1. Limited time only! Sign up for a free 30min personal 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!

Formula for calculating the distance traveled by accelerating object

  1. Nov 14, 2013 #1
    I don't understand what the logic is behind the formula for the distance traveled by an object under the influence of a constant accelerating force, namely [(Vf+Vi)/2].t

    How come it equals the distance?

    I mean, I understand that you can derive the distance based on a constant velocity by solving for d = V.t

    Velocity means that with each elapsed fraction of time, a certain distance is covered, so you naturally add up as many amounts of distance as there are fractions of time.

    But when the body accelerates, the distance per fraction of time always changes. It doesn't make any sense to me that one could still derive the distance despite that.

    My teacher said that since the velocity isn't constant, you take the average, hence the fact that "a" is substituted for "[(Vf+Vi)/2]" in the original formula "v.t equals distance". But by adding the initial velocity (which often equals 0 anyway) and the final one to have an averga, you may have a rough idea of the distance, but this shouldn't be accurate, isn't it?

    To me it's like taking the age of the eldest (16) and the youngest (2) in a group of people where most are aged 10 and then use that outcome (16 years old) as though it represents the age in that group. Clearly, it doesn't.

    You need to sum up all the elements of a group and devide by the amounts of elements to get an average and The velocity of an accelerating object changes every second, every 1/10th of a second, 1/100th of a second, 1/1000th... so it's a group an infinite number of values between which the difference is an infitesimal change. You can't possibly add that up and devide...

    And yet this formula is widely used in physics... I'm no Einstein, so I won't contradict the world, but is the distance obtained with it just a rough approximation or is it accurate after all? How come it is accurate if it is? What's the logic behind it? What's the flaw in my logic?

    I'm a person who absolutely needs to understand where formulas come from to memorize them, just studying them by heart is not an option if I don't want to fail my exam (and I don't), so this is really important to me.

    You can stop reading here and help me with an answer (please) if you don't have much time, but here are some more details that might help you understand my problem a little more easily if you have.

    I think my lack of understanding of this formula may be related to the fact that I have trouble conceptualizing acceleration. I will lay it out here so as to let you glimpse inside my mind when I try to imagine what "acceleration" means:

    To me, an acceleration of, let's say, 3 m/s² means that, with each second passed, the velocity of the body increases with 3 m/s right.

    So if I drop the ball in my hand and it falls in a straight line to the ground with an initial velocity of 0 under the influence of the accelerating force of gravity (which equals 10 m/s² in this exercise) during the first 4 seconds, the following happens:

    After 0 second: distance traveled = 0, no increase in velocity

    In this interval [0s , 1s]: velocity = 0 m/s + infinite amount of infinitesimal changes

    After 1 second: distance traveled = ?m, velocity + 10 m/s

    In this interval [1s , 2s]: velocity = 10 m/s + infinite amount of infinitesimal changes

    After 2 seconds: distance traveled ?m + 10 m, velocity + 10 m/s

    In this interval [2s , 3s]: velocity = 20 m/s + infinite amount of infinitesimal changes

    After 3 seconds: distance traveled = ?m + 20 m , velocity + 10 m/s

    In this interval [3s , 4s]: velocity = 30 m/s + infinite amount of infinitesimal changes

    After 4 seconds: distance traveled = ?m + 30 m, velocity + 10 m/s

    "?" is a number that I can't figure out because acceleration only tells me what happens each second, so I can only calculate starting from the first second, right? I just don't see how one can possibly work with something like that... Maybe I sound really stupid right now, but it's worth succeeding my exam, I guess... :(

    Thanks a lot for reading and (hopefully helping)!
    Last edited: Nov 14, 2013
  2. jcsd
  3. Nov 14, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    The equation: $$d=\frac{1}{2}(v+u)T$$ ... is for displacement, not distance.

    Draw a velocity time graph for the motion.
    The area between the v-t curve and the time axis is the displacement.

    If the velocity starts at some speed u in some direction and does not change in time T, then the area under the graph is ##d=vT## the displacement in that direction - which is the area of a rectangle.

