How is the Equation for Steady Acceleration Derived?

  • Thread starter AcousticBruce
  • Start date
In summary, the equation s=(\frac{v_i+v_f}{2})t is derived from the equation a=\frac{\Delta v}{\Delta t} and v_f=v_i+at. Substituting in v_i=v_f+at, you get s=v_it+\frac{1}{2}at^2.
  • #1
AcousticBruce
37
0
I am working on a section called "Speeding up steadily". I am wondering how this equation is made.
distance traveled (s)
time elapsed (t)
acceleration (a)

Homework Statement



How far does a car travel if it is going 4 meters per second and accelerates at 3.5 meters per second squared for 5 seconds.


Homework Equations




[tex]s=v_it+ \frac {1}{2}at^2[/tex]


The Attempt at a Solution



[tex]s=(4 m/s)(5.0 s)+ \frac{1}{2}(3.5 m/s^2)(5.0 s)^2[/tex]
s =64 m






How is that equation derived from

[tex]a=\frac{\Delta v}{\Delta t}[/tex]
[tex]\frac{\Delta v_1}{\Delta t_1}=\frac{\Delta v_2}{\Delta t_2}[/tex]
[tex]v_f=v_i+at[/tex]
[tex]s=(\frac{v_i+v_f}{2})t[/tex]
Where does the 1/2 come from?
[tex]v_f^2=v_i^2+2as[/tex]
[tex]s=v_it+ \frac {1}{2}at^2[/tex]

I want to see a clearer picture instead of just remembering an equation.
 
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  • #2
AcousticBruce said:
I am working on a section called "Speeding up steadily". I am wondering how this equation is made.
distance traveled (s)
time elapsed (t)
acceleration (a)

Homework Statement


How far does a car travel if it is going 4 meters per second and accelerates at 3.5 meters per second squared for 5 seconds.

Homework Equations



[tex]s=v_it+ \frac {1}{2}at^2[/tex]

The Attempt at a Solution



[tex]s=(4 m/s)(5.0 s)+ \frac{1}{2}(3.5 m/s^2)(5.0 s)^2[/tex]
s =64 m

How is that equation derived from

[tex]a=\frac{\Delta v}{\Delta t}[/tex]
[tex]\frac{\Delta v_1}{\Delta t_1}=\frac{\Delta v_2}{\Delta t_2}[/tex]
[tex]v_f=v_i+at[/tex]
[tex]s=(\frac{v_i+v_f}{2})t[/tex]
Where does the 1/2 come from?
[tex]v_f^2=v_i^2+2as[/tex]
[tex]s=v_it+ \frac {1}{2}at^2[/tex]

I want to see a clearer picture instead of just remembering an equation.

The equation
[tex]s=(\frac{v_i+v_f}{2})t[/tex]

is an expression of the fact that if acceleration is uniform, the distance traveled is average velocity times time.
Average velocity is (initial + final) divided by two. It's a simple mean value.
It is also equal to the area under the velocity time graph between those two values.

The equation you used to solve the problem is what you get if you substitute
vf = vi + at
in [tex]s=(\frac{v_i+v_f}{2})t[/tex]
 
  • #3
[tex]s=v_it+ \frac {1}{2}at^2[/tex]

This is the one I really don't understand. Is there a visual way to see this?

Why 1/2 acceleration x time squared?



This one makes more sense.

[tex]s=(\frac{v_i + v_f}{2})t[/tex]
 
  • #4
Ok I had to do some basic algebra and I figured it out. Thanks for taking the time.

I never did much algebra II in school so I am also going through a algebra textbook. I started with electronics and it started getting into lots of algebra and trig and physics seemed to be a topic also.

But I am successful because of forums and youtube. Thanks everyone.
 
  • #5
Well I thought I got it... I just do not understand the algebra going from the 2nd to the 3rd equation.


[tex]v_f = v_i + at[/tex]

substitution in this

[tex]s = (\frac{v_i + v_f}{2})t[/tex]

creates this

[tex]s = v_it + \frac{1}{2}at^2[/tex]
 
  • #6
AcousticBruce said:
Well I thought I got it... I just do not understand the algebra going from the 2nd to the 3rd equation.


[tex]v_f = v_i + at[/tex]

substitution in this

[tex]s = (\frac{v_i + v_f}{2})t[/tex]

creates this

[tex]s = v_it + \frac{1}{2}at^2[/tex]

What do you get when you subsitute it in?

[tex](\frac{v_i + v_i + at}{2})t[/tex]

Yes?
 
  • #7
TaxOnFear said:
What do you get when you subsitute it in?

[tex](\frac{v_i + v_i + at}{2})t[/tex]

Yes?

[tex]\frac{v_i}{2} + \frac{v_i}{2} = v_i[/tex] duh! haha
Sheesh... I am factoring and simplifying trinomials right now, perhaps i need to do some fraction algebra problems a bit more. I am doing very well in my algebra trig book so far, but yet I think I need some more basic algebra I practice. I am so close to being very good at algebra. I just tend to miss some things like this and just NOT see it!

Thanks again.
 

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