Kinematics

Introduction

Kinematics is a branch of classical mechanics whose purpose is to describe how physical objects move without making any assumptions as to why they move the way they do. Essentially, it strives to give a description of how the position and motion of physical objects evolves through time.

Position

Describe the motion of a physical object boils down to describing its position and how it changes over time. To do this, we seek to find an expression for its position as a function of time.

Definition: Position

Let $\mathcal{R}$ be a reference frame with origin $O$.

At a given moment $t$, the position of a point particle $p$ is the vector assigned to the point in space where $p$ is located.

NOTATION

$$\mathbf{r}(t)\mathbf{r}(t)\vec{r}(t)$$

Tip: Visualizing Position

The position of $p$ can be visualized as an arrow extending from $O$ to its location.

Definition: Path

As $p$ moves, its position draws out a curve in space which we call $p$‘s path.

Note: Position as Parametric Curve

Thus the function $\mathbf{r}:[t_{\text{initial}},t_{\text{final}}]\subset \mathbb{R}\to {\mathbb{R}}^3$ which to each moment $t$ assigns $p$‘s position at $t$ is a parametric curve whose endpoints are the initial and final position of $p$ during its journey.

From this definition, it follows that the position of a point particle depends on the reference frame. This is due to the fact that there are infinitely many possible reference frames and no single one is objectively superior to the rest. Physically, this manifests as the following phenomenon: Imagine two cars heading down the same road, right next to each other, with the same speed. We consider two reference frames:

• In the first one, you are standing on the side of the of road. This reference frame is such that you are at its origin. In this case, the positions of both cars appear to be changing, while the objects in the background appear stationary.
• In the second one, you are you sitting in one of the cars. In this case, the origin of the reference frame is attached to you. If looked out the window, the other car would appear stationary, while the positions of the objects in the background would appear to be changing.

Displacement

Definition: Displacement

Let $\mathcal{R}$ be a reference frame with origin $O$ and let $t_1$ and $t_2\ge t_1$ be two moments in time.

The displacement of a point particle $p$ during the interval between $t_1$ and $t_2$ is the difference in $p$‘s position at the latter moment ($t_2$) and at the earlier moment ($t_1$):

$$\mathbf{r}(t_2)-\mathbf{r}(t_1)$$

NOTATION

$$\Delta \mathbf{r}\Delta \mathbf{r}\Delta \vec{r}$$

Tip: Visualizing Displacement

Displacement can be visualized as an arrow pointing from $p$‘s location at the moment $t_1$ to $p$‘s location at the moment $t_2$.

Displacement gives us a very rough idea of the overall direction in which $p$ has moved from time $t_1$ to time $t_2$ and the distance between its starting and end locations.

Just like position, displacement depends on the reference frame. Imagine again the two cars heading down the same road, right next to each other, with the same speed:

• In the reference frame where you are on the side of the road, if you looked at the cars at one moment and then at a later moment, they would both appear to have a displacement pointing in the direction of their motion. Meanwhile, the objects in the background will appear to have no displacement, since they are stationary.
• In the reference frame where you are in one of the cars, if you looked outside the window at one moment and then at a later moment, the other car would appear to have no displacement, while the objects in the background would appear to have a displacement in the direction opposite the direction of the displacement of the cars in the other reference frame.

Velocity

Displacement gives us a rough description of how a point particle’s position differs between two moments. However, it does not factor in the amount of time between these two moments. In other words, it tells us nothing about how fast the particle’s position changes. This is why we need the notion of velocity.

Definition: Average Velocity

Let $\mathcal{R}$ be a reference frame and let $t_1$ and $t_2\ge t_1$ be two moments in time.

The average velocity of a point particle $p$ during the interval $\Delta t$ between $t_1$ and $t_2>t_1$ is the displacement of $p$ during $\Delta t$ divided by the length of the interval:

$${\mathbf{v}}_{\text{avg}}\stackrel{\text{def}}{=}\frac{\Delta \mathbf{r}}{\Delta t}=\frac{1}{t_2-t_1}\Delta \mathbf{r}$$

NOTATION

$$\bar{\mathbf{v}}{\mathbf{v}}_{\text{avg}}{\mathbf{v}}_{\text{avg}}{\vec{v}}_{\text{avg}}$$

Definition: Average Speed

The average speed of $p$ is the magnitude of its average velocity.

NOTATION

$$\bar{v}$$

Average velocity gives us a rough idea of how rapid the change in $p$‘s position between $t_1$ and $t_2$, but we are often interested in how fast this change is during very short intervals.

Definition: Instantaneous Velocity

Let $\mathcal{R}$ be a reference frame with origin $O$ and let $t^{\ast }$ be a moment in time.

The instantaneous velocity of a point particle $p$ at $t^{\ast }$ is the derivative of its position at $t^{\ast }$ with respect to time:

$$\mathbf{v}(t^{\ast })\stackrel{\text{def}}{=}\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t^{\ast })$$

Tip: Instantaneous Velocity and Average Velocity

Using the definition of the derivative, we see that the instantaneous of $p$ at $t^{\ast }$ is just its average velocity within an infinitesimally small interval after $t^{\ast }$:

$$\mathbf{v}(t^{\ast })=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t^{\ast })=\underset{\Delta t\to 0}{\lim }\frac{\mathbf{r}(t^{\ast }+\Delta t)-\mathbf{r}(t^{\ast })}{\Delta t}$$

Definition: Instantaneous Speed

The instantaneous speed of $p$ is the magnitude of its instantaneous velocity.

