Introduction

Welcome to the start of a bunch of lessons on "real" physics! Are you excited? Let's jump right in!

Position is defined as the vector (wake up call if you were dozing off in lesson 1!) from the origin to the point where an object is on a coordinate plane. Its SI unit is the meter ($\textrm{m}$).

We can actually define the origin however we wish and position will still be valid, so there is no absolute way to measure position. Still confused? Let's walk through an example! So, let's say my friend and I were standing in a room. I could be $5$ meters from a certain object in the room, but my friend standing at a different spot in the room is only $2$ meters from the same object. The object is still in the same place, so both points of view must be correct! So if position is not constant for each observer, then what is the point of defining it? Well, the change in position, or displacement, is constant. From my point of view, my friend is still $2$ meters from the object. This is true for ANY point of view in the room!

Position is typically defined with variables. Now if you haven't learned variables yet, don't worry! We got your back! Remember how vectors were defined using letters with an arrow on the top? Variables are just like that, except without the arrow! You can think of it as if a value/quantity was stored in the letter, and we wanted to refer to it without actually saying the value/quantity. The variable $x$ is primarily used, though $y$ and $d$ are sometimes used.

Displacement

Do you remember when we just said that displacement is far more useful than just position by itself? Well, given your original position and your final position as two separate points, we can draw a vector connecting them. That vector is what is called displacement, and how it is calculated from your original and final position. We can also do this mathematically, $$\Delta x = x_f-x_0$$Hold on! If you didn't sign up for this complex mathematical brain torture, we won't put you through too much of it! I promise! Remember when I said variables store values and quantities? That's what's in play here! $\Delta x$ is just a math-y way to say "displacement", and $x_f$ stands for final position, while $x_0$ (also sometimes seen as $x_i$) is initial position! You can think of them like a super mathematical (and maybe nerdy) way to use abbreviations.

Now let's do some simple math. Let's say you started at a dot on the floor, which let's call the origin (remember, we can define origin however we like!), and walked $10$ meters to the right. Your initial position is $0$, since you started from the origin, and your final position is $10$, since you are $10$ meters away from the origin (and that's the definition of position!). Thus, your displacement, is $10-0=10$ meters! While we did this math super simply, remember that displacement is a vector, and this number is just the length, or more formally, the magnitude. The direction of the vector would be to the right, since you walked to the right.

Having done this, now let's work through a more interesting example. Say you were to start at your house, which let's call the origin for this example. If you took a trip all around the world, and returned to your house, what was your displacement? Well, we know that displacement is final position minus initial position. So if your final position is back to the origin, and you started at the origin... That can only mean one thing... Your displacement is $0$! Do you see why? If your final position is $0$ meters from the origin, but so is your initial position, $0-0=0$!

Distance vs Displacement

Moving on to a different (but related) topic, you may also have heard of distance, which does actually hold some significance in physics, and it is defined as the the length of the path taken from one point to another, represented like this: $\Delta s$ (not to be confused with the unit seconds, $\textrm{s}$ !!).

Yet this raises one very pressing question: What's the big difference, between distance and displacement and all these fancy whatnots??

Well, let's work through an example! For instance, let's say I drive $50$ kilometers in one direction and then back home, my distance traveled is $100$ kilometers but my displacement is $0$ kilometers, since my final position is the same as my initial (and displacement is calculated from your final position to your initial position). Note that displacement is a vector quantity, but distance is a scalar (simply remember this as a term for regular numbers, or non-vectors). These terms are not interchangeable and will come up again!! Welcome to the start of a bunch of lessons on "real" physics! Are you excited? Let's get right into it, shall we?

Position is defined as the vector (wake up call if you were dozing off in lesson 1!) from the origin to the point where an object is on a coordinate plane. Its SI unit is the meter ($\textrm{m}$).

We can actually define the origin however we wish and position will still be valid, so there is no absolute way to measure position. Still confused? Let's walk through an example! So, let's say my friend and I were standing in a room. I could be $5$ meters from a certain object in the room, but my friend standing at a different spot in the room is only $2$ meters from the same object. The object is still in the same place, so both points of view must be correct! So if position is not constant for each observer, then what is the point of defining it?

Displacement

Well, the change in position, or displacement, is constant. From my point of view, my friend is still $2$ meters from the object. This is true for ANY point of view in the room! (Well, only in Newtonian mechanics, which thankfully is what we're learning about [you'll probably learn about this when you learn about relativity]).

