warehouse/content/physics/kinematics.md

title, date

title	date
Vectors and Kinematics	2024-08-19

One must always start a study into the heavily crippled IB editions of the glorious subject of Physics with the initial understanding that the road ahead leads to pain immeasurable.

-> Prime of the Faith

Motion in 1D

In this rendering of motion, you will never need to use vector quantities to describe movement due to the scalar nature of all quantities discussed.

Reference Frames and Displacement

• Any measurement about motion is taken in terms of a reference frame.

Note

Example: A train that moves with respect to the ground being held stationary, is not moving with respect to a stationary person inside that train. If a person were to walk, at let's say 5 km/s toward the back of the train while it moves forward at 80 km/h, the person is moving at 75 km/h with respect to the stationary ground.

• In one dimension, only one axis, the $x$-axis of a coordinate plane is used.
• the net distance an object has traveled is known as displacement
• total distance is the overall distance traveled by the object/particle regardless of reference frame or initial/final positions
• change in positions is described using \Delta x

Average Velocity

It is important to note that this equation was derived from the more complicated calculus variant of the velocity equation. There are 2 important terms here.

\text{average speed}=\frac{\text{distance traveled}}{\text{time elapsed}}=\frac{\Delta x}{\Delta t}

For any form of velocity, the speed is simply calculated by using the magnitude of the velocity vector ||\vec{v}||, but since we exist in one dimension now, we will use |v(t)|

Instantaneous Velocity

If the position equation is defined as x(t), then:

\text{instantaneous velocity} = \tfrac{\mathrm{d}x}{\mathrm{d}t} = v(t) = \lim_{\Delta t \to 0}{\tfrac{\Delta x}{\Delta t}}

Average Acceleration

The acceleration of an object is the rate at which the velocity of said object changes. Average Acceleration is defined as the change in velocity from 2 distinct points divided by the change in time between those 2 distinct points.

\text{average acceleration} = \frac{\text{change of velocity}}{\text{time elapsed}}

Or more mathematically:

a=\frac{v_2-v_1}{t_2-t_1}=\tfrac{\Delta v}{\Delta t}

Instantaneous Acceleration

If the velocity function is defined as v(t) (this notation only applies to 1 dimension):

a(t)=\tfrac{\mathrm{d}v}{\mathrm{d}t}=\lim_{\Delta t \to 0}{\tfrac{\Delta v}{\Delta t}}

Please take calculus/study calculus if you want a neuron or two to function during the course of this, well, course.

Motion at Constant Acceleration

During this section and any point you use these formulas, always take t_0 to be equivalent to 0=t_1. This allows for the following line of reasoning:

\bar{v}=\frac{x-x_0}{t-t_0}=\frac{x-x_0}{t}

This clearly makes sense because x_0 is taken to be the initial position of the particle in 1D, and t_0 is taken to be simply 0, when the clock starts. Therefore acceleration must be a constant like so:

a=\frac{v-v_0}{t}

Firstly, it is important to note that all variables without subscripts are taken to be definitions of that variable completely. In contrast, when you saw that v had a little bar on top of it, \bar{v} meant to be the average of v. Moving on, if you are required to find the velocity of an object after elapsed time t:

v=v_0+at 

Note

Example: You are given that the acceleration of a motorcycle is constantly 4.0 \mathrm{m}/\mathrm{s}^2. You need to find the speed of that motorcycle (note that we haven't given you any sense of position, \therefore did not ask for velocity) after elapsed time t=6.0\mathrm{s}. The motorcycle starts from total rest, and when t_1=t_0=0.

