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Math problem regarding kinetic motion

  1. Sep 19, 2007 #1
    Sorry, i have problem in this question which comes from my textbook. I couldn't solve it and i would like someone to guide me.

    (Question)
    A body travels 30 metres in the sixth second of motion and 24 meters in the ninth second. Find:
    i) the initial velocity (ans: 41 m/s )
    ii) the retardation, assumed uniform (ans: -2 m/s/s )
    iii) the time elapsing before coming to rest (ans: 20.5s)


    From my thinking is that, the body is actually moving in a curve so it changes its displacement from 30 meters to 24 meters within the time [6~9] seconds,
    but this problem is in charge of kinetic problem as well, as it's in the chapter 'kinetic'.
    And since it's not moving in a straightline, i assume that it cannot apply the ordinary equation : S = u * t * 1/2 a*t^2;
    But by only reconstructing the equation from a = dv/dt;

    Plus, the retardation and initial velocity seems to be dependantly constant variable. So my guess is that, since it involves friction, i would simply guess that it's actually having altering of work as :
    Sum(W) = Fn * d(Sn);
    where W is the total Work,
    Fn is the changing of Forces within the time interval [0,t];
    dSn is the delta of distance traveled within time interval [0,t];


    Thanks in advance.
    Daniel.
     
  2. jcsd
  3. Sep 19, 2007 #2

    CompuChip

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    From my thinking, this means that the velocity changes from 30 m/s to 24 m/s in the [6, 9] s time interval.

    Also, "the retardation is assumed uniform" sounds to me like "the acceleration is constant", which means the formula
    [tex]s(t) = s_0 + v_0 t + \tfrac12 a t^2[/tex]
    will certainly hold (it follows from solving
    [tex]s''(t) = a[/tex]
    with the appropriate boundary conditions).

    "Kinetic" basically just means "movement". E.g. "kinetic energy" is just "movement energy", in a quite literal translation.
    But perhaps what is meant is, that you should solve it using energies instead of dynamic equations?
     
  4. Sep 19, 2007 #3

    HallsofIvy

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    Why do you assert "it's not moving in a straightline"? I can find no where in the problem statement that it is said it is not moving in a straight line. "Kinetic" simply means "moving"- not necessarily moving in a curve.

    Assuming constant acceleration, a, (although that's not mentioned until (b)), a body will move (1/2)at2+ v0t where v0 is the initial speed. After 5 seconds, it will have gone (25/2)a+ 5v0. After 6 seconds, it will have gone (36/2)a+ 6v. During the "6th second", it will have gone (36/2)a+ 6v0- ((25/2)a+ 5v0= (11/2)a+ v0= 30. Do the same for the distances traveled in 8 and 9 minutes, setting the difference equal to 24 and you have two equations to solve for a and v0.
     
  5. Sep 20, 2007 #4
    oh yes! i misunderstood the question by assuming it has "travelled along". Thanks everyone for the answer and detailed explanation. I got it solved. ^^
     
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