Position is typically defined with the variable $x$, though $y$ and $d$ are sometimes used. I personally dislike using $d$, and any calculus student will understand why. To calculate displacement, we use the formula $x=x_f-x_i$, where $x_f$ and $x_i$ are the final and initial positions, respectively.

Distance vs Displacement

A quantity called distance also exists, and it is the length of the path taken from one point to another, represented by the variable $s$ (not to be confused with the unit seconds, $\textrm{s}$ !!). What's the big difference? Well, let's look at an example. For instance, if I drive $50$ kilometers in one direction and then back home, my distance traveled is $100$ kilometers but my displacement is $0$ kilometers, since my final position is the same as my initial. Note that displacement is a vector quantity, but distance is a scalar. These terms are not interchangeable!!

Velocity

Now, velocity ($v$), in its formal definition, is defined as the instantaneous rate of change of position (but we won't be dealing with the formal definition). The units of velocity is meters per second, $\frac{ \textrm{m} }{\textrm{s} }$

However, instantaneous velocity is not the only way we can express velocity. Another useful definition is average velocity, which is denoted by $\bar{v}$, and it is the displacement divided by the time interval in which it took place. So, from that, we know that: $$\bar{v} = \frac{\Delta x}{\Delta t}$$Or using the previous equation from the section on displacement, we know that: $$\Delta x = x_f-x_i$$So...$$\bar{v} = \frac{x_f-x_i}{\Delta t}$$The math can be a bit confusing if you are new to this, but just try to read and follow along.

If your math is to that level, you may have noticed we can directly related displacement ($\Delta x$) and average velocity ($\bar{v}$). Rearranging give us: $$\bar{v} \cdot \Delta t = \Delta x$$ Which tells us the average velocity multiplied by the change in time is the same as your displacement! It is important to remember that a $v$ will always denote instantaneous velocity, or velocity at a certain point in time, while $\bar{v}$ will always denote average velocity, or velocity over a certain period of time.

Do note that while your instantaneous velocity could be equal to zero, but your overall average velocity is not zero! For example, if I walk to my friend's house, but stop temporarily at a traffic light, my average velocity is positive, since my displacement was positive, but my velocity at the instant I was at the traffic light is zero!

Now the inverse is also true: my average velocity could be zero, but my instantaneous velocity could be nonzero. Let's say I run a lap around a park and stop at where I started from. My average velocity is zero, since if you look at the previous equation, $\bar{v} = \frac{x_f-x_i}{t}$, my final position and initial position are zero, and zero divided by any number will always result in zero. But let's take a look at my velocity at the instant I'm halfway around the park, that is definitely not zero. I am running as fast as I can because I want to finish my lap!

All this talk about instantaneous vs average velocity is very important to remember. Calculating the instantaneous velocity requires higher level math than what we expect that you have, but there is one equation you can use to find the initial and final velocities, involving acceleration later on.

Speed vs Velocity

As with distance and displacment, you may have heard of speed. Speed and velocity are two non interchangeable terms as well. You might be able to predict, if average velocity is displacement divided by time, then average speed must be distance divided by time! Let's go back to example of me running around the park. My average velocity is zero, since my final and initial positions are the same, but my average speed is not zero, since my distance traveled is the length of the path around the park!
However, unlike distance and displacement, speed can technically refer to velocity. Let me explain: Average velocity and average speed are calculated differently, but your instantaneous speed at any point in time is simply equivalent to the magnitude of your instantaneous velocity. Instantaneous speed in problems is usually stated as itself or just as "speed", while average speed is almost always stated as itself. Just keep this in mind whenever you encounter the word speed, to be able to distinguish which type it is referring to.

Velocity

Now, velocity ($v$) is defined as the instantaneous rate of change of position. This typically requires calculus to calculate, so we won’t worry too much about that definition. Another useful definition is average velocity, denoted by $\bar{v}$, the displacement divided by the time interval in which it took place. We also realize that this is the slope of a position versus time ($x\textrm{-}t$) graph ($x$ divided by $t$, which is rise over run!).

From this, we easily derive $x=\bar{v}t$ or, if you please, $x_f=x_i+\bar{v}t$ (although we omitted the capital letter delta, it is important to remember that "$t$" still refers to "$\Delta t$", or change in time, and the same goes for "$x$" and "$\Delta x$").

It is important to remember that a $v$ will always denote instantaneous velocity, or velocity at a certain point in time, while $\bar{v}$ will always denote average velocity, or velocity over a certain period of time. Do note that while your instantaneous velocity could be equal to zero, but your overall average velocity is positive! (can you see why?)