    If the velocity starts at zero and goes to v in time T, then the area is ##d=\frac{1}{2}vT## ... the area of a triangle. The triangle has that formula because it is a rectangle base T and height v that has been cut in half corner-to-corner.

    If you started, instead, at some non-zero speed u in the same direction, then you would go further. The velocity-time graph for this motion is a trapezium.

    The area of a trapezium is the sum of the areas of a rectangle height u and a triangle height (v-u).
    Sketch it out, you'll see.

    But if the initial velocity was speed u in the opposite direction to the final velocity v, then the graph would be two triangles and $$d=\frac{1}{2}(v-u)T$$

    Notice that if ##u=v##, then ##d=0##?
    The displacement is zero - the distance traveled is not zero: the object went away and came back.
    This is why we make a distinction between distance and displacement.

    The area rule works because velocity is the rate of change of displacement.
    i.e. it is the slope of the displacement-time graph.
    Last edited: Nov 14, 2013
  4. Nov 14, 2013 #3
    The equation (Vf+Vi)/2 * t is derived from the kinematics equation s = s0 + vi*t + (1/2)*a*t2. We know that a = Δv/Δt, so we can substitue it into a in the initial kinematics equation. s0 is 0 if we're starting from the origin(let's assume we are) and ti starts from 0) so that gives us

    s = vit + (1/2)*((vf - vi)/t)*t2
    the vf - vi comes from the Δv, which is final - initial

    now we can simplify the equation by multiplying vit by 2/2, which is perfectly valid due to the fact that mathematically we're multiplying a term by one which changes nothing. You can also see that the t's in the 1/2at2 term cancel, leaving one t in the upper portion of the fraction, and upon expansion results in

    s = (2vit + vft - vit)/2
    collect like terms and factor.

    s = ((vi + vf) * t)/2

    Would this also be a valid explanation? I'm not sure, but that's how my physics prof explained it. We rarely use it, as we usually stick with the standard d = vi*t + 1/2*a*t^2 and derive our equations from it. And everywhere were I put s, I actually meant d for displacement
    Last edited: Nov 14, 2013
  5. Nov 14, 2013 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    s is displacement in the "suvat equations" - I just got into the habit of using d and T for displacement and a time period back in high school. OTOH: I learned vi instead of u for initial velocity and I managed to drop that habit due to a loathing of subscripts.

    I prefer geometric proofs/demonstrations - The v-t graph approach removes the need for special equations by expressing the relationships geometrically and paves the way for the calculus approach later, but I realize that many students are uncomfortable with v-t diagrams (and have memorized the equations.)

    The problem I have with the algebraic answer is that it kinda begs the question - since you can express any kinematic equation as a combination of two other kinematic equations, all it says is that the equation works because it is a kinematic equation. How it is a valid kinematic equation is the question. Of course, if the problem is one of discomfort then starting from equations one is comfortable with can help.

    Between us we should have got it :)

    Of course we could be misreading the question and it is actually one of those "why" questions.
  6. Nov 14, 2013 #5
    One of those "why" questions, as it seems to me needs some sophisticated knowledge about integration, which I don't have yet. That comes next semester :)
  7. Nov 14, 2013 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    For constant acceleration - you only need the geometry.
    You know how to find the areas of rectangles and triangles already right?

    Calculus ends up back to that anyway.
    You are doing well for pre-calc btw.
  8. Nov 14, 2013 #7
    For pre-cal? I'm in first year calculus for engineers! I must say I do actually find calculus more interesting than precalculus, as crazy as it sounds.
  9. Nov 14, 2013 #8

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Oh that explains it - you sounded like you were confessing to not having done calculus before - i.e. at secondary school?

    @yhPscis: any of this any use?
  10. Nov 17, 2013 #9
    Thank you guys a lot. I appreciate your help, but this is actually one of those "why" questions... I don't understand the logic behind the area rule either. Not that I don't understand that the diagram has geometrical shapes of which one calculates the area, I do, but I ask myself how come one can be sure that the area equals the displacement?

