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Mechanics: Accleration as a function of position

  1. Jan 27, 2008 #1
    A particle is moving along a straight line with accelerated motion such that a=-ks, where s is the distance from the starting point and kis the proportionality constant to be determined. For s=2 ft the velocity is 4 ft/s, and for s=3.5 ft, the velocity is 10 ft/s. What is s when the velocity is zero?

    Here is a description of my approach, but I keep running into dead ends. I'd like to verify that the approach is correct before I go further.

    My Approach: Establish the relationship between velocity and acceleration as a differential equation. v dv=a*ds. I know that I'm going to need to define my constant of integration as well as my k value to get anywhere. So, I want to integrate using boundaries, with the v dv integration having boundaries of 4 and V, whereas the ds boundaries will be 2 and s.

    After Isolating for V, the function is:

    v = sqrt(-ks^2-12). My guess in regards to determining K was to plug in the other given S and V values into this equation to find K. However, this does not work, as my K value does not come out to be correct.

    Any clues?
  2. jcsd
  3. Jan 28, 2008 #2

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    Integrate the differential eqn vdv/ds = -ks directly, and you'll get a const of integration, say C. Using the two given conditions, you can find k and C. After that, put v=0 to find the reqd s.
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