[!faq]- Answer 4.0*6.0=24\mathrm{m}/\mathrm{s}

Next, calculating the position:

x=x_0+\bar{v}t 

Because of the symmetric and uniform growth of the velocity over time:

\bar{v}=(\frac{v_0+v}{2})t=x_0+t(\frac{v_0+v_0+at}{2})=x_0+t(\frac{2v_0+at}{2})

This finally results in the formula for position in these scenarios:

x=x_0+tv_0+\tfrac{1}{2}at^2

Objects in free fall

Just remember that g=9.80\mathrm{m}/\mathrm{s}^2

Vectors

You haven't seen any vectors to this point yet, so here we go! Vectors are essentially made of 2 components as long as they are two dimensional. You can either specify the relative coordinates they meet or the direction and magnitude. For example, you can write it out as either: \vec{p}=\langle a,b\rangle or as \vec{p}=\langle m;\theta \rangle. To convert between them: \vec{p}=\langle m\sin{\theta},m\cos{\theta}\rangle

Adding and Subtracting Vectors:

Now that you understand that, I can say that displacement in 2D is all about a certain amount of movement in \hat{i} direction or \hat{j}, then adding or subtracting them to compute the same dimensional cumulative vector. Anywyas, Adding them is as simple as arranging the vectors tail to tip, and then the next tail to tip. Then you compute the vecor formed between the exposed tail and tip. Algebraically, it looks like this:

\langle a,b \rangle + \langle c,d \rangle = \langle a+c,b+d \rangle

Magnitude

For the rectangular component vector form:

||\vec{p}||=\sqrt{a^2+b^2}

In more than two dimensions, you would just sum more components in the same fashion as a and b. With the circular vector form:

||\vec{p}||=m 

Refer to the stuff above to make sense of this.

Multiplying a vector and a scalar

c\vec{p}=c\langle a,b \rangle=\langle ca,cb \rangle

2D Kinematics

Calculating Displacement in 2D

The displacement vector \vec{D} is simply the sum off all movement vectors \vec{m_{0 \ldots n}}. You can the calculate the magnitude of that vector to get the distance of the displacement as ||\vec{D}||

Projectile motion

Not that this section on projectile motion is very different from the application of the constant acceleration velocity equation to gravity using -g as the acceleration. Here, motion is 2D, and is propelled. For an object to be propelled means for that object to have an initial velocity. Additionally, the object will also usually have a launch angle, starting from \theta=0 on the positive x axis. There are generally 2 methods to go about this, with one being a substitution of gravity into another.

General Projectile Vectors for Constant Acceleration

Without any further definitions, other than \vec{v} is the velocity vector, \vec{a} is the acceleration vector, and \vec{s} is the position vector. The velocity vector is a function of any given point of time t given as such:

\vec{v}(t)=\vec{v}_0+\vec{a}t=\langle v_{x0}+t(a_x), v_{y0}+t(a_y)\rangle

This can also be visually represented as such:

You can also calculate the square of the velocity vector independent of of the time t. This can low-key be helpful when you only have the position vector \vec{s}.

\vec{v^2}=\langle v^2_{x0}+2a_x(x-x_0),v^2_{y0}+2a_y(y-y_0) \rangle

The position finally being:

\vec{s}(t)=\vec{s}_0+t\vec{v_0}+\frac{1}{2}\vec{a}*t^2

Standard Projectile motion with Gravity

First it is defined that \vec{a}=\langle 0,-g\rangle. The composite velocity vector is therefore defined as:

\vec{v}(t)=\langle v_{x0}, v_{y0}-gt\rangle

That g is for gravity. It is also important to remember that \vec{v}=\langle v_0\cos\theta,v_0\sin\theta\rangle. You can also get the position vector by integrating the velocity vector.

\vec{s}=\langle s_{x0}+tv_{x0},s_{y0}+tv_{y0}-\frac{1}{2}gt^2 \rangle

Note

An artillery shell is fired at an angle of 67.9\degree above the horizontal ground with an initial speed of 1890 \mathrm{m}/\mathrm{s}. The gravity acceleration constant is the standard value for g. What is the total flight time of the shell in minutes?

[!faq]- Answer If you want the long answer just message me, I'm too tired to do this now.$5.9563$ minutes. Essentially, produce the position vector, then solve for the y component to be 0.

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