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Fee fall question

  1. Jun 6, 2010 #1
    Here are the tabulated t = time (sec), v = velocity (m/s) = g(t), and d = displacement (m)=4.9(t)^2 for an object if dropped (free fall) from rest position.

    When t=0, v=0, d=0
    When t=1, v=9.8, d=4.9
    When t=2, v=19.6, d= 19.6
    When t=3, v=29.4, d= 44.1
    When t= 4, v=39.2, d=78.4
    When t=5, v=49, d=122.5

    When t= 3600, v= 35280, d=63504000

    Question:

    1- At the end of first second; velocity of a free fall object is 9.8 m/s while its displacement is 4.9m. Why its displacement less than 9.8 m?

    2- Similarly, displacement w.r.t velocity is increases after 2nd second (time) – not identical. Because 9.8 m/s means an object cover a distance of 9.8 m in one second.

    3- Is it wrong to say that there exiting an initial acceleration (POTENTIAL) of 9.8 m/s/s if an object dropped from rest (free fall) – just like a mass rest on ground?
     
  2. jcsd
  3. Jun 7, 2010 #2
    1)
    9.8 m/s is its instantaneous velocity at the end of one second. At 0 seconds its velocity was 0 and at 0.5 seconds its velocity was 4.9 m/s. This means that it was not traveling at 9.8 m/s for a whole second, and its average velocity over the first second was 4.9 m/s, and so it went 4.9 m in that first second.
    2)
    Same idea
    3)
    I apologize but I am having trouble understanding the question.
     
  4. Jun 7, 2010 #3

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    could you elaborate on how this data was tabulated?
    since you mention free fall, no other forces except for gravity should be acting on the object
     
  5. Jun 7, 2010 #4
    Your questions 1 and 2 are fully answered with calculus. The function [tex]v(t) = -9.8m/s^2 \cdot t[/tex] represents the velocity of the freefalling object at a given time [tex]t[/tex].

    Integrating the equation, we get
    [tex]d(t) = \frac{-9.8m/s^2}{2} t^2 \,,[/tex]
    which describes the function of displacement. In other words, the area under the graph of the equation [tex]v(t)[/tex] represents the displacement at a given time. In this graph, the instantaneous velocity at 1 second is -9.8 m/s, but the shaded area from t=0 to t=1 is only half that: -4.9 m. Intentionally, the average velocity is also -4.9 m/s.

    [PLAIN]http://dl.dropbox.com/u/977994/PhysicsForums/freefall.png [Broken]

    Sorry about the overkill, Zarmewa. I'm just having a little fun with TeX and Mathematica tonight. ;)
     
    Last edited by a moderator: May 4, 2017
  6. Jun 9, 2010 #5
    So you mean 9.8 m/s is instantaneous velocity and 4.9 m/s is the average velocity= displacement/ time elapsed = (xf – xi)/ (tf – ti) = (4.9 – 0)/(1 -0) =4.9 m/s and went on 4.9 m in first second.

    Is this also true for time between 0 to 5 sec; ti =0, tf =5 sec, v= 49 m/sec and xi =0 and xf=d=122.5m - its average velocity= (122.5 – 0)/ (5 -0) = 24.5 m/sec so went on 24.5 m in five second. So which displacement is correct for five second; 122.5 m or 24.5m?

    Here are the instantaneous velocities for the duration of one second (one tenth of second)
    When

    t=0 sec, v=0 m/sec , d=0 m
    t=0.1, v= 0.98, d=0.049,
    t=0.2, v=1.96, d=0.196
    t=0.3, v=2.94, d=0.441
    t=0.4, v=3.92, d=0.784
    t=0.5, v=4.9, d=1.225
    t=0.6, v=5.88, d=1.764
    t=0.7, v=6.86, d=2.401
    t=0.8, v=7.84, d=3.136
    t=0.9, v=8.82, d=3.969
    t=1.0, v=9.8, d= 4.9

    Below are the average velocities = displacement/ time elapsed = (xf – xi)/ (tf – ti)

    1- Between 1 and 0 sec : where =xf= 4.9, xi= 0, tf=1.0, ti= 0 hence Av velocity = 4.9 m/s and total displacement occur in one second =4.9 m

    2- Between 1 and 0.1 sec : where =xf= 4.9, xi= 0.049, tf=1.0, ti= 0.1 hence Av velocity = 5.39 m/s and total displacement occur in 0.9 second =5.39 m

    3- Between 1 and 0.2 sec : where =xf= 4.9, xi= 0.196, tf=1.0, ti= 0.2 hence Av velocity = 5.88 m/s and total displacement occur in 0.8 second =5.88 m

    4- Between 1 and 0.3 sec : where =xf= 4.9, xi= 0.441, tf=1.0, ti= 0.3 hence Av velocity = 6.37 m/s and total displacement occur in 0.7 second =6.37 m

    5- Between 1 and 0.4 sec : where =xf= 4.9, xi= 0.784, tf=1.0, ti= 0.4 hence Av velocity = 6.86 m/s and total displacement occur in 0.6 second =6.86 m

