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Mathematics-Kinematics help

  1. Jan 24, 2009 #1
    The question goes like this:
    A body moves in a straight line from a fixed point O . Its distance from O, s meters , is given by s=t-[tex]\frac{1}{9}[/tex]t³ , where t is the time in seconds after passing through O. Fubd

    (a)the time when the body returns to O.

    I have done this question by plugging s=0 and i get t=3

    (b)the velocity at this instant.

    I found the [tex]\frac{ds}{dt}[/tex] and i plug in t=3 and i got -2m/s

    (c)the value of t when the body is instantaneously at rest.

    I assumed [tex]\frac{ds}{dt}[/tex]=0 and i've gotten myself t=[tex]\sqrt{3}[/tex]

    (d) the distance moved by the body in the 2nd second.

    I dont know what is the question asking , but the answer should be 0.304.
     
  2. jcsd
  3. Jan 24, 2009 #2

    CompuChip

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    Re: Mathematics-Kinematics

    Your approach for the first three looks correct, although for b and c I didn't check the numbers (so assuming that you did the math correctly as well you should have found the correct answers).

    For d, I suppose you would just argue like: the first second is the time between t = 0 and 1, the 2nd second is between t = 1 and t = 2, etc. - then find the distance that is travelled between those times.
     
  4. Jan 24, 2009 #3
    Re: Mathematics-Kinematics

    Thanks JIANKAI!!
    Thank you chip (:

    I know how its works le.

    Let f(x)=t-[tex]\frac{1}{9}[/tex]t³

    f(root3)-f(1) + f(root3)-f(2)
     
  5. Jan 24, 2009 #4

    CompuChip

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    Re: Mathematics-Kinematics

    What or who is Jiankai?

    And are you sure about that answer? The signs look a bit off to me. How did you get that?
     
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