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Math Help - Kinematics (Deriving Equation)

  1. #1
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    Kinematics (Deriving Equation)

    Question:
    A particle starts with speed 20 \ ms^{-1} and moves in a straight line. The particle is subject to a retardation which is initially 5 \ ms^{-2} and which increases uniformly with the distance moved, having a value of 11 \ ms^{-2} when the particle has moved 12 \ m. Given that the particle has speed v \ ms^{-1} and has moved a distance of x, show that, while the particle is in motion

     v  \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right).


    My Problem with this question:
    I can't derive it. Help please? Thanks in advance.
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    Quote Originally Posted by Air View Post
    Question:
    A particle starts with speed 20 \ ms^{-1} and moves in a straight line. The particle is subject to a retardation which is initially 5 \ ms^{-2} and which increases uniformly with the distance moved, having a value of 11 \ ms^{-2} when the particle has moved 12 \ m. Given that the particle has speed v \ ms^{-1} and has moved a distance of x, show that, while the particle is in motion

     v  \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right).


    My Problem with this question:
    I can't derive it. Help please? Thanks in advance.
    the retardation force is initially -5m N and is increasing uniformly with distance.

    so the retardation force is of the form -(5+kx)m. Your given when x = 12 and the force has magnitude -11m

    Finding k is pretty simple. then you have m \ddot{x} = -(5+kx)m

    the rest follows.

    Bobak
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    Quote Originally Posted by Air View Post
    Question:
    A particle starts with speed 20 \ ms^{-1} and moves in a straight line. The particle is subject to a retardation which is initially 5 \ ms^{-2} and which increases uniformly with the distance moved, having a value of 11 \ ms^{-2} when the particle has moved 12 \ m. Given that the particle has speed v \ ms^{-1} and has moved a distance of x, show that, while the particle is in motion

     v  \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right).


    My Problem with this question:
    I can't derive it. Help please? Thanks in advance.
    We know the speed decreases uniformly with the distance, so
    \frac{da}{dx} = \text{constant} = \frac{-11~m/s^2 - -5~m/s^2}{12~m} = -\frac{1}{2}~s^{-2}

    Thus
    a = a_0 + \frac{da}{dx} x

    a = -5 - \frac{1}{2}x = - \left ( 5 + \frac{1}{2}x \right )

    Now for a little variable work:
    a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}
    so we are done.

    -Dan
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    ^Thanks.

    Also, a follow on question is:
    Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

    I worked it out to get two answers, -40m and 20m. I accepted 20m which is correct but realised that they are asking for distance so doesn't that mean that it could also be |-40|m = 40m?
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    Quote Originally Posted by Air View Post
    ^Thanks.

    Also, a follow on question is:
    Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

    I worked it out to get two answers, -40m and 20m. I accepted 20m which is correct but realised that they are asking for distance so doesn't that mean that it could also be |-40|m = 40m?
    So you found the value of x when v = 0?

    While the particle continues to move in the same direction (slowing down of course) the value of x when v = 0 will be the distance travelled (assuming the initial condition x = 0 at t = 0). So x = -40 is a spurious solution.
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    Quote Originally Posted by mr fantastic View Post
    So you found the value of x when v = 0?

    While the particle continues to move in the same direction (slowing down of course) the value of x when v = 0 will be the distance travelled (assuming the initial condition x = 0 at t = 0). So x = -40 is a spurious solution.
    Could it have been that it travelled -40m then +20m (displacement) so the displacement is -20m. But as they are asking for distance, it would be the modulus of |-20m| hence the answer being 20m?
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  7. #7
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    Quote Originally Posted by Air View Post
    Could it have been that it travelled -40m then +20m (displacement) so the displacement is -20m. But as they are asking for distance, it would be the modulus of |-20m| hence the answer being 20m?
    No.

    I'll take moving right as the positive direction and moving left as the negative direction. There are two cases to consider: v = 20 m/s (the particle moves to the right with an intial speed of 20 m/s) and v = -20 m/s (the particle moves to the left with an initial speed of 20 m/s). You have v^2 = 400 - 10x - \frac{x^2}{2}.

    1. v = 20 m/s. Then v = + \sqrt{400 - 10x - \frac{x^2}{2}}, x > 0.

    Draw a graph of v versus x. Note that it starts at v = 20 and x = 0 (corresponding to t = 0) and x > 0. It's very clear that when v = 0, x = 20. That is, the distance travelled is 20 m.

    2. v = -20 m/s. Then v = - \sqrt{400 - 10x - \frac{x^2}{2}}, x < 0.

    Draw a graph of v versus x. Note that it starts at v = -20 and x = 0 (corresponding to t = 0) and x < 0. It's very clear that when v = 0, x = -40. That is, the distance travelled is 40 m.


    The origin of your two solutions should now be clear. Clearly the book has assumed v = 20 m/s .......


    By the way, the graph of v^2 = 400 - 10x - \frac{x^2}{2} is an ellipse.
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