Understanding Equations

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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
berkeman
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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.
It comes from integration and differentiation. Are you familiar with basic calculus?
 
  • #3
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Some "Equations" conform to meet the "reality". Nature is not interested in conforming to "nice looking" equations.
 
  • #4
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@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
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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
berkeman
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@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
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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
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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
rcgldr
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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
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"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
rcgldr
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"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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