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Acceleration of the particle at t = 1s

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    The position of a particle as it moves along the x axis is given by x = 15e-2t m, where t is in s. What is the acceleration of the particle at t = 1 s?



    2. Relevant equations
    d = vi.t + 1/2 a ts
    since I don't know how distance is implemented.

    3. The attempt at a solution
    I couldn't do it
    = 15e-2t = 0 + 1/2 a 1 s2
     
  2. jcsd
  3. Jan 29, 2014 #2
    Can't you find a value for x with the equation you have? That should be all you need!
     
    Last edited: Jan 29, 2014
  4. Jan 29, 2014 #3

    collinsmark

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    Per the forum rules, you must at least try and show your effort before help is allowed.

    And at the very least, please put some effort into making it easy to read what the problem actually is. Do you really mean that

    x = 15e - 2t [m]?

    Or do you mean,

    x = 15e-2t [m]?

    Or is it,

    x = 0.15t [m]?

    One is much more likely to get help if one does the due diligence to make the original post easy to read.
     
  5. Jan 29, 2014 #4

    haruspex

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    If you are given the position as a function of time, x = f(t), what operation do you need to perform to find the velocity function? And what to get the acceleration function?
     
  6. Jan 29, 2014 #5
    To find the velocity = delta x/ delta t
     
  7. Jan 29, 2014 #6

    collinsmark

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    What operation does [itex] \frac{\Delta x}{\Delta t}[/itex] become as [itex] \Delta t [/itex] and the corresponding [itex] \Delta x [/itex] become smaller and smaller to the point of being infinitesimally small?

    In other words, if you graph x(t) vs. t (x on the vertical axis and t on the horizontal axis), what is the slope of the curve at any point in time, t?
     
    Last edited: Jan 29, 2014
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