    You said that it works because velocity is the rate of change of the deplacement, Simon Bridge, but... why does the fact that it's the rate of change lead tot the conclusion that the area rule works?

    I don't mean to sound too philosophical, but I'd like to be able to solve that problem starting from ground zero, without any pre-established tactics, just pure logic... otherwise I won't completely understand the logic behind it.

    Zeus5966, it's funny because my teacher taught me the other way around ( we derived s+vt+(at^2)/2 from (Vi+Vt)/2 ), so I guess this reflects Simon Bridge's comment on justifying one kinematics equation with another kinematics equation pretty well. :p
  11. Nov 17, 2013 #10
    The definition of an integral is basically the net signed area under a curve. So you know that a derivative gives you the rate of change of a function like, the derivative of displacement with respect to time is velocity, and then derivative of velocity with respect to time, and the derivative of velocity with respect to time is acceleration. The integral is the anti derivative so you go backwards, so integral of acceleration = velocity, since you know the definition of an integral you can say that the area under a curve works, kind of a circular argument but that is the best you're going to get without some understanding of calculus.
  12. Nov 17, 2013 #11


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    One should say that a "non-calculus physics course" is more difficult than a calculus-based one. Newton invented calculus to solve mechanical problems and in this way started the whole business of theoretical physics. Of course, using brandnew mathematics was not a good idea to present his physics and that's why his Principia used a quite complicated geometrical line of arguments instead of his new mathematical machinery, which in fact was calculus.

    The best thing you can do, when you are forced to avoid calculus is to give some arguments why the area under the acceleration-time graph is the velocity change of a particle between the two times under consideration. In calculus language is says that the integral of the acceleration with respect to time is the velocity, and this follows from the definition of acceleration as the time derivative of the velocity, because integration is the inverse of differentiation.

    To make this "area rule" clear in a non-calculus way, you argue as in the definition of the (Riemann) integral in calculus, but don't tell your students that in fact you are doing calculus ;-). This argument goes as follows:

    Let the acceleration of the particle be given as a function of time, [itex]a(t)[/itex]. Now you want to get the velocity as a function of time, [itex]t[/itex] given the velocity at the "initial time" [itex]t_0<t[/itex]. Since the acceleration is changing with time, you cannot simply write "velocity = acceleration times time diference + initial velocity", but you can do so for very tiny time steps, [itex]\Delta t=(t-t_0)/N[/itex], where [itex]N[/itex] is some (large) integer number. Then the velocity is given to some accuracy as
    [tex]v(t) \approx v_0 + a(t_0) \Delta t + a(t_0+\Delta t) \Delta t + a(t_0+2 \Delta t) \Delta t + \cdots a[t_0+(N-1) \Delta t] \Delta t.[/tex]
    If you now consider the graph of the function [itex]a(t)[/itex] ([itex]t[/itex] along the horizontal and [itex]a[/itex] along the vertical axis), the above formula tells you that the change of [itex]v[/itex] from its value [itex]v_0[/itex] at [itex]t=t_0[/itex] is approximately given by adding the areas of rectangles, whose widths are given by [itex]\Delta t[/itex] (that's because we have chosen equidistant time intervals) and the height is given by the value of the acceleration at the left end of the time interval. If you draw this, you see that the larger [itex]N[/itex] is made, i.e., the smaller the time steps become, the more accurately you get the area under the curve [itex]a(t)[/itex] and the more accurate your velocity becomes! That's why the area under the curve is the change in velocity.

    Another nice thing is, that this idea is even correct, if [itex]a(t)[/itex] becomes negative at some parts of the time interval under consideration. In the graph this means the [itex]a(t)[/itex] curve gets below the horizontal axis. Then the velocity becomes smaller. This leads to the rule to count such pieces of the area negative, and again you get the correct velocity difference by calculating the area but taking those parts which are below the horizontal axis negative.

    As I said, that's nothing else than integration (in the Riemann sense).

    The same arguments can be used to get the rule that the position change of the particle is given as the area under the velocity-time graph, where again parts below the horizontal axis have to be counted negative, because a negative velocity simply means you move in the opposite direction.

