1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to interpret the equation of position with constant acceleration?

  1. May 14, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    x = x0 + v0 * t + 1/2 * constant acceleration * t^2

    So this is supposed to be very very simple physics, but I still feel like there's a part of this equation I don't fully understand. The first term is the initial position of the body at t=0. The second term is the initial velocity at t=0, multiplied by the time elapsed since then. So far so good. Now, it's the third term I have trouble interpreting. I know it comes from integrating the acceleration function, but why exactly is it so?

    I tried to reason it the following way: Let's assume a=2 m/s^2, and for the sake of simplicity make both the initial position and velocity zero. Ok, so if I want to know the position of this body at t=1, then given that constant acceleration, velocity at t=1 will be 2 m/s. So the body should have had a displacement of 2 metres, right?

    But if I plug in the values in the formula, I'd get x = 1/2 * 2 m/s^2 * 1 s^2 = 1 metre. Why is this the case?

    I know it must be pretty stupid, but I can't figure out what's wrong with my reasoning.
    Thanks in advance!
     
  2. jcsd
  3. May 14, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Right.

    Wrong. The velocity at t = 1 would be 2 m/s, but it starts out with v = 0. So the average speed over that first second is only 1 m/s.
     
  4. May 14, 2016 #3
    I see. But what happens when t=2, for example? Because then the velocity at time 2 would be 4 m/s (the average over those two seconds will be 2 m/s), while the total displacement is 8 metres.
     
  5. May 14, 2016 #4
    Acceleration is change of velocity with respect to time. In case of constant acceleration ##a##, it is simply ##a = \frac{ v_{final}-v_{initial}}{t_{final}-t_{initial}} \Rightarrow v_{final} =v_{initial} + a (t_{final}-t_{initial})##. If ##v_{initial} =0, t_{initial} = 0## and ##t_{final} = t##, then ##v_{final} = at##. But calculation displacement is a challenge here. Because here velocity is not constant. It varies with time (##v=at##). So you cannot write that displacement=##v_{final}t = at^2## or something like this. To counter this problem you have to assume that within every sufficiently small interval of time ##\Delta t##, the velocity remains constant. So within an interval ##\Delta t## displacement will be ##v\Delta t = at\Delta t##. Now if you add all the displacements at all intervals you will get total displacement which is ##\Sigma at \Delta t##. This is nothing but integration and the result is ##\frac{1}{2} at^2##.
     
  6. May 14, 2016 #5

    rcgldr

    User Avatar
    Homework Helper

    For constant acceleration, v = v0 + a t, and the average velocity at time t is ( (v0) + (v0 + a t) ) / 2 = v0 + 1/2 a t. Then the distance traveled is v0 t + 1/a 2 t^2 .
     
  7. May 14, 2016 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you plot a velocity vs. time graph you get a straight line with slope ##a##, the acceleration. The distance moved from times ##0## to ##t## is the area under the ##v## vs. ##t## graph between times ##0## and ##t## (do you see why?) So, the distance travelled between times ##0## and ##t## is the area of a triangle with base ##t## and height ##at##.
     
  8. May 14, 2016 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I think you forgot the factor if ½ .

    (½)(2)(22) = ½⋅2⋅4 = 4, not 8 .
     
  9. May 14, 2016 #8

    Doc Al

    User Avatar

    Staff: Mentor

    At t = 2 the velocity will be 4 m/s. For those first 2 seconds the velocity went from zero to 4, for an average speed of 2 m/s. So the distance traveled would be 4 m, not 8.
     
  10. May 14, 2016 #9
    I think this is what I was looking for. So the term ##\frac{1}{2} at^2## is really the area of a triangle given by the change in velocity with respect to the initial velocity.

    Relating it to Doc Al's answer above (please correct me if I'm wrong), another way to look at it would be as the area of half the rectangle given by base ##t## and height ##at## (which would be the average velocity over that time interval, times the interval). Right?

    Right, my bad. It's just I was trying to make sense of that 1/2 factor. Though I think I got it now with Ray's answer.

    Thanks to all of you guys!
     
  11. May 14, 2016 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Right. And that half rectangle is the triangle that Ray was talking about.
     
  12. May 14, 2016 #11
    Exactly. Thanks for the help, I really appreciate it :smile:
     
  13. May 14, 2016 #12
    Since a picture is worth a thousand words.

    yP9R4ez.png
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How to interpret the equation of position with constant acceleration?
Loading...