# Third Equation Of Motion Question

1. Jan 24, 2010

### IronBrain

I am wondering how is the third equatio of motion derived, I was reading a text on my physics course which it is very unclear how they exactly they arrived to this equation, knowing this from the book, using the equations for velocity and position, you can combine them to get 3 new equations of motion, I know this equations are important to know because I recently got stuck on a question and would have never know to use the third equation to find an correct answer

Third Equation Of Motion
$v^2 = v_{0}^2 + 2ax$

v = final velocity, v_0 = initial velocity, a = acceleration x = position

Said equations from text to be "combined" and with eliminated the variable, t, time to arrive at the velocity equation above

$v = v_0 + at$

$x = v_{0}(t) + \frac{at^2}{2}$

I know there just some simple algebra being used here, but someone who has just general knowledge of the motion of along a straight plane equations of the sort, more specifically, acceleration, velocity, position, going into a physics course for the first time how do you know when/which "extra" equation to use?

2. Jan 24, 2010

### Stonebridge

I would derive that equation from
v = u + at
and
x = (u+v)t/2 (distance = average velocity times time)
eliminate t

3. Jan 24, 2010

### IronBrain

Thanks for the input, I am just a very trivial person, I like to know how things are derived, kind of biting me in the read in my subjects of interest, maths, etc. something are the way they are. I just want to know when/how to derive these things with proper analytically explanation.

4. Jan 24, 2010

### diazona

OK, then: equation (1):
$$v = v_0 + at$$
can be solved for $t$ as
$$t = \frac{v - v_0}{a}$$
Then substitute into equation (2):
$$x = v_{0}t + \frac{at^2}{2}$$
to get
$$x = v_{0}\biggl(\frac{v - v_0}{a}\biggr) + \frac{a}{2}\biggl(\frac{v - v_0}{a}\biggr)^2$$
Expanding those products gives
$$x = \frac{v v_0 - v_0^2}{a} + \frac{v^2 - 2v v_0 + v_0^2}{2a}$$
Multiply the first term by 2 on top and bottom to get a common denominator,
$$x = \frac{2v v_0 - 2v_0^2}{2a} + \frac{v^2 - 2v v_0 + v_0^2}{2a}$$
$$x = \frac{v^2 - v_0^2}{2a}$$
and rearrange into
$$v^2 = v_0^2 + 2ax$$

Any time you're confused about the derivation of an equation like that, try to do it yourself. If necessary, look up how it's done in some reference (like a book), and then once you've seen it, close the book and try to do it yourself. The more you practice, the better you'll get.

5. Jan 24, 2010

### IronBrain

Thanks! The exact derivation I was looking for. Also, thanks for the tips, I tried deriving the equation myself ,and, looking up google for some quick references, but that was a no go just gave the equations themselves.I was unsure of what to substitute, however, is this equation only limited to finding the speed/velocity at some arbitrary position value that is set as the maximum height that has a constant acceleration with disregards to time, or, can this equation if need be manipulated to find other unknown variables?

6. Jan 24, 2010

### diazona

The equation relates four quantities: the velocity of an object at one point, the velocity of the same object at another point, the distance between the two points, and the average acceleration of the object as it travels from one to the other. Any time you have any three of those values, you can use this equation to find the fourth. For example, given the initial and final velocity and the average acceleration, you can find the net distance traveled.

This is true of any equation: when you know all quantities in the equation except one, you can use the equation to find that unknown quantity.

P.S. As far as the derivation: you'll notice that time appears in both of the original equations, but does not appear in the final equation that you were trying to derive. That's a big clue that you should solve one of the equations for time and substitute it into the other.