NOTE

When one says “velocity”, they usually mean the instantaneous velocity.

Velocity (both average and instantaneous) is dependent on the reference frame, just like position and displacement. Using the previous example:

• In the reference frame where you are on the side of the road, both cars will appear to have the same non-zero velocity. Meanwhile, the objects in the background will appear to have no velocity, since they are stationary.
• In the reference frame where you are in one of the cars, the other car would appear to have no velocity, while the objects in the background would appear to have the same non-zero velocity in the direction opposite the direction of the velocity of the cars in the other reference frame.

Theorem: Position from Velocity

Let $\mathcal{R}$ be a reference frame and let $t_0$ and $t^{\ast }\ge t_0$ be two moments in time.

The position of a point particle $p$ at $t^{\ast }$ is given by the integral of $p$‘s instantaneous velocity from $t_0$ to $t^{\ast }$:

$$\mathbf{r}(t^{\ast })={\int }_{t_0}^{t^{\ast }}\mathbf{v}(t)\mathrm{dt}$$

PROOF

TODO

Path Length

Knowing the speed of a point particle at each moment during a specific time interval allows us to give a sensible definition of the total distance through space it has covered within said interval.

Definition: Path Length

Let $\mathcal{R}$ be a reference frame with origin $O$ and let $t_1$ and $t_2\ge t_1$ be two moments in time.

The path length which a point particle $p$ covers during the interval between $t_1$ and $t_2$ is the value of the Riemann integral of its instantaneous speed from $t_1$ to $t_2$:

$${\int }_{t_1}^{t_2}||\mathbf{v}(t)||\mathrm{dt}$$

Path length is also dependent on the reference frame. Again, using the previous example:

• In the reference frame where you are on the side of the road, both cars would traverse the same path length in a given time interval because they are moving with the same non-zero velocity. Meanwhile, the objects in the background would appear have zero path length, since they would be stationary.
• In the reference frame where you are in one of the cars, the other car would appear to have cover zero path length, while the objects in the background would appear to cover some non-zero path length, since they would appear to be moving with the same non-zero velocity. In this case, the path length covered by the background objects would be the same as the path length covered by the cars in the other reference frame.

Acceleration

Just in the same way we use velocity to get an idea of how the position of a point particle changes, we almost always need a description of how velocity itself changes as well.

Definition: Average Acceleration

Let $\mathcal{R}$ be a reference frame with origin $O$ and let $t_1$ and $t_2\ge t_1$ be two moments in time.

The average acceleration of a point particle $p$ during the interval $\Delta t$ between $t_1$ and $t_2>t_1$ is the difference in the instantaneous velocity of $p$ between $t_1$ and $t_2>t_1$ divided by the duration of $\Delta t$:

$${\mathbf{a}}_{\text{avg}}\stackrel{\text{def}}{=}\frac{\Delta \mathbf{v}}{\Delta t}=\frac{\mathbf{v}(t_2)-\mathbf{v}(t_1)}{t_2-t_1}$$

NOTATION

$${\mathbf{a}}_{\text{avg}}{\mathbf{a}}_{\text{avg}}{\vec{a}}_{\text{avg}}\bar{\mathbf{a}}$$

Definition: Instantaneous Acceleration

Let $\mathcal{R}$ be a reference frame and let $t^{\ast }$ be a moment in time.

The instantaneous Acceleration of a point particle $p$ $p$ at $t^{\ast }$ is the derivative of $p$‘s instantaneous velocity at $t^{\ast }$ with respect to time:

$$\mathbf{a}(t^{\ast })\stackrel{\text{def}}{=}\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}(t^{\ast })$$

Tip: Instantaneous Acceleration and Average Acceleration

Using the definition of the derivative, we see that the instantaneous acceleration of $p$ at the moment $t^{\ast }$ is just its average acceleration within an infinitesimally small interval after $t^{\ast }$:

$$\mathbf{a}(t^{\ast })=\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}(t^{\ast })=\underset{\Delta t\to 0}{\lim }\frac{\mathbf{v}(t^{\ast }+\Delta t)-\mathbf{v}(t^{\ast })}{\Delta t}$$

NOTE

When one says “acceleration”, they usually mean the instantaneous acceleration.

Although it is possible to consider changes in acceleration and changes in the changes in acceleration and so on, we rarely need to concern ourselves with them.

Theorem: Velocity from Acceleration

Let $\mathcal{R}$ be a reference frame and let $t_0$ and $t^{\ast }\ge t_0$ be two moments in time.

The velocity of a point particle $p$ at $t^{\ast }$ is given by the integral of $p$‘s instantaneous acceleration from $t_0$ to $t^{\ast }$:

$$\mathbf{v}(t^{\ast })={\int }_{t_0}^{t^{\ast }}\mathbf{a}(t)\mathrm{dt}$$

PROOF

TODO

Theorem: Motion with Constant Acceleration

Let $\mathcal{R}$ be a reference frame and let $t_0$ and $t\ge t_0$ be two moments in time.

If a point particle $p$‘s instantaneous acceleration $\mathbf{a}$ does not change as $p$ moves, then the velocity and position of $p$ at $t$ are given by the following equations:

$$\begin{array}{rl}& \mathbf{v}(t)=\mathbf{v}(t_0)+t\mathbf{a}\\ & \mathbf{r}(t)=\mathbf{r}(t_0