    6- Between 1 and 0.5 sec : where =xf= 4.9, xi= 1.225, tf=1.0, ti= 0.5 hence Av velocity = 7.35 m/s and total displacement occur in 0.5 second =7.35 m

    7- Between 1 and 0.6 sec : where =xf= 4.9, xi= 1.764, tf=1.0, ti= 0.6 hence Av velocity = 7.84 m/s and total displacement occur in 0.4 second =7.84 m

    8- Between 1 and 0.7 sec : where =xf= 4.9, xi= 2.401, tf=1.0, ti= 0.7 hence Av velocity = 8.33 m/s and total displacement occur in 0.3 second =8.33 m

    9- Between 1 and 0.8 sec : where =xf= 4.9, xi= 3.136, tf=1.0, ti= 0.8 hence Av velocity = 8.82 m/s and total displacement occur in 0.2 second =8.82 m

    10- Between 1 and 0.9 sec : where =xf= 4.9, xi= 3.969, tf=1.0, ti= 0.9 hence Av velocity = 9.31 m/s and total displacement occur in 0.1 second =9.31 m

    So are aforementioned calculation are wrong for average velocities and actual displacement?

    As we know velocity at t=0.1 sec is 0.98 m/s. Assume it is constant for one second duration (0.98 +0.98 +0.98 +0.98+0.98 +0.98+0.98 +0.98+0.98 +0.98 =9.8 m/s). Thus an object cover a distance of 9.8 m in one second while in falling mass case 0.98 is not constant but increases at every instant so should it cover more than 9.8 m in one second or 4.9 m is still OK.

    FOR Q3: pls click on the following link.

    http://regentsprep.org/regents/physics/phys01/accgravi/index.htm

    There are two masses rest on the wooden plank before falling. Thus both have weight w = mg where g = acceleration due to gravity. Should their initial acceleration due to gravity (which is potential) be considered in their free falling?
     
    Last edited by a moderator: Apr 25, 2017
  7. Jun 11, 2010 #6
    The equation for the distance an object travels at a constant acceleration is given by multiplying the average speed by the time traveled:

    [tex]d = \frac{V_i + V_f}{2} \cdot t[/tex]

    where Vi is the initial velocity (in this case, 0 m/s) and Vf is the final velocity (which, after 1 second is 9.8 m/s since the object gains 9.8 m/s per second)

    Therefore you have:

    [tex]d = \frac{0 + 9.8}{2} \cdot 1= 4.9[/tex]
     
    Last edited: Jun 12, 2010
  8. Jun 13, 2010 #7
    Thank you so much.

    D 0= (0 + 0.98) x 0.05= 0.049 meters
    D 1= (0.98 + 1.96) x 0.05= 0.147 meters
    D 2= (1.96 + 2.94) x 0.05= 0.245 meters
    D 3= (2.94 + 3.92) x 0.05= 0.343 meters
    D 4= (3.92 + 4.9) x 0.05= 0.441 meters
    D 5= (4.9 + 5.88) x 0.05= 0.539 meters
    D 6= (5.88 + 6.86) x 0.05= 0.637 meters
    D 7= (6.86 + 7.84) x 0.05= 0.735 meters
    D 8= (7.84 + 8.82) x 0.05= 0.833 meters
    D 9= (8.82 + 9.8) x 0.05= 0.931 meters

    Total = 0.049+0.147+0.245+0.343+0.441+0.539+0.637+0.735+0.833+0.931 = 4.9 meters
    Hence proved
     
  9. Jun 13, 2010 #8
    Not proved, but close enough for your purposes. Mathematics is the only branch of science that is not considered proven using experimental means.

    For a bonus, imagine you have an object accelerating at a changing rate. The velocity could be expressed as, say [tex]\inline{V(t) = A\sin(\omega t)}[/tex]. By calculating the acceleration at each interval and summing the results, you can come to an approximation of the actual answer. To show how this is done, let's let A = 1, [tex]\inline{\omega = 2 \pi}[/tex], and [tex]\inline{\Delta t = 0.1}[/tex]. The interval of calculation will be [0, 0.5].

    sin(2 pi 0) = 0
    sin(2 pi 0.1) = 0.587785...
    sin(2 pi 0.2) = 0.951057...
    sin(2 pi 0.3) = 0.951057...
    sin(2 pi 0.4) = 0.587785...
    sin(2 pi 0.5) = 0

    If we total the results, we obtain roughly 3.07768. Since we incremented by [tex]\inline{\Delta t}[/tex] each time, we must multiply the total by 0.1. (The same could be done for each individual calculation, but to solve the problem quicker we can remember that [tex]\inline{ax + bx \equiv (a + b)x}[/tex].) So our final answer is approximately 0.307768.

    However, there is a quicker and exact way of solving this problem. Using knowledge of calculus, you can rewrite the problem with an integral.

    [tex]\int_0^{0.5} A\sin(\omega t) \,dt[/tex]

    Solving this, you get [tex]\frac{-A \cos(\omega t_f)}{\omega} - \frac{-A \cos(\omega t_i)}{\omega}[/tex], which evaluated at fi = 0, tf = 0.5, and the values of the previously defined constants, simplifies to [tex]\inline{1 / \pi}[/tex] or 0.318310...

