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Acceleration problem.

  1. Jan 24, 2014 #1
    1. The problem statement,
    In 2.0 s, a particle moving with constant acceleration along the x axis goes from x = 10 m to x = 50 m. The velocity at the end of this time interval is 10 m/s. What is the acceleration of the particle?

    all variables and given/known data

    t = 2.0 s, constant acceleration , disttance = 50 - 10 = 40 m, Vf = 10 m/s(I'm not sure if this would be cosidered as fenal velocity or not)


    2. Relevant equations

    Vf = Vo + at

    3. The attempt at a solution
    I don't know the Vo.
    I've tried doing it using distance( d = vi.t + (1/2 . a.(t^2)) formula didn't work.
    Thank you
     
    Last edited: Jan 24, 2014
  2. jcsd
  3. Jan 24, 2014 #2
    The particle did not move a distance of 10m, look at what you did, I think you just made a silly mistake. You are right when you say the final velocity is 10m/s.

    If I say another SUVAT equation:

    s=[(Vf+Vo)/2]*t

    where s is the distance

    Do you think you would be able to calculate Vo? From this you can just use the equation you stated to find the acceleration.
     
    Last edited: Jan 24, 2014
  4. Jan 24, 2014 #3

    BvU

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    You are doing well!
    You have two equations:
    one: x(t) = x0 + v0 * t + 1/2 * a * t^2 with x0 = x(0) = 10 m, given is x(2) = 50
    two: v(t) = v0 + a * t and given is v(2) = 10 m/s

    As you say, v0 is unknown. And a is another unknown. Two equations with two unknowns generally can be solved !
     
  5. Jan 24, 2014 #4

    BvU

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    I missed your response because it came to me as a private message:
    Well, the formulas are there in #3 and the strategy is to interpret them correctly and then solve them.

    That probably isn't helping you enough, so let me try it a bit more elaborate:

    Something is described that takes two seconds. Let's say from t=0 s to t=2 s. This sets the time scale, just as you have done already.

    At t=0 s, (t0) the position on the x-axis is x(0) = x0 = 10 m This is where we start when we start.
    We don't know the speed at this point, so we call it v0. v0 = v(0) .

    At t=2 s, the position on the x-axis is x(2) = 50 m
    We know the speed at this point and time, v(2) = 10 m/s .

    We have expressions for linear motion with constant acceleration:
    1) for the position as a function of time
    x(t) = x0 + v0 * t + 1/2 * a * t^2
    2) for the speed in the x-direction as a function of time
    v(t) = v0 + a * t

    I have used symbols x0 and v0 for the known x0 and the (still) unknown v0

    We fill in what we know at time t = 2 s:
    x(2) = 50 m = 10 m + v0 * 2 s + 1/2 * a * (2 s)^2
    v(2) = 10 m/s = v0 + a * 2 s

    This is two equations with two unknowns. The exercise asks for the acceleration a, so we eliminate v0 from the first equation by using the second equation: The latter gives us
    v0 = 10 m/s - a * 2 s
    which gives us one equation with one unknown:
    50 m = 10 m + (10 m/s - a * 2 s) * 2 s + 1/2 * a * (2 s)^2

    writing it out and bringing all knowns to one side gives us a !
     
  6. Jan 24, 2014 #5

    BvU

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    I notice Doc Al has helped you nicely with the balcony exercise. I'm going to sleep now. You are doing well, keep it up!
     
  7. Jan 25, 2014 #6
    Thank you BvU.
    Both of your help is appreciated.
     
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