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

Distance formula and addition of meters per s

  1. Jan 3, 2010 #1
    Hello guys.

    I have issues understanding a rather simple physics concept and I hope you can help me out. I've tried asking help from several people, but nobody was able to isolate the problem.
    For constant acceleration, it is said that d=(Vi+Vf)/2*t. d being distance, Vi is initial velocity, Vf is final velocity and t is time of the travel.

    The problem occurs when I try to add - what seems intuitively for me atleast - the distance of object traveled each second with the added velocity from the constant acceleration.

    An example, object starts at 10m/s and after 10 seconds reach a velocity of 20m/s.
    150 = (10+20)/2*10, and (20-10)/10 = 1, so acceleration is 1m/s/s.

    Now when I try to add the distance traveled of each second over a duration of 10 seconds, I get 10+11+12+13+14+15+16+17+18+19 = 145. I don't add 20 because to my understanding the object doesn't actually travel any seconds with 20m/s, it merely reaches that velocity after going a full second at 19m/s.

    So I can't the results to match eachother, only if I do something like (10+19)/2*10.
     
  2. jcsd
  3. Jan 3, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Use the same formula for each 1-second interval that you used for the 10-second interval:
    In the first second, d = (10 + 11)/2 * 1 = 10.5 (not 10). And so on, for each interval.
     
  4. Jan 3, 2010 #3
    I see. So just to pinpoint where my intuition is wrong; that by adding the way I did, I assumed the acceleration only occurs after each second. It's more of a constant "push" toward the object, that happens during the seconds itself.

    Thanks for the help.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook