiation of arcsinh(x) = 1/sqrt(...)

    • one year ago
  19. saifoo.khan Group Title
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    I guess no, then. :S

    • one year ago
  20. hba Group Title
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    hmm No need to worry about that it's easy,you can do that later. What about finding the area from a diag ?

    • one year ago
  21. saifoo.khan Group Title
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    We have to learn that parallelogram thing. But i can find the area/vol of a curve and stuff

    • one year ago
  22. hba Group Title
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    Lets see. find the Area A+B |dw:1356722864399:dw|

    • one year ago
  23. saifoo.khan Group Title
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    NOOO! Give me more difficult one.

    • one year ago
  24. hba Group Title
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    Ok.

    • one year ago
  25. hba Group Title
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    find the area A |dw:1356723204131:dw|

    • one year ago
  26. saifoo.khan Group Title
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    |dw:1356723438919:dw|

    • one year ago
  27. saifoo.khan Group Title
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    Let me grab a pen and paper.

    • one year ago
  28. hba Group Title
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    Hmm :)

    • one year ago
  29. saifoo.khan Group Title
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    Is calc allowed?

    • one year ago
  30. hba Group Title
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    No not allowed+You changed the question lol :P

    • one year ago
  31. saifoo.khan Group Title
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    Sinx graph starts from zero. lol.

    • one year ago
  32. hba Group Title
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    Oh okay then :/

    • one year ago
  33. saifoo.khan Group Title
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    Haha.

    • one year ago
  34. hba Group Title
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    Now solve it .

    • one year ago
  35. saifoo.khan Group Title
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    isk what's sin(pi/4). i can only solve with the help of calc.

    • one year ago
  36. hba Group Title
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    180/4

    • one year ago
  37. saifoo.khan Group Title
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    area under cos is 1/sqrt2

    • one year ago
  38. hba Group Title
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    Sin45=0.707

    • one year ago
  39. saifoo.khan Group Title
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    0.707-0.293 ?

    • one year ago
  40. hba Group Title
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    @saifoo.khan Solve it and let me know :)

    • one year ago
  41. saifoo.khan Group Title
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    0.414

    • one year ago
  42. hba Group Title
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    Don't do it in decimals.

    • one year ago
  43. saifoo.khan Group Title
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    dang.

    • one year ago
  44. saifoo.khan Group Title
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    1/sqrt2 - (-1+sqrt2)/sqrt2

    • one year ago
  45. saifoo.khan Group Title
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    Making: (2-sqrt2)/sqrt2

    • one year ago
  46. saifoo.khan Group Title
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    Making: \[\frac{2\sqrt2-2}{2}\]

    • one year ago
  47. saifoo.khan Group Title
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    sqrt2-1 final answer?

    • one year ago
  48. hba Group Title
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    \[\sqrt{2}-1 \ sq.unit\] Yeah right :D

    • one year ago
  49. saifoo.khan Group Title
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    Wheo! B)

    • one year ago
  50. hba Group Title
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    One more. Find the area enclosed by x=y^2 and y=x-2

    • one year ago
  51. saifoo.khan Group Title
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    umm okay. solving.

    • one year ago
  52. hba Group Title
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    k.

    • one year ago
  53. saifoo.khan Group Title
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    1,4

    • one year ago
  54. saifoo.khan Group Title
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    area under line = 2 squnits

    • one year ago
  55. saifoo.khan Group Title
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    trying to figure out about x=y^2

    • one year ago
  56. saifoo.khan Group Title
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    1/2 final answer?

    • one year ago
  57. hba Group Title
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    Let me give you options lol A)9/2 sq.units B)9/4 // C)9/7 // D)7/9 // E)4/9 //

    • one year ago
  58. saifoo.khan Group Title
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    Can't fig out.

    • one year ago
  59. hba Group Title
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    I also couldn't figure this out lol :P

    • one year ago
  60. hba Group Title
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    @hartnn Can you solve it ?

    • one year ago
  61. hba Group Title
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    Find the area enclosed by x=y^2 and y=x-2

    • one year ago
  62. hartnn Group Title
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    this is double integration problem.. right ?

    • one year ago
  63. hba Group Title
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    i don't think so.

    • one year ago
  64. hartnn Group Title
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    plot both curves..

    • one year ago
  65. saifoo.khan Group Title
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    |dw:1356725537575:dw|

    • one year ago
  66. hba Group Title
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    @hartnn Yes ?

    • one year ago
  67. hartnn Group Title
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    sorry, got disconnected.

    • one year ago
  68. hartnn Group Title
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    |dw:1356727367179:dw| \(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy|dw:1356727499321:dw|\)

    • one year ago
  69. hartnn Group Title
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    \(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy\)

    • one year ago
  70. hartnn Group Title
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    find y2 and solve the integral

    • one year ago
  71. hartnn Group Title
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    sorry, y1 and y2 limits should be |dw:1356727654740:dw|

    • one year ago
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