For calculus students, we realize that we decrease the time interval, the expression reduces to $\frac{dx}{dt}$, the time derivative of position. This result can be used to derive the velocity given a position function, and vice versa (by integrating $dx = v~dt$). Much of the physics right now isn't necessarily calculus heavy and while we can technically do all of these things with calculus, it would simply be overkill. Calculus plays a much more vital role in terms of graphical analysis.

As with distance and displacment, you may have heard of speed. Speed and velocity are two non interchangeable terms, just like distance and displacement! You might be able to predict, if velocity is displacement divided by time, then speed must be distance divided by time! Let's go back to example of me running around the park. My average velocity is zero, since my final and initial positions are the same, but my average speed is not zero, since my distance traveled is the length of the path around the park!

Speed vs Velocity

However, unlike distance and displacement, speed can technically refer to velocity. Let me explain: Average velocity and average speed are calculated differently, but your instantaneous speed at any point in time is simply equivalent to the magnitude of your instantaneous velocity. Instantaneous speed in problems is usually stated as itself or just as "speed", while average speed is almost always stated as itself. Just keep this in mind whenever you encounter the word speed, to be able to distinguish which type it is referring to.

Thus, for the remainder of this unit, we will do our best to refer to the magnitude of velocity as "the magnitude of velocity". There may be times though, where we mess up, and there may be cases where you're working on physics from a separate source! It is up to you to decide, using your best judgement and the context, whether "speed" in that scenario refers to its actual definition or just a shorthand for the magnitude of the velocity vector.

However, without any context, speed and velocity are two noninterchangeable terms! Speed will always refer to $\frac{\Delta s}{\Delta t}$ and is always a nonnegative scalar, while velocity is $\frac{\Delta x}{\Delta t}$ and a vector, a quantity with a magnitude and direction! Remember that from now on!

Acceleration

Acceleration ($a$) is our final kinematic quantity, and it is the rate of change of velocity. As with velocity, there is instantaneous acceleration, and average acceleration. We will deal with average acceleration, since it is more of use to us!

Likewise, average acceleration is equal to the change in velocity divided by change in time: $$a=\frac{\Delta v}{\Delta t}$$ From the equation, we can see the units acceleration is equal to $\frac{\frac{\textrm{m}}{\textrm{s}}}{\textrm{s}}$, or simply $\frac{\textrm{m}}{\textrm{s}^2}$, or meters per second squared (for those that don't know, it just means $\frac{\textrm{m}}{\textrm{s} \cdot \textrm{s}}$.)

We can also write an equation with velocity and acceleration, just like we did before with displacement and velocity, like shown: $$a \cdot \Delta t = \Delta v$$ But wait! Change in velocity is equal to final velocity minus initial! So we can plug this equation $\Delta v = v_f-v_i$ into that equation to get: $a \cdot \Delta t = v_f-v_i$, and we can rearrange it to get: $$v_i + a \cdot \Delta t = v_f$$ We now have means of calculating your initial and final velocity, which are instantaneous velocities! However, given this equation, there is one pitfall of using this. This equation is only valid under the circumstance that your acceleration is constant, or you are using the average acceleration. If the acceleration is not constant, you would have to use the average acceleration, or use more advanced calculus tools.

Conclusion

Now before we close off this lesson, there is something conceptually that is important to realize. All these quantities listed are VECTOR quantities. Need a refresher? A vector is defined by its length and direction. Now, if a vector's length changes, then it is a different vector! That's what the math above was dealing with. It dealt with the length of those vectors. However, it is also important remember that if the vector's direction changes, then wouldn't it also be a different vector?

So now let's look at an example. Let's say you have a toy car moving at a constant speed (not velocity!) around a circular track. Does the car have acceleration? Your intuition will instantly jump to the answer, NO! The car isn't speeding up or slowing down, so it isn't accelerating! However, that just isn't true. Remember that the definition of acceleration is the change in velocity, and velocity is a vector. If you are moving around a circle, your direction is changing basically every instant, if you think about it! Thus, the car DOES have acceleration, since the direction of your velocity is changing, even though it may not be speeding up or slowing down! This is a very important concept that you must know! (It will come up again)