    By the way: It's very important to ask such kinds of "why questions", because to answer such questions is precisely what's science about :-).
  13. Nov 17, 2013 #12

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Put it this way, I can be sure that gravity acts downwards because that is the definition of the word "down". Similarly the velocity is defined in relation to the displacement. That definition means that, when the velocity is plotted against time, the area will give the displacement and when displacement is plotted against time then the slope is the velocity.

    ... Because those are two ways of saying the same thing.
    Perhaps you should look at the displacement time graph for a constant velocity?

    Logic is not the only tool needed to understand the World... You also need stuff like arithmetic and geometry (this is philosophy 101)... However, to get a hold of the logic, you need to understand the definitions.

    Can you see why the slope of the displacement-time graph is the velocity for starter? From the definition of velocity?
    Last edited: Nov 17, 2013
  14. Nov 17, 2013 #13

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    I should probably clear up a bit about the aceleration in relation to velocity and displacement.

    For your example: falling from rest from under 10m/s^2 gravity ... taking positive direction "downwards"-

    By the end of the first 8th of a second, the velocity was 1.25m/s and displacement 0.072m

    By the end of the second 8th of a second, the velocity is 2.50m/s and displacement 0.313m

    If the ball drops for a full five seconds, the velocity and displacement by the end of each 8th pf a second interval is given in the following table:

    Code (Text):

            t           d           v
         0.00000     0.00000     0.00000
         0.12500     0.07812     1.25000
         0.25000     0.31250     2.50000
         0.37500     0.70312     3.75000
         0.50000     1.25000     5.00000
         0.62500     1.95312     6.25000
         0.75000     2.81250     7.50000
         0.87500     3.82812     8.75000
         1.00000     5.00000    10.00000
         1.12500     6.32812    11.25000
         1.25000     7.81250    12.50000
         1.37500     9.45312    13.75000
         1.50000    11.25000    15.00000
         1.62500    13.20312    16.25000
         1.75000    15.31250    17.50000
         1.87500    17.57812    18.75000
         2.00000    20.00000    20.00000
         2.12500    22.57812    21.25000
         2.25000    25.31250    22.50000
         2.37500    28.20312    23.75000
         2.50000    31.25000    25.00000
         2.62500    34.45312    26.25000
         2.75000    37.81250    27.50000
         2.87500    41.32812    28.75000
         3.00000    45.00000    30.00000
         3.12500    48.82812    31.25000
         3.25000    52.81250    32.50000
         3.37500    56.95312    33.75000
         3.50000    61.25000    35.00000
         3.62500    65.70312    36.25000
         3.75000    70.31250    37.50000
         3.87500    75.07812    38.75000
         4.00000    80.00000    40.00000
         4.12500    85.07812    41.25000
         4.25000    90.31250    42.50000
         4.37500    95.70312    43.75000
         4.50000   101.25000    45.00000
         4.62500   106.95312    46.25000
         4.75000   112.81250    47.50000
         4.87500   118.82812    48.75000
         5.00000   125.00000    50.00000
    However - the ball passes through every velocity and displacement in between each entry in the above table. At each instant of time, the velocity and displacement are in the process of changing.
    This is probably why you are having trouble with the concepts - it's the continuous change that keep tripping you up.

    At every instant of time, the velocity is changing - the rate that it is changing at that instant is the instantaneous acceleration.

    Here: ##a=10\text{m/s}^2##
    The velocity and displacement are related to the t axis like: (1) ##v=at##, and (2) ##d=\frac{1}{2}at^2##.
    ... can you see how this is?

    The rest is a matter of math: ##a=v/t## [from (1) since a is a constant], which means that (2) becomes ##d=\frac{1}{2}(v/t)t^2=\frac{1}{2}vt## (3) which is the equation that puzzles you.

    The math here is basically just an exercise in tautology - it is saying that if expressions (1) and (2) are both true, then (3) is also true.

    That's just logic - so it all boils down to if you know why (1) and (2) must be true.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted