I did advanced geometry after alg 2, I don’t know if that was the right order now that I think about it

I actually disagree.

While the calculus itself is super super super simple and easy, it’s much harder than trig because the more abstract concepts and applications are much harder for people to grasp. While there is a lot of memorization and algebra-style stuff going on with derivatives and derivative rules, there’s SO much more to calc than that, and it’s ultimately just a small part of it. You’ll see that eventually as you learn and explore calc more.

Trig and geometry may be a lot of info to take in, but it’s super super easy to understand and you can memorize the rules and when to apply what and you’re all set. There’s not necessarily much of a puzzle there

Ohhh so that’s why advanced geometry was so much easier than algebra

Yep

Typically you take geo first and build on what you learn there in alg 2

I don’t know the exact curriculum but the class is called algebra 8r (the r is for advanced). I’m in seventh grade doing the 8th grade class

Yeah, I see your point, and I agree with it.

In school, all we have learned are derivatives, simple integrals, and simple limits. These are pretty easy, and they don’t really teach the ideas behind them (I did a lot of research for this paper), and I don’t think we will, which kind of is a bummer.

It’s just that calc (at least for derivatives) is pretty straight forward. Yeah, the idea behind it is really cool and complex, and much harder to understand and grasp than trig, but most calc questions (that I’ve had to solve) were usually easier than trig, because all it took were really straightforward applications of the rules of calc. It definitely gets harder, but I would say for me, ignoring the need to understand calc, calc is easier than trig (at least with derivatives).

Did you right that entire paper?

Good for you!

Yep!

Thanks.

your paper is amazing

Thank you!

Hopefully the next one will be more understandable.

That’s the problem (persay) lol. imo, that’s the boring stuff . You’re just kinda scratching the surface and missing out on all of the amazing stuff below. That’s where all of the complexities are. I’m glad you’re looking into that stuff on your own.

Yeah. It is a bummer, and I’m in an accelerated course, so I would’ve hoped I would be learning things more in-depth. Do you know of any website that I could learn this stuff from? (Not khan academy, that is really slow on my computer)

Well, I just realized that everything after Forces (including) will be basically impossible to teach without pictures. This isn’t a goof solution though, because I would have to post a *ton* of pictures all in all for these topics. So, if someone has a good solution (that doesn’t involve pictures), I would really like to hear it!

Thanks!

I guess I have an update regarding the next “lecture”:

In the next “lecture” we’ll discuss free fall, too (as opposed to what I originally planned on - it being in BASIC MOTION IN TWO DIMENSIONS)

Also, the first draft is done, I’m just waiting for some friends to review it.

Okay, I’m here with another update:

I’m going to start this off by apologizing. I’ve been away from the Forum for a couple of days and I haven’t worked on the lecture either. But the lecture itself is finished, I just need to convert it to a format that is forum-friendly. I’m hoping on doing this today. So, the lecture should be posted later on tonight (my time, so afternoon in the east coast).

And… It’s here! After a lot of procrastinating, I took the time to convert the document to a forum-friendly format. In fact, I procrastinated so much, that the next two lectures are already ready. That doesn’t mean that they’ll be out soon, though. Because they deal with trigonometry and vectors, I used quite a bit of pictures. So, I’m not sure how the next one’ll come out.

The format here isn’t amazing, because the forum doesn’t really support math, so the fractions here might look a bit weird. The way I wrote them is by doing < sup > numerator < /sup > / < sub > denominator < /sub >, so they’ll look like this: ^{ numerator } / _{ denominator }

To be honest, I didn’t realize how much people actually cared about these, and I was about to notify you guys that I wasn’t going to be continuing with the lectures. But SilverStar posted about how she was looking forward to the next one, so I decided to stop procrastinating and got down to work. So thank you Silvie!

## LECTURE 1.2

**LECTURE 1.2**

**BASIC MOTION/MOTION IN ONE DIMENSION**

If you read the last lecture, you’ll at least have a grasp on what calculus is. This will be important for this lecture.

In this lecture we’ll be discussing motion in one dimension.

**Units for motion:**

The SI (International Standard, it’s French) units for motion are:

Position (x) : meters (m)

Velocity (v) : meters per second (m/s)

Acceleration (a) : meters per second squared (m/s2)

… (For derivatives of position)

Time (t) : seconds (s)

All of these units have precise definitions, and if you’re interested, look them up.

**Graphing motion**

Before we begin discussing motion, we need to come up with a way to graph it (like in a coordinate plane). To do this, we usually have position (or one of its derivatives) on the y axis and time (t) on the x axis. So instead of: y = mx + b, we have: x = vt + x0 (remember how velocity is the derivative of position and the derivative of a function at some point gives the slope of the function at that point. Also x0 means the position at time 0.)

**Constant-velocity motion**

Velocity is the change in position over a unit time. Usually, velocity and speed are used synonymously, but this is incorrect. Velocity is a *vector* while speed is a *scalar* (we’ll get to that next lecture). In fact, position is also a vector. But in one dimensional motion, you can ignore vectors.

Velocity in constant velocity motion is defined as:

**I.** v^{-} = Δx÷Δt

In physics, Δ means “difference”. So if at t=1 seconds you measure the position of some object to be 10 meters, and at t=5 seconds you measure the position to be 18 meters, the velocity would be **thing** or 2 meters/second. What is that “meters a second”? Well, you’re taking the difference in position (meters) divided by the difference in time (seconds), so the SI unit of measurement (SI is the standard system of units) for velocity is meters/second.

**Constant-accelerated motion**

Constant-velocity motion is simple and nice and all, but it isn’t realistic. Most of the time we’re changing our velocity. This change in velocity (with respect to time) is called acceleration. Acceleration can be defined as:

**II.** a^{-} = Δv÷Δt

So the SI unit for acceleration is meters/second squared ( m÷s^{2} ).

Now, accelerated motion is more complicated than constant-velocity motion. Acceleration, you may have noticed, is the *rate of change* of velocity, and velocity is the *rate of change* of position. The acute reader may have realized that this is the same as the *derivative* of a function.

Because of this, the equations I wrote for velocity and acceleration is the *average* velocity and acceleration over time Δt (that’s why they have a line over them).

So, to find an equation for x, let’s start at the beginning. Because this is constant-accelerated motion, a(t) (the function for acceleration) is constant. Let’s write that as:

a(t) = a

Acceleration, as we now know, is the derivative of velocity, so velocity is the integral of acceleration.

v(t) = ∫a(t)dt

**III.** v(t) = at + v_{0}

What is that v0? Well, remember that constant c that showed up when we integrated a function? V0 is that constant c, but written in a different form. It is the velocity at time 0, just like how c is the y value at x=0.

Position is also the integral of velocity, so:

x(t) = ∫v(t)dt

x(t) = v_{0}t + 0.5at^{2} + x_{0}

Like velocity, x(t) also has x0. Another way of writing this function is:

**IV.** Δx = v_{0}t + 0.5at^{2}

All of these functions (x(t), v(t), and a(t)) don’t need the (t) after them, as they are already written with respect to t. Think about how you write y = … and not y(x) = … . y is already written with respect to x. So, I will only be adding the (t) if I need to integrate or differentiate the function.

Let’s try and make more equations! Let’s start by simplifying the equation for Δx.

Δx = t(v_{0} + 0.5at) = t(^{2v0+at} / _{2}) = t(^{v0 + v0 + at } / _{ 2 } )

**V.** Δx = t(^{v +v0 } / _{ 2 })

v is the velocity at time t. This is pretty neat. We have an equation for Δx without the need for acceleration! Let’s try and combine III and V for an equation without t

III. t = ^{v - v0}/_{a}

V. Δx = t(^{v +v0 } / _{ 2 }) => Δx = ^{v +v0 } / _{ 2 } × ^{v - v0}/_{a} = (v^{2} - v_{0}^{2}) / (2a)

**VI.** v^{2} = v_{0}^{2} + 2aΔx

This is one of the more important equations in motion, and I highly encourage you to remember it. As it doesn’t take t as input, it is a very useful equation.

Now, another important thing to remember is that velocity and acceleration are the derivatives of position and velocity, respectively. This means that given a graph of, say, velocity with respect to time, to find the difference in position over some time starting at t=a and ending at t=b, all you need to do is find the area under the function v(t) between t=a and t=b. (The same goes with finding the change in velocity given the function a(t))

**Free fall**

In Newtonian mechanics, free fall is the motion of a body when the only force acting upon it is gravity (we’ll get to forces soon). Neglecting air resistance, dropping or throwing a ball results in it being in free fall. This is true for any initial angle, but for now we’ll only focus on dropping a ball (no initial velocity) and throwing a ball straight up or down.

Because throwing an object so that the only force acting on it is gravity results in a constant acceleration downwards, an object in free fall obeys these laws of motion that we have found. Let’s try splitting up the motion of an object in free fall:

*Initial throw:* The object is thrown in some direction with velocity v0. I usually use positive v0 to denote an upwards throw, but either way is fine as long as you orient your force correctly.

