Third Equation Of Motion Question

In summary, the third equation of motion, v^2 = v_{0}^2 + 2ax, can be derived by eliminating the variable t from the equations for velocity and position. This equation relates the velocity of an object at two points, the distance between those points, and the average acceleration of the object. It can be used to find any unknown quantity if the other three are known. It is important to practice deriving equations to better understand their meaning and applications.
  • #1
IronBrain
52
0
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
[itex]v^2 = v_{0}^2 + 2ax[/itex]

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

[itex] v = v_0 + at[/itex]

[itex] x = v_{0}(t) + \frac{at^2}{2}[/itex]

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?
 
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  • #2
I would derive that equation from
v = u + at
and
x = (u+v)t/2 (distance = average velocity times time)
eliminate t
 
  • #3
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
OK, then: equation (1):
[tex]v = v_0 + at[/tex]
can be solved for [itex]t[/itex] as
[tex]t = \frac{v - v_0}{a}[/tex]
Then substitute into equation (2):
[tex]x = v_{0}t + \frac{at^2}{2}[/tex]
to get
[tex]x = v_{0}\biggl(\frac{v - v_0}{a}\biggr) + \frac{a}{2}\biggl(\frac{v - v_0}{a}\biggr)^2[/tex]
Expanding those products gives
[tex]x = \frac{v v_0 - v_0^2}{a} + \frac{v^2 - 2v v_0 + v_0^2}{2a}[/tex]
Multiply the first term by 2 on top and bottom to get a common denominator,
[tex]x = \frac{2v v_0 - 2v_0^2}{2a} + \frac{v^2 - 2v v_0 + v_0^2}{2a}[/tex]
add the fractions,
[tex]x = \frac{v^2 - v_0^2}{2a}[/tex]
and rearrange into
[tex]v^2 = v_0^2 + 2ax[/tex]

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
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
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.
 

1. What is the third equation of motion?

The third equation of motion, also known as the kinematic equation, is a mathematical representation of the relationship between an object's displacement, velocity, acceleration, and time. It is expressed as d = v0t + ½at2, where d is displacement, v0 is initial velocity, a is acceleration, and t is time.

2. How is the third equation of motion derived?

The third equation of motion is derived from the basic laws of motion, specifically Newton's second law of motion (F = ma) and the definition of average velocity (v = d/t). By combining these two equations and rearranging terms, the third equation of motion can be derived.

3. What are the units of measurement for the variables in the third equation of motion?

The units of measurement for the variables in the third equation of motion are as follows:

  • Displacement (d): meters (m)
  • Initial velocity (v0): meters per second (m/s)
  • Acceleration (a): meters per second squared (m/s2)
  • Time (t): seconds (s)

4. How is the third equation of motion used in real-world applications?

The third equation of motion is used in a variety of real-world applications, including engineering, physics, and sports. It can be used to calculate the displacement, velocity, or acceleration of an object in motion, as well as to predict the future motion of an object based on its initial conditions.

5. Are there any limitations to the third equation of motion?

The third equation of motion is based on the assumption of constant acceleration, which may not always be the case in real-world scenarios. It also does not take into account external factors such as air resistance, friction, or other forces acting on the object. Therefore, it may not always accurately represent the motion of an object in complex situations.

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