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Constant acceleration and time

  1. Sep 11, 2008 #1
    Is there a formula that will calculate time when I know the following...?

    Initial velocity = 0 Final velocity = 28.14 feet per second and distance traveled = 3 feet.

    I would like to know the formula and the time it would take to reach a velocity of
    28.14 feet per second if I only have 3 feet to reach that velocity. I would like to
    alter the velocity and re-calculate the time.

    Thanks for any help...I have no physics or calculus background.
  2. jcsd
  3. Sep 11, 2008 #2


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    Welcome to PF!

    Hi Photon713! Welcome to PF! :smile:

    With constant acceleration, the average speed is half the maximum speed:

    vaverage = vfinal/2.

    Then time = distance/vaverage. :smile:
  4. Sep 11, 2008 #3


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    Here's a trick that will help. If an object is moving with a constant acceleration, it "average velocity"- total distance moved, divided by time moved- is equal to the arithmetic average of the beginning and final velocities. In this case, Your average velocity 3/T, where T is the time required and since your final velocity is 28.14 and initial velocity is 0, the average velocity is (28.14+ 0)/2= 14.07 ft/sec. Now you have
    3/T= 14.07 ft/sec so T= 3/14.07 seconds.

    More generally, to go from velocity 0 to V in distance D, you have an average velocity of D/T= V/2 so T= 2D/V. Of course, that must be done at acceleration V/T.
  5. Sep 11, 2008 #4
    Gentlemen...thank you for your responses. It's exactly what I wanted. Appreciate the welcome and the formula. Regards
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