Acceleration

Acceleration is our final kinematic quantity, and it is the instantaneous rate of change of velocity. Likewise, average acceleration is equal to the change in velocity over time ($a=\frac{\Delta v}{\Delta t}$). Obviously, we realize that acceleration is the time derivative of velocity from this, but also the second-order time derivative of displacement. A derivation is shown at the bottom of the next page. the equation, we can see the units acceleration is equal to $\frac{\frac{\textrm{m}}{\textrm{s}}}{\textrm{s}}$, or simply $\frac{\textrm{m}}{\textrm{s}^2}$

For nearly all scenarios in our study of physics, we will consider the case of constant acceleration. Why is this? Constant acceleration is the simplest case of Newtonian mechanics that we can learn. There are other ways to express nonconstant acceleration, but that requires calculus. Since we know calculus, you might be excited to work on nonconstant acceleration. However, sadly, this will come at a much later lesson.

Now let's derive why the acceleration is constant: We know that acceleration is the time derivative of velocity, and so this is what we can do: $$a=\frac{dv}{dt}$$ $$a~dt=dv$$ $$\int_{t_0}^{t_f}a \,dt=\int_{v_0}^{v_f}\,dv$$ Since $a$ is a constant, we can factor it out, and completely integrating both sides yields: $$v_f-v_0=a(t_f-t_0)~\textrm{or}~v_f=v_0+at$$ Another important thing is that the kinematics equations (that we will learn about shortly) only hold true for cases that deal with constant acceleration, so we must use that.

Since we know acceleration is the slope of the velocity graph, we have $\Delta v=at$ or $v_f = v_i + at$.

Now, you might think that from the previous equation, $x=\bar{v}t$, that $x=at^2$. This is NOT true. One important thing to realize here is that $\bar{v}$ is average acceleration, and the previous equation, $\Delta v=at$, finds change in velocity. To find the true equation, we can use one of two methods: algebra/graphical analysis, or calculus.

Deriving Equations: Algebraic Analysis

First, let's take a look at how algebra and graph analysis can lead us to the right equation. Notice that we can calculate our average velocity using this equation (If you ever see $v_i$ and $v_0$, just remember they both refer to initial velocity.): $$\bar{v} = \frac{v_f+v_0}{2}$$ Do you see how this works? Taking the average of our initial and final velocities gives us, well, average velocity!

But hold on! What is $v_f$? If you remember from just a few paragraphs ago, $v_f$ is equal to $v_i + at$! So when we plug this in, we get:

$$\bar{v} = \frac{v_0 + at + v_0}{2} =$$ $$ \frac{2v_0 + at}{2} = v_0 + \frac{1}{2}at$$ So, now we have an equation that we can plug into $x=\bar{v}t$! Plugging this in will give us:

$$x = (v_0 + \frac{1}{2}at) \cdot t = v_0 t + \frac{1}{2}at^2$$ So this is the equation that involves acceleration! We'll show examples of more equations later on, but this is part of five equations that you MUST memorize.

That's pretty much all there is to position, velocity, and acceleration. Like with the vectors chapter, we're going to introduce something pretty cool that is a little beyond the scope of the lesson and this math level, just to give you something to chew on. Warning! This section has fancy notation and symbols and requires algebra skills! If you did the Basic Algebra lesson earlier, you should be fine though. However, I do recommend you try your best and read it and follow along. If it's not for you, great! Keep moving forward. If you're interested, stick around!

Now you're probably thinking, is there a way to relate acceleration and displacement? Given the equations that we have, you'll see that we have $\bar{v} \cdot \Delta t = \Delta x$, and $a \cdot \Delta t = \Delta v$. However, we can't directly plug this equation into our first one! Notice that we need $\bar{v}$, and not $\Delta v$! Let's try and solve this task at hand.

First, we need to derive an equation for average velocity, $\bar{v}$. For those that don't know, taking the average of some amount of quantities is to add all the quantities together, and divide by the number of quantities there are. There are a few ways to calculate average velocity, but how about we check out this one: $$\bar{v} = \frac{v_f+v_i}{2}$$ The average of your final velocity and your initial velocity must be, well, your average velocity! So now we have this equation, right?

Taking an equation from the acceleration chapter, we know what $v_f$ is equal to! $v_f = v_i + a \cdot \Delta t$! So if we plug this into that equation for $v_f$, we get: $$\bar{v} = \frac{v_i + v_i + a \cdot \Delta t}{2}$$ $$ = \frac{2 \cdot v_i + a \cdot \Delta t}{2}$$$$ = \frac{\cancel{2} \cdot v_i}{\cancel{2}} + \frac{a \cdot \Delta t}{2}$$$$ = v_i + \frac 1 2 a \cdot \Delta t$$ Wow, now you can see that average velocity is equal to this! Now we can plug this into $\Delta x = \bar{v} \cdot \Delta t$!