*Upwards motion:* The object will continue to move upwards, its velocity decreasing (in the downward direction) by g m/s2. Usually you round g up to 10 m/s2, or 9.8 if you want to challenge yourself, but 10 is good enough.

*Maximum height:* As velocity is the derivative of position, when the position reaches its maximum value, its derivative, velocity, will be 0.

*Downwards motion:* The object will begin to move downwards, its velocity increasing (in the downward direction) by g m/s2.

*Return to initial height:* At some time t, the object will return to its initial height. Its velocity will be equal to its initial velocity (we’ll prove that later)

*…And beyond:* The ball may continue downward (say if it is thrown off of a building)

Let’s try and find a few things:

Max height, time until max height, total time in the air, final velocity, the time until the objects reaches a given height.

Max height:

As we discussed earlier, when the object reaches its maximum height, its velocity is 0. So, without finding tmax (the time until the object reaches max height), let’s find the max height. We can use equation VI for this:

0^{2} = v_{0}^{2} -2gΔy

Pretty straight forward.

Time until max height:

We can use equation III for this:

0 = v_{0} – gt

t = ^{v0} / _{g}

Let’s check on this:

Δy = ^{0 + v0 } / _{2} × ^{v0} / _{g} = (v_{0}^{2}) / 2g

This supports our conclusion from before, great!

Total time in the air:

When the object reaches its initial height, Δy will be 0. So, using equation IV:

0 = v_{0}t - ^{g}/_{2}t^{2} = t(v_{0} - ^{g}/_{2}t)

This gives us two solutions, one of which is t = 0. This is the starting time, so it isn’t the solution we are looking for. So:

t = ^{2v0} / _{g}

Interestingly, this is twice the amount of time it took to reach the maximum height.

Final velocity:

When the object reaches its initial height, Δy will be 0. So using equation VI:

v^{2} = v_{0}^{2} - 2g×0 => v = ±v_{0}

We know that v cannot be facing the same direction as v0, it must be in the opposite direction (you don’t fall up), so v = -v0.

Let’s check this by using equation III:

-v_{0} = v_{0} - gt => t = ^{2v0} / _{g}

This is the same conclusion we came to as before, great!

The time until the object reaches a given height:

Let’s take the height we want to get to to be h. Using equation IV, we get:

h = v_{0}t - ^{g}/_{2}t^{2}

Using the quadratic formula, we get:

a = 0.5g b = -v_{0} c = h

t_{1/2} = (v_{0} ± √(v_{0}^{2} - 2gh)) / g

Say we wanted to find when the object gets to some point *lower* than the initial height (maybe it was thrown off of a roof?). This equation will also give us that. Here we are taking the initial height to be 0, so to find when it reaches some point *under* the initial height, h will be less than 0.

**Relative motion**

Say you have an object moving with velocity va and another object moving with velocity vb. What would be be object a’s velocity relative to object b (va,b)?

Well, let’s take a look at position first: what would xa,b be? This is pretty simple, it is basically asking if b was the origin ((0,0)), what would a be? All this is what the difference between these two positions are. So x_{a,b} = x_{a} - x_{b}.

We know that velocity is the derivative of position, so:

v_{a,b} = ^{dxa,b}/_{dt} = ^{d(xa - xb) }/_{dt} = ^{dxa}/_{dt} - ^{dxb}/_{dt} = v_{a,b}

Remember, d is not a number, we are not multiplying by d, it is a stand in for the limit of x(t) as t approaches 0 (dt is an infinitesimal t).

The same thing can be proven for acceleration in the same way, so:

a_{a,b} = a_{a} - a_{b}

The same holds for jerk, snap, crackle, and pop (the third, fourth, fifth, and sixth derivative of position, respectively).

Really xa is also a relative position, it is measured relative to an *inertial* frame of reference. An inertial frame of reference if a frame of reference that is not undergoing acceleration (it is in constant-velocity motion). Usually we use the Earth as an inertial frame of reference, but it is not. As it is rotating around its axis and orbiting the sun, objects on its surface are subject to acceleration. But we ignore this because A) not ignoring it *really* complicates things, B) the effect is negligible, and C) there is no such thing as a perfect inertial frame of reference, everything is subject to acceleration (as turning requires acceleration, even massless particles are non inertial frames of reference, as they travel through *curved* spacetime, but we’re straying from the subject here.)

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Do notify me if something doesn’t make sense, please. Thanks and enjoy!

Ooh very nice

Great content and super nice job with the formatting haha

Thanks!

Unfortunately, I’ve been busy and I haven’t been able to read them yet. Could you put all your lectures in one spot? Thanks.

I think it would also help if you coded it with examples