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hba

  • one year ago

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.

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  1. hba
    • one year ago
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    @saifoo.khan

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

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

  4. saifoo.khan
    • one year ago
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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.

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

  6. hba
    • one year ago
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    @saifoo.khan

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

  8. hba
    • one year ago
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    Yeah

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

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

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

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

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

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

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

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

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

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

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

  20. hba
    • one year ago
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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 ?

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

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

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

  24. hba
    • one year ago
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    Ok.

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

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

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

  28. hba
    • one year ago
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    Hmm :)

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

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

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

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

  33. saifoo.khan
    • one year ago
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    Haha.

  34. hba
    • one year ago
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    Now solve it .

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

  36. hba
    • one year ago
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    180/4

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

  38. hba
    • one year ago
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    Sin45=0.707

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

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

  41. saifoo.khan
    • one year ago
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    0.414

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

  43. saifoo.khan
    • one year ago
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    dang.

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

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

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

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

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

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

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

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

  52. hba
    • one year ago
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    k.

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

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

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

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

  57. hba
    • one year ago
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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 //

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

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

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

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

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

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

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

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

  66. hba
    • one year ago
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    @hartnn Yes ?

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

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

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

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

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

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