Personal_Knowledge_Base/Math_and_Physics/20240825114712-physics_231_kinematics.org

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:ID:       2a02b387-4fd0-4f83-9eac-131887b7abc8
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#+title: Physics 231: Kinematics
#+filetags: :physics:textbook_notes:
#+STARTUP: latexpreview



Before diving into the topics of kinematics, we must ponder the questions that are raised when we think of an object all alone in the universe.
Imagine a rock in space that is far enough away from any star or planet, that it can interact with nothing. Answering the following questions will allow us to understand what will happen when the object is /not/ alone:

1. Could the rock speed up or slow down on its own?
2. Could it turn?
3. If the rock is observed to have motion, what would the properties of this motion be?

The laws of nature say that the rock's motion will be constant. Where it is going now, it will continue to go forever, unless acted upon.
This is referred to as the /inertia/ of an object.
This is Newton's first law of motion: *To the extent that objects are unaffected by external influences, objects at rest remain at rest, and objects in motion remain in constant-speed, straight-line motion.*
A deep implication that this hints towards is that there is fundamentally no difference between an object at rest and one that is in motion. This provides some of the underpinnings of Einstein's Theory of General Relativity.
*Motion can only be defined relative to another object.*


* Position

A definition of position: *The position of an object is a vector that describes the object's location with respect to some agreed-upon coordinate system.  Both the location of the origin and the orientation of the axes must be known to make any sense of a position vector.*

The position vector is usually denoted as $\vec{r}$
The change in position is referred to as *displacement* and is denoted by $\Delta \vec{r}$

For more information regarding the basics of vectors see[[id:6b3d2dbb-7c50-4ebf-b66c-4b0d1f20cdd1][ Physics 231 Calculus and Vectors]]

Oftentimes, position vectors are functions of time, t.


* Velocity

Velocity is the derivative of the position vector with respect to time.

\begin{equation*}
\vec{v} = \frac{d\vec{r}}{d t}
\end{equation*}

To determine the direction, just divide the vector by its magnitude to find the *unit vector*.
To find the velocity vector from position, simply take the derivative of each component.

It is also very important to realize that the integral of a velocity vector does not result in the position vector but the *change in position* or the displacement.

We can find the latter position by adding the change to the first position, quite simply:

\begin{equation*}
\vec{r_2} = \vec{r_1} + \Delta \vec{r}
\end{equation*}

But notice that the first position is a piece of information that cannot be attained from the velocity vector!


To find the average velocity over a specific range, simply employ this calculus technique of dividing the result of an integral by its range:

\begin{equation*}
\vec{v}_{avg} = \frac{\int_{t_1}^{t_2} \vec{v} dt}{t_2 - t_1}
\end{equation*}


It is also important to note that the magnitude of the average velocity can be /different/ than the average speed. This is because the average velocity vector takes into account the direction of the straight-line path from point A to point B, while the average speed is dependent on the actual path that is different.
For example, if you run one lap around a running track, you should end up right where you started. Thus, the magnitude of your average velocity is 0, but of course, your average speed was definitely not 0 (hopefully).


* Acceleration

Acceleration can then be difined as the derivative of the velocity vector.

\begin{equation*}
\vec{a} = \frac{d \vec{v}}{d t}
\end{equation*}

And, like with integrating velocity, we can get:

\begin{equation*}
\vec{v}_2 = \vec{v}_1 + \Delta \vec{v}
\end{equation*}

We can continue to find information about the displacement as well:

\begin{equation*}
\Delta \vec{r} = \int \vec{v} dt = \int (\vec{v}_1 + \Delta \vec{v}) dt
\end{equation*}

Now, describing acceleration in terms of both change in speed /and/ direction is very useful and is given by (derived by the first equation in this sections and using the product law of derivatives):

\begin{equation*}
\vec{a} = \frac{d v}{d t} \hat{v} + \frac{d \hat{v}}{d t} v
\end{equation*}

We can see that the first term describes a change in speed only, which we know from natural observation, takes place in the direction of the velocity, such as a car speeding up or slowing down without changing direction.
The second term describes acceleration as a change in the direciton of the unit vector with time (radians per unit time) at constant speed.

The direction of the acceleration is perpendicular to the velocity vectors, just like the velocity vectors are perpendicular to the position vectors. This gives acceleration a direction in the $-\hat{r}$



* Eliminating the Time Variable

We now seek an expression that relates velocity to position, but contains no time variable.
If we want to eliminate the time variable, we must use the chain rule:

\begin{equation*}
\frac{dv_x}{d x} = \frac{d v_x}{d t} \frac{d t}{d x} = a_x \frac{1}{v_x}
\end{equation*}

Simplifyling, we have:

\begin{equation*}
\frac{d v_x}{d x} = \frac{a_x}{v_x}
\end{equation*}

From this simple equation we can create a host of kinematic equations that model physical systems with:
1. Constant acceleration
2. Position-dependent acceleration
3. Velocity-dependent acceleration


For *constant acceleration* applications, we can simply separate the terms and then integrate, ultimately leaving us with this equation (after some simplification):

\begin{equation*}
v_x^2 = v_{x0}^2 + 2 a_x(x - x_0)
\end{equation*}


However, it is very important to realize that these equations do not have a built-in sense of the correct passing of time (after all, we have eliminated the time variable). So, according to these equations, there are two values for time. However, only one is correct given the context. For example, the two time values express the time it takes for a ball to fall from a point and hit the ground, and the time that it would take for the ball to climb upwards from the ground to the point at which it was thrown, respectively.


For *Postion-dependent* accelerations, we can have an expression that depends on the slope:

\begin{equation*}
a_x = -g(slope)
\end{equation*}

Where g is the gravitational acceleration.

If we simply plug this expression in for the acceleration and separate and integrate like we have been doing before, we get an equation like this (after simplification):

\begin{equation*}
v_x^2 = v_{x0}^2 - (slope(x))g(x - x_0)
\end{equation*}

Where slope is a function of x, the position.



For *velocity-dependent* accelerations, let us give an example. Let's say that a car is rolling to a stop and its acceleration is a function of its velocity (since friction and air drag depend on velocity). Let's say this is:

\begin{equation*}
a_x = -0.2 v_x
\end{equation*}

\begin{equation*}
\frac{d v_x}{d x} = \frac{-0.2 v_x}{v_x}
\end{equation*}

From this, we see how easy this calculation can be, especially after cancelling out the velocity term.
Then, after separating and integrating, we arrive at this simple equation for velocity-dependent accelerations (ignoring air drag).

\begin{equation*}
v_x - v_{x0} = -0.2(x - x_0)
\end{equation*}



* Drag-free Projectile Motion

Using what we know about position, velocity, and acceleration, and what it means to take their derivatives and integrals, we can combine all of these things to create a glorious, complete equation.

\begin{equation*}
\vec{r} = (r_{x0}+v_{x0}t)\hat{i} + (r_{y0} + v_{y0}t-\frac{a_g t^2}{2})\hat{j}
\end{equation*}


/This equation does not account for landing or impact with the ground./

After solving for the time of flight using the y-component of the position equation given above, one can use it to find the range, or the distance in the x-direction the object goes before it hits the ground. The velocities in the x and y directions are just simply the components of the overall velocity vector. After doing this (substiuting the time of flight into the equation) we find that the range is:

\begin{equation*}
r_x = \frac{v_0^2 sin(2\theta)}{a_g}
\end{equation*}