Unraveling the Mystery of One-Half Coefficient in Kinematic Equations

[Tanahagae]

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.

 

[berkeman]

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?

 

[jmatejka]

Some "Equations" conform to meet the "reality". Nature is not interested in conforming to "nice looking" equations.

 

[Tanahagae]

@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.

 

[WindScars]

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...

 

[berkeman]

@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:

$$v(t) = v_0 + \int a dt = v_0 + at$$

To get position, you integrate velocity:

$$y(t) = y_0 + \int (v_0 + at) dt = y_0 + v_0 t + \frac{1}{2} a t^2$$

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?

 

[Tanahagae]

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.

 

[meichenl]

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

$a = \frac{1}{2}base*height = \frac{1}{2}t*at = \frac{1}{2} at^2$

 

[rcgldr]

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):
v_avg = 1/2 (v₀ + v₁)

distance:
d₁ = d₀ + v_avg 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

 

[meichenl]

"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.

 

[rcgldr]

"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
t_mid = 1/2 (t₀ + t₁)
v_mid = 1/2 (v₀ + v₁)

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

For constant acceleration, the midpoint of the line {t₀, v₀} to {t₁, v₁} occurs at {t_mid, v_mid}. The geometric area under the line {t₀, v₀} to {t₁, v₁}, can be rerranged by moving the triangle shaped area above the horizontal line v_mid to the triangle shaped gap under below the horizontal line v_mid, to create a rectangle of area = v_mid x Δt = distance. Since average velocity = distance / time, then average velocity = v_mid = 1/2 (v₀ + v₁).