    As you can see, the exact answer is slightly different than the previous approximation. I encourage you to repeat the approximation with a smaller value of [tex]\inline{\Delta t}[/tex], as you may find the answer converges to [tex]\inline{1 / \pi}[/tex].
     
  10. Jun 13, 2010 #9
    Sorry it’s me again but I recant because aforesaid equation distracted me from my original question. You will notice how, in the following instantaneous velocities between 0.9 and 1.0 seconds.

    t=0.9, v=8.82, d=3.969

    t=0.91, v= 8.91, d=4.05769,
    t=0.92, v=9.016, d=4.417
    t=0.93, v=9.114, d=4.557
    t=0.94, v=9.212, d=4.32964
    t=0.95, v=9.31, d=4.655
    t=0.96, v=9.408, d=4.51584
    t=0.97, v=9.506, d=4.61041
    t=0.98, v=9.604, d=4.70596
    t=0.99, v=9.7, d=4.80249

    t=1.0, v=9.8, d= 4.9

    As you can see instantaneous velocities are increasing between the duration of 0.9 and 1.0 second therefore how it possible for dispalcement between 0.9 and 1.0 second which = 0.931 m.

    Common sense:

    Let an object travel with the constant velocity of 8.82 m/s (viz vi at 0.9 second) between 0.9 and 1.0 second. The final average value = (8.82 +8.82+8.82 +8.82+8.82 +8.82+8.82 +8.82+8.82 +8.82)/10 = 8.82 m/s and thus cover a distance of 88.2 m in the said duration.

    Just imagine if an object is travelling with initial velocity of 8.82 m/s and then increases its velocity then how come dispalcment between 0.9 and 1.0 second duration = (8.82+9.8)0.1/2 = 0.931 m. I hope I have explained things clearly enough.

    same idea for the rest of any intervals
     
    Last edited: Jun 13, 2010
  11. Jun 13, 2010 #10
    I'm not sure I understand the question you are asking. If you restate it, I would gladly help you.
    Also, do you have knowledge of summation [tex]\inline{\sum}[/tex] or calculus?
     
  12. Jun 14, 2010 #11
    Yes, it’s not clear. I flub with displacement, average velocity and tenth of second because I was tired and edit that reply many times. I reply you asap but if it is clear then I will repent. Thanks
     
  13. Jun 14, 2010 #12
    Ah, okay. :) But to quickly answer your first question in your first post, take a look at the graph I posted a few messages back. As your initial model suggests, the object is accelerating 9.8m/s^2 downward. This means that for every second, the velocity is being deducted by 9.8m/s. You can generalize that after t seconds, the velocity is -9.8m/s t. This concept is visible in the graph.

    The point I wanted to show, however, is the area of the graph. In a specific interval (ie. 0s to 1s), the total displacement will equal the area of the graphed function. You can either calculate this using calculus or the formula for area of a triangle.
     
  14. Jun 16, 2010 #13
    It’s all ok. I messed up with calculating displacements from average velocities by not multiplying with their respective times (d= v x t); I thank you and all others. Sorry for not responding promptly.
     
  15. Sep 16, 2010 #14
    It's me after a long hiatus.

    Am I wrong to flip-flop over the aforementioned equation [v=4.9*t^2] in which MEETING point of the instantaneous velocity "v" and time "t" is missing?

    As "v" is directly proportional to "t" therefore "v" increases when "t" increases. Since both "v" and "t" start from zero therefore there must be a point where "v" meets and overtakes "t" but "v" always ahead of "t" at 4.9 fold "t" and this an axiomatic truth.

    Similarly for any velocity [length/time] to occur, distance must be travelled with chugging speed in some duration but an object here in is case is under the influence of constant acceleration due to gravity "g" therefore should the instantaneous velocity be replaced by "potential instantaneous*velocity"? * *
     
  16. Sep 18, 2010 #15
    1. The body is not moving with constant velocity of 9.8 m/s. It is accelerating with that rate and that means it's velocity "is moving" (increasing) with that rate. You can see, after every second velocity increases by 9.8.

    2. In first (data) initial velocity was 0 and velocity was increasing with a magnitude of 9.8 every second. In second initial velocity was 9.8m/s and again it was increasing with a magnitude of 9.8 every second. so what depends is the initial velocity. More the initial velocity more will be the displacement.

    3. yes, 9.8 m/s/s will always act. Even when the body rests on the ground but since the ground exerts a normal reaction force of same magnitude with which earth pulls the object toward its center so the resultant force and acceleration is zero.
     
  17. Sep 18, 2010 #16
    Sorry, i messed up in my previous post therefore I want to add the following correction.*

    Instantaneous velocity = g *t =9.8 * t

    Both v and t do not meet but I meant*should the*magnitude of both v and t become equal after some time at the very inception of time, like 1 micro meter/ micro second.

    Sorry about that*
     
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