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==Footnote to Velocity Calculation==
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==Footnotes==
 
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Revision as of 15:45, 7 March 2017

Formula for a Final Velocity for a Constant Acceleration[1]

"Math is dank" - Issac Pascal[2]

There are situations where you are interested in finding the final velocity, but velocity is not a constant number. There is an algebraic equation to find this final velocity without involving calculus.

What you do need to know is:

  1. Your initial velocity
  2. Your acceleration(must be constant)
  3. Your change in position


Step 1: Time

Consider the following equation:
<math>V_t = V_o + at</math>

Using basic algebra, t can be solved as
<math> t = \frac{V_t-V_o}{a}</math>

Step 2: Average Velocity

Also consider what the definition of "average velocity" is: your chance in distance over your change in time
<math>\bar{V} = \frac{\Delta{x}}{\Delta{t}}</math>
The change in distance is also known as velocity, making average velocity

A previous equation we learned for x(t) was
<math>x(t)= x_0+v_0t+\frac{1}{2}at^2</math>
If we consider that <math>\Delta{x} = x(t_f) - x(t_i)</math> then we can plug those into the above equation and find the following:
<math>\Delta{x} = V_o(t_f-t_i)+\frac{1}{2}a(t_f^2-t_i^2)</math>
Do one step of factoring and we get:
<math>\Delta{x} = V_o(t_f-t_i)+\frac{1}{2}a(t_f+t_i)(t_f-t_i)</math>
Now consider that <math>\Delta{t} = t_f-t_i</math> and we can go back to the original equation for average velocity.

<math>\frac{\Delta{x}}{\Delta{t}} = \frac{V_o(t_f-t_i)+\frac{1}{2}a(t_f+t_i)(t_f-t_i)}{t_f-t_i}</math>
Once you simplify by canceling the bottom with the top, you are left with
<math>\frac{\Delta{x}}{\Delta{t}} = V_o+\frac{1}{2}a(t_f+t_i)</math>
Now let us consider just one part of the equation for a moment.
<math>a(t_f+t_i)</math>
We can transform this information into something more usable knowing that <math>V(t) = V_o+at</math>
Note that because a distributes to two different times, we will do the next operation twice.
<math>V(t_f) = V_o + at_f </math>
<math> V_f - V_o = at_f</math>
Repeat for <math>V_i</math> and get
<math>V_i-V_o = at_i</math>
So, finally,
<math> a(t_f+t_i) = at_f+at_i = V_f+V_i - 2V_o</math>
Plugging this back in above we get
<math>\frac{\Delta{x}}{\Delta{t}} = V_o+\frac{1}{2}( V_f+V_i - 2V_o)</math>
And as the last step, simply by removing the <math>V_o</math>s and end up with
<math>\frac{\Delta{x}}{\Delta{t}} = \bar{V} = \frac{V_f+V_i}{2}</math>

Step 3: Putting it all together

Taking another look at average velocity,
<math>\bar{V} = \frac{\Delta{x}}{\Delta{t}}</math>

<math>\bar{V} = \frac{x_f-x_i}{t_f-t_i}</math>

Solving for final x we get
<math>x_f =\bar{V}(t_f-t_i)+x_i</math>

Now for the fun part! Using the equations for t and for <math>\bar{V}</math> we found in step one and two, we plug those into this equation.

<math>x_f = x_i + (\frac{V_f+V_i}{2})(\frac{V_f-V_o}{a}-\frac{V_i-V_o}{a})</math>

<math>x_f= x_i+(\frac{V_f+V_i}{2})(\frac{V_f-V_i}{a})</math>

<math>x_f = x_i + \frac{V_f^2-V_i^2}{2a}</math>

<math>x_f-x_i = \frac{V_f^2-V_i^2}{2a}</math>

<math>2a(x_f-x_i) = V_f^2-V_i^2</math>

<math>V_f^2 = 2a(x_f-x_i)+V_i^2</math>

This is not the most fun to derive, but the equation ends up being useful when you have a linear type of velocity.


Footnotes

Projects

  1. Deriving the final velocity equation

Sorry but I couldn't resist vandalizing this page--Guy vandegrift (talk)

P͏̛́͏͎̗͙͡ͅẖ̨̖̩̱̟̣̮̪͎̺͉̲̫͜͜͜ͅy̡҉̶̸̥͚̞͉̗̯͈͟s̷̶̡̥̘̲͇͙̭̝̱͚͙̻͘͠i̷̛͚͕͕̘̩͓̮̲̕͜ͅͅc̵̨̤̼̫͈͚̪̪̞̥̦̯̤̼͉̟̖͚̪̀ͅṣ͇̺͚̼͜͠ ҉̶̥͍̭̤̟i̶͠͏̜̠̟̞̹̠̮͈̺̻̫̭͚̟͟ş̡̝̱̫͔͡͝ ̶̡̯̻͖͓̺̳͘͜a̵̢̧̛͉̩̺͈̤̖̬͖̯͚͠ ̴̵͍̯̱͚͎͇̟̯͡͝͠g̶͘͡͏͉͚̭̫͉̝͟ͅŗ̶̝̲͚̼̼̭̩̗̞̙͠e̶̯̺͓̫͈̜̻̩̕à̶̳̯̦͇̻̤͍͇̹̲t̛̀͘͠͏̱̯̥͓̰̘͕̯̭̟̘̬̼̝͈̩̻̣̲ ̷͞҉҉̤̳̹̳͉̫͉͕̻̹̺̲̰ͅc̷̨͡͏̤͍͖̜l̨͇͍̝̥͓̩̼͇̗̪͉͚̲a̸̕͏̶̶̫̳͔͎͍̦͈s̛̛̝̫͖̗̝̳̭̕͟s͢͟͟҉͓̮̺̭̺͍͍̳͕

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Spring 2017 PHY2400

I am Tom

Notes

  1. Notes 1/11
  2. Notes 1/17

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