Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook