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Calculating the final velocity?

  1. Apr 18, 2012 #1
    If an object starts with an initial velocity of 0, and has a velocity of 1.07 m/s after 20 cm, what is the velocity at 1 meter, assuming a constant acceleration? I do not know how long it takes to travel the full 1 meter. All I know is that after 20 cm the velocity was 1.07 m/s and it took 177 miliseconds to travel the 20 cm.
     
  2. jcsd
  3. Apr 18, 2012 #2
    Firstly, recall the general relationship between a constant force and a change in kinetic energy: $$F\cdot\Delta x = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$$

    In this case you have ##v_1=0##, so that comes out to simply $$F\cdot\Delta x = \frac{1}{2}mv_2^2$$

    Now find out what ##F## is by rearranging this to

    $$F = \frac{\frac{1}{2}mv_2^2}{\Delta x}$$

    Plug in any available velocity and corresponding location into the above, and you'll find the constant force.

    Next rearrange to instead solve for the velocity as a function of distance

    $$\sqrt{\frac{2F\cdot\Delta x}{m}} = v_2$$

    And then flip that, because it just looks better flipped:

    $$v_2 = \sqrt{\frac{2F\cdot\Delta x}{m}}$$

    Plug in a value of 1 meter for the distance, and you'll find your velocity at that location.

    Your textbook didn't explain this?
     
  4. Apr 18, 2012 #3

    HallsofIvy

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    Equivalently, if an object moves with constant acceleration a m/s^2, starting with 0 velocity, after t seconds it will have velocity at m/s and will have traveled (1/2)at^2 m.

    It will have traveled .2 m when (1/2)at^2= .2 so t= sqrt(.4/a) seconds. If, at that time, it has velocity 1.07 m/s, we have at= a(sqrt(.4/a))= 1.07. Squaring both sides, a^2(.4/a)= .4a= 1.07^2. Solve that for a.

    Once you know a, use (1/2)at^2= 1 to determine the time the object has traveled to 1 m, then put that time and a into at to find the velocity
     
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