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Determining initial velocity and instant velocity

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data

    The acceleration of a particle is given by ax(t) = 2.90t - 2.06

    Q. Find the initial velocity v0x such that the particle will have the same x-coordinate at time t = 4.06 as it had at t = 0.

    2. Relevant equations

    vx = v0x + [tex]\int[/tex]axdt evaluated from 0 to t.
    x = x0 + [tex]\int[/tex]vxdt evaluated from 0 to t.

    This is defined for straight-line motion with varying acceleration, which appears to fit my scenario.

    3. The attempt at a solution

    vx = v0x + [tex]\int[/tex](2.90t - 2.06)dt evaluated from 0 to t.
    vx = v0x + 1.45t2 - 1.03t

    x = x0 + [tex]\int[/tex](v0x + 1.45t2 - 1.03t)dt evaluated from 0 to t.
    x = x0 + [tex]\int[/tex](1.45t2 - 1.03t)dt + v0x[tex]\int[/tex]dt

    x = x0 + v0xt + (1.45t3/3) - (1.03t2/2)

    Unfortunately this is as far as I got. I don't understand how to incorporate the values I was given for t. Could someone point me in the right direction please?
  2. jcsd
  3. Jan 22, 2010 #2


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    Homework Helper

    Well from where, when t=4.06, what is x equal to ? (in terms of v0x and x0)

    And when t=0, what is x equal to ? (in terms of v0x and x0)

    Then just put those equal to each other and solve for v0x
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