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Thread: Add maths: Kinematics

  1. #1

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    Add maths: Kinematics

    Q1.A particle moves in a straight line and its displacement ,x metres, from a fixed point O, t seconds after passing O, is given by x= 12t - t^3.

    (a)calculate the velocity of the particle when it is next at 0. [Ans: -24m/s]
    (b)the distance travelled by the particle during the first 3 seconds. [ 23m]

    Q2. A particle P travels in a straight line so that its displacement, x metres, from a fixed point O is given by x = 3t^2 - 4t^3 + 60, where t is the time in seconds measured from the start of the motion. Calculate:

    the average speed of P over the first 2 seconds.[Ans: 10 1/4 m]



    [ Formula to apply; velocity= ds/dt and acceleration = dv/dt]

    Please help me on this, i'm so stressed out with this questions.Worse of all it kept repeating in the papers my teacher gave us,caused me to blank all these questions.....TT^TT so please help me on this

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    vbuggy krtxmrtz's Avatar
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    Re: Add maths: Kinematics

    Quote Originally Posted by yumiryuiko
    Q1.A particle moves in a straight line and its displacement ,x metres, from a fixed point O, t seconds after passing O, is given by x= 12t - t^3.

    (a)calculate the velocity of the particle when it is next at 0. [Ans: -24m/s]
    (b)the distance travelled by the particle during the first 3 seconds. [ 23m]
    To the first question:

    The particle moves in such a way that first x increases, then the velocity inverts its sign and x decreases and goes back past 0. So, at t=0, the particle is at the origin, but after a time t1:

    0 = 12t1 - t13
    Solving:
    t1 = 2Sqrt(3)
    The velocity is:
    v = dx/dt = 12 - 3t2
    and its value at that time is:
    v(t1) = 12- 3t12 = 12 - 3(2Sqrt(3))2 = -24

    The distance travelled during the first 3 seconds is the sum of the distance traveled from t=0 until the velocity inverts its sign plus the distance traveled from that time until t=3. So, first you determine when the velocity inverts its sign, i.e. when it is 0:

    0 = 12 - 2t2 -> t = 2

    Distance traveled from t=0 to t=2 (away from the origin):
    x(+) = x(2) - x(0) = 16

    Distance traveled from t=2 to t=3 (toward the origin):
    x(-) = ABS[x(3) - x(2)] = ABS(9 - 16) = ABS(-7) = 7

    Total = x(+) + x(-) = 16 + 7 = 23
    Last edited by krtxmrtz; May 4th, 2007 at 10:50 AM.
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