Unraveling the Mystery of One-Half Coefficient in Kinematic Equations

  • Thread starter Tanahagae
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In summary, the equation for velocity comes from integration and differentiation. It comes from taking a constant acceleration and using that to get the velocity. The equation for position comes from integrating the velocity. The equation for distance is just the area under the velocity vs. time graph.
  • #1
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I have been in Physics for a month now and have been plagued by one simple detail, why in most Kinematic equations is acceleration given the coefficient of one-half? I have been researching lately and have not found any understand reasoning, in most explanations I just see the one-half appear as it had always been there.

Granted we do not have to understand how to derive one equation into another, I cannot have something like this on my conscience. I guess I like to understand rather than follow blindly. If anyone has explanation for the phenomenon I will greatly appreciate it. My math background goes as far as Calculus 1 which stopped at Integrals so hopefully these equations do not require anything beyond that.

Thank you. I love the forums, have been an abundance of help with any questions I come across in my efforts to do homework.
 
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  • #2
Tanahagae said:
I have been in Physics for a month now and have been plagued by one simple detail, why in most Kinematic equations is acceleration given the coefficient of one-half? I have been researching lately and have not found any understand reasoning, in most explanations I just see the one-half appear as it had always been there.

Granted we do not have to understand how to derive one equation into another, I cannot have something like this on my conscience. I guess I like to understand rather than follow blindly. If anyone has explanation for the phenomenon I will greatly appreciate it. My math background goes as far as Calculus 1 which stopped at Integrals so hopefully these equations do not require anything beyond that.

Thank you. I love the forums, have been an abundance of help with any questions I come across in my efforts to do homework.

It comes from integration and differentiation. Are you familiar with basic calculus?
 
  • #3
Some "Equations" conform to meet the "reality". Nature is not interested in conforming to "nice looking" equations.
 
  • #4
@berkeman: Yes, I am familiar with basic Calculus. I know the majority of Physics deals with Calculus and honestly I am upset I am not allowed to take Physics 207 which is a Engineering physics course that revolves around Calculus.

@jmatejka: You are quite right, the equation is not "nice looking." I am more curious at why it is that way. I have a feeling my stay in Physics this semester will be much more enjoyable if I understand rather than memorize.
 
  • #5
Well I probably didn't understand what you want but if I did, it comes from the fact that position = average speed * time, and average speed = (initial_speed + final_speed)/2. As final_speed is acceleration * time, after multiplying you get that a*t/2 in the formula...
Sorry if I said something stupid, anyway...
 
  • #6
Tanahagae said:
@berkeman: Yes, I am familiar with basic Calculus. I know the majority of Physics deals with Calculus and honestly I am upset I am not allowed to take Physics 207 which is a Engineering physics course that revolves around Calculus.

So start with a constant acceleration a.

To get velocity, you intetrate acceleration:

[tex]v(t) = v_0 + \int a dt = v_0 + at[/tex]

To get position, you integrate velocity:

[tex]y(t) = y_0 + \int (v_0 + at) dt = y_0 + v_0 t + \frac{1}{2} a t^2[/tex]

And in the case where a = -g (pointing down in the -y direction), you can see the equations that you normally use for projectile motion.

Does that help?
 
  • #7
Yes, that was basically all I wanted to find out. I had no clue what the first step was, I knew how to get to certain parts but a lot of steps are skipped when people explain things. By taking the integral it certainly makes more sense now.
 
  • #8
If an object starts from rest and accelerates, then a plot of its velocity over time is a line, starting at zero with a slope equal to the acceleration.

The distance traveled is the area under the velocity vs. time graph. That area under the sloped line is a triangle. The base of the triangle is the time, the height is the final velocity, or the acceleration*time, and so the area is

[itex] a = \frac{1}{2}base*height = \frac{1}{2}t*at = \frac{1}{2} at^2[/itex]
 
  • #9
If acceleration is constant, then you can derive the formula using algebra

Δt = t1 - t0
v1 = v0 + a Δt

average velocity (if acceleration is constant):
vavg = 1/2 (v0 + v1)

distance:
d1 = d0 + vavg Δt
d1 = d0 + 1/2 (v0 + v1) Δt
d1 = d0 + 1/2 (v0 + (v0 + a Δt)) Δt
d1 = d0 + 1/2 (2 v0 + a Δt)) Δt
d1 = d0 + v0 Δt + 1/2 a Δt2
 
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  • #10
"If acceleration is constant, then you can derive this using algebra"

To be a derivation starting from the fact of constant acceleration, you'd have to prove the equation about average velocity rather than taking it for granted.
 
  • #11
meichenl said:
"If acceleration is constant, then you can derive this using algebra" To be a derivation starting from the fact of constant acceleration, you'd have to prove the equation about average velocity rather than taking it for granted.
This can be done geometrically using a graph, time on the x-axis, velocity on the y-axis.

define delta t and midpoints:
Δt = t1 - t0
tmid = 1/2 (t0 + t1)
vmid = 1/2 (v0 + v1)

Assume zero acceleration, constant velocity. Draw the graph from {t0, vc} to {t1, vc}. Then area_under_line = vc x Δt = velocity x time = distance.

For constant acceleration, the midpoint of the line {t0, v0} to {t1, v1} occurs at {tmid, vmid}. The geometric area under the line {t0, v0} to {t1, v1}, can be rerranged by moving the triangle shaped area above the horizontal line vmid to the triangle shaped gap under below the horizontal line vmid, to create a rectangle of area = vmid x Δt = distance. Since average velocity = distance / time, then average velocity = vmid = 1/2 (v0 + v1).
 
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What is the one-half coefficient in kinematic equations?

The one-half coefficient, often denoted as 1/2 or 0.5, appears in kinematic equations to account for the average value of a quantity over a given time period. It is used to simplify the calculation of average values in equations that involve changing values over time.

Why is the one-half coefficient used in kinematic equations?

The one-half coefficient is used in kinematic equations because it represents the average value of a quantity over a given time period. This is important in kinematics, as many physical quantities, such as velocity and acceleration, can change over time. By using the one-half coefficient, we can simplify the calculation of average values and make kinematic equations more manageable.

How does the one-half coefficient affect the equations?

The one-half coefficient affects the equations by representing the average value of a quantity over a given time period. In kinematic equations, this coefficient is typically multiplied by the quantity that is changing over time, such as velocity or acceleration. This allows us to calculate average values more easily and accurately.

Is the one-half coefficient always necessary in kinematic equations?

No, the one-half coefficient is not always necessary in kinematic equations. It is typically used when we want to calculate the average value of a quantity over a given time period. If we are only interested in instantaneous values, the one-half coefficient may not be needed.

Can the one-half coefficient be used in other types of equations?

Yes, the one-half coefficient can be used in other types of equations, not just kinematic ones. It is commonly used in equations that involve changing values over time, such as in physics, engineering, and finance. It is a useful tool for simplifying calculations and finding average values in these types of equations.

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