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Thread: Kinematics (Complex)

  1. #1
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    Kinematics (Complex)

    Question:
    A particle moves on the positive x-axis. The particle is moving towards the origin $\displaystyle O$ when it passes through the point $\displaystyle A$, where $\displaystyle x=2a$, with speed $\displaystyle \sqrt{\left(\frac{k}{a}\right)}$, where $\displaystyle k$ is constant. Given that the particle experiences an acceleration $\displaystyle \frac{k}{2x^2} + \frac{k}{4a^2}$ in a direction away from $\displaystyle O$, show that it come instantaneously to rest at a point $\displaystyle B$, where $\displaystyle x=a$. Immediately the particle reaches B the acceleration changes to $\displaystyle \frac{k}{2x^2} - \frac{k}{4a^2}$ in a direction away from $\displaystyle O$. Show that the particle next comes instantaneously to rest at $\displaystyle A$.

    My Problem with this question:
    I have no idea on how to approach it. Please help. Thanks in advance.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Air View Post
    Question:
    A particle moves on the positive x-axis. The particle is moving towards the origin $\displaystyle O$ when it passes through the point $\displaystyle A$, where $\displaystyle x=2a$, with speed $\displaystyle \sqrt{\left(\frac{k}{a}\right)}$, where $\displaystyle k$ is constant. Given that the particle experiences an acceleration $\displaystyle \frac{k}{2x^2} + \frac{k}{4a^2}$ in a direction away from $\displaystyle O$, show that it come instantaneously to rest at a point $\displaystyle B$, where $\displaystyle x=a$. Immediately the particle reaches B the acceleration changes to $\displaystyle \frac{k}{2x^2} - \frac{k}{4a^2}$ in a direction away from $\displaystyle O$. Show that the particle next comes instantaneously to rest at $\displaystyle A$.

    My Problem with this question:
    I have no idea on how to approach it. Please help. Thanks in advance.
    HAve you tried taking the derivative and realizing that there is a max or min when v(t)=0?
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