Substituting this heap of variables and numbers for $\bar{v}$ gives us:

$$\Delta x = (v_i + \frac 1 2 a \cdot \Delta t) \cdot \Delta t $$$$= v_i \cdot \Delta t + \frac 1 2 a \cdot (\Delta t)^2$$
Wow, the final equation involving displacement and acceleration! Isn't that amazing? Like with the previous equation, though, this is only valid under a constant acceleration. Another way you could have done this is by rearranging the original equation to get $v_i=v_f-at$, and this would have yielded something like this:

$$\Delta x=v_f \cdot \Delta t-\frac 1 2 \cdot a \cdot (\Delta t)^2$$
Now if you haven't learned algebra yet, this could be difficult to grasp. However, I promise, for the next few chapters, we will focus a little less on the equations.

This was just a fun experiment to see how we can come up with an equation for displacement and acceleration. If you genuinely are having a hard time comprehending and understanding, we recommend you go back to the Basic Algebra lesson, read it and come back here. That should help you a lot.

OK! Are you ready to learn more about kinematics? Then let's go to it!

Deriving Equations: Graphical Analysis

Now let's take a look at the same thing but via graphical analysis. We know that acceleration is the slope of the velocity versus time (v-t) graph.

The area under the curve is displacement, since it is computed by multiplying velocity and time interval (the axes) which gives the units of displacement (what is $\frac{\textrm{m}}{\textrm{s}} \cdot \textrm{s}$). Don't worry! If you don't know what area under the curve is, it is exactly what it sounds like. You calculate the area under the line of the graph for a certain interval. The reason why this gives us displacement is because we just mentioned, if you look at the units for the x-axis, and the y-axis, if you multiply the units, you will get the units of displacement (believe it or not, you are learning fundamental calculus right now!).

For the case with initial velocity $v_0$ and constant acceleration $a$, we have a graph that looks like this:

For any $v_f$, the area is:

$$x=\frac{v_0+v_f}{2} \cdot t$$
Using the formula for the area of a trapezoid (I will assume you know the trapezoid formula, if not, you can search it up). You can visualize this trapezoid by drawing a vertical line from the tip of the arrow to the x-axis, which you will see via the dotted line is $v_f$. You can see that the two "bases" are $v_0$ and $v_f$, while the "height" is the change in time (because the x-axis is time, and the y-axis is velocity).

(Again, note that the use of $\Delta t$ is often replaced with just $t$ for convenience.)

You can see that this implies that $\bar{v}=\frac{v_0+v_f}{2}$

But...
$$v_f = v_0 + at$$
So...
$$x=\frac{2v_0+at}{2}t=v_0t+\frac{1}{2}at^2$$
Which is what we exactly just derived!! It is highly recommended (maybe almost required) that you memorize these equations, because you will likely be using them a lot for the next two lessons.

This is much easier to derive from a calculus perspective:

$$v = v_0 + at$$
$$\frac{dx}{dt} = v_0 + at$$ $$dx = (v_0 + at)dt$$ $$\int_{x_0}^{x_f}dx = \int_{0}^{t}(v_0+at)dt$$ $$x_f-x_0 = \Delta x = v_0 t + \frac{1}{2}at^2$$

Another interesting thing you can do is remember when we taught you dimensional analysis? You can go back to the equations we mentioned and double confirm with us, are the equations also dimensionally accurate? Just something for you to investigate upon.

From this, we can graph an $x$-$t$ graph for constant acceleration scenarios. A desmos example is shown below (you can play around with the sliders):

Conclusion

That's pretty much all there is to position, velocity, and acceleration. Usually, when we say $x$, we do mean displacement, but occasionally, if you see $x$ denoted with subscripts $f$ or $i$ or $0$, it will refer to instantaneous position.

However, as we mentioned at the start of the lesson, it is vastly more useful and advantageous (in various different ways) to find the change in position, because we can utilize it and draw connections with it to other quantities, like velocity and acceleration. If there is one key takeaway from this lesson, it would be how the three values are related. Remember: velocity is the (instantaneous) rate of change of displacement, and acceleration is the (instantaneous) rate of change of velocity.

Are you ready to learn how to use what you just learned to do some cool calculations? Let's go to it!