Motion
- Page ID
- 238
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)1-Dimensional Motion
Consider a car moving driving down the freeway, and suppose you were given the velocity of a car as well as the car's constant acceleration. Well it might strick you as odd that someone would even supply you with this information, what if they asked you to find how far the car would be down the freeway after 60 seconds (1 minute) of time had passed? Would you be able to solve the said problem presented to you?
This presents the world of beginning to understand motion, and how we can use equations given to us to predict how scenarios turn out (for instance determining the position of an object after a certain amount of time as elapsed). We will start off dealing with only 1-dimensional motion which involves simply traveling along a straight line, and then expand from there.
Deriving The Equations
To start off with finding our fundamental equations for motion along a straight line, let us define the word acceleration. You have probably heard it before many times in terms of a car's acceleration or a runner's acceleration, but what exactly is it? To put it simply: acceleration is the change of an object's velocity over a certain time interval.
This gives us a very easy formula just from the definition if we assume the acceleration of the object to be constant:
\[a = \frac{\Delta \! v}{\Delta \! t}\]
The \(\Delta \!\) symbol simply means a change in something. Hence, acceleration is equal to the change in velocity over a change in time. Thus we can change the formula to a different form:
\[a = \frac{v_{\textrm{final}}- v_{\textrm{initial}}}{t_{\textrm{final}} - t_{\textrm{initial}}}\]
Usually the \(t_{\textrm{initial}}\) is considered to be 0 sec, and as a result we usually just ignore that term when writing our equation.
Let's rearrange this formula though now for an even better form by multiplying both sides by the final time and then adding the inital velocity to both sides:
\[v = v_{\textrm{inital}} + a*t\]
Now if we know the acceleration of the object and it's initial velocity, we can determine the object's velocity at any time \(t\) that we would like. We will consider this to be our first equation of one dimensional motion!
Let's see what else we can come up with to use for 1-dimensional motion. Perhaps let's come up with a formula to determine the position of object.
Let's establish the defintion of what velocity is. Like acceleration in a way, it is the change in position of an object over a change in time. This gives the objects average velocity:
\[\overline{v} = \frac{(x - x_{\textrm{initial}})}{t}\]
Average velocity though is very easy to derive a formula though as well because the average of anything is just the sum of the things you want to average divided by the number of things you are taking the average of:
\[\overline{v} = \frac{(v + v_{\textrm{initial}})}{2}\]
Which if you set the two \(\overline{v}\textrm{'s}\) equal to each other it leads to...
\[\frac{(x - x_{\textrm{initial}})}{t} = \frac{(v + v_{\textrm{initial}})}{2}\]
Before we simply solve this equation tho.ugh for \(x\) (i.e. the position of the object), let's do something. Let's assume that we are still given our constant acceleration for the problem and an intial velocity of \(v_{\textrm{initial}}\) for the problem. Then let's plug in our first fundamental equation for velocity in order to get an equation for \(x\) where we would know all of the other variables in the equation and can just solve for \(x\):
\[\frac{(x - x_{\textrm{initial}})}{t} = \frac{(v_{\textrm{initial}} + a*t + v_{\textrm{initial}})}{2}\]
Which simplifies to:
\[\frac{(x - x_{\textrm{initial}})}{t} = \frac{2v_{\textrm{initial}}}{2} + \frac{a*t}{2}\]
Then:
\[x - x_{\textrm{intial}} = v_{\textrm{initial}}*t + (\frac{1}{2})a*t^{2}\]
Finally leading to the form we want:
\[x = x_{\textrm{initial}} + v_{\textrm{initial}}*t + (\frac{1}{2})a*t^{2}\]
This is our second fundamental equation of 1-dimensional motion!
Suppose though now that we wanted to have a motion formula to be able to determine an objects velocity without having to know the travel time. If we knew the initial position, final position, and beginning velocity, can we derive a formula to help us find the end velocity? Let's try it:
First step: rewrite our first fundamental equation of motion with \(t\) as the dependent variable
\[t = \frac{v - v_{\textrm{initial}}}{a}\]
Then substitute that into our second fundamental equation, and then solve for \(v\)
\[x = x_{\textrm{initial}} + v_{\textrm{initial}}*\frac{v - v_{\textrm{intial}}}{a} + (\frac{v - v_{\textrm{intial}}}{a})^{2}*\frac{1}{2}*a\]
\[x = x_{\textrm{initial}} + \frac{v*v{\textrm{initial}} - v^{2}_{\textrm{intial}}}{a} + \frac{v^{2} - 2v*v_{\textrm{intial}} + v^{2}_{\textrm{initial}}}{2a}\]
Then if we subtract \(x_{\textrm{initial}}\) from both sides of the equation and multiply both sides by \(2a\), we get:
\[a*(x - x_{\textrm{initial}}) = 2({v*v{\textrm{initial}} - v^{2}_{\textrm{initial}}) + v^{2} - 2*v*v_{\textrm{initial}} + v^{2}_{\textrm{intial}}}\]
Which then simplifies down to:
\[a*(x - x_{\textrm{initial}}) = v^{2} - v^{2}_{\textrm{initial}}\]
Solved for \(v\) it turns into:
\[v^{2} = v^{2}_{\textrm{initial}} + 2*a*(x - x{\textrm{initial}})\]
This gives us our final fundamental equation of one dimensional motion! This means that in total, we have three equations at our disposal to use when trying to determine an object's position or velocity depending upon which values we know or are given. The actual solving of these problems is usually quite basic as it involves some simple algebra and "plugging and chugging" with the equation.
2-Dimensional Motion
Since we have our equations for determining 1-D motion, we now need to expand our equations to incorpate more dimensions of movement. We as humans move about in a three dimensional world, and, as such, need equations that can be utilized in more than just one dimension of motion. Some classic examples of problems that younger physics students encounter when just beginning to learn about this involve cannons shooting a cannon ball on an open field, or an airplane dropping a crate while flying. The 1-D equations are just not adequate enough right away to tell us where the ball/crate would land or even when the box/crate would land. The way to derive these equations that we are looking for though simply employs the use of vectors and a little bit of basic trigonometry.
Deriving Some More Equations
Let's consider the cannon problem I had presented earlier. Let the cannon have an incline of some \(\Theta \!\), and also let the cannonball that is shot have an inital velocity of \(v_{\textrm{initial}}\). If you look at the velocity vector of the cannonball after it is shot, you can see how it moves diagonally. This diagonal movement though than be represented by two vectors of one with just vertical velocity and one with just horizontal velocity. Once we do that, it becomes very easy to solve these equations right? because then we can just apply our simple 1-Dimensional equations of motion again. So let's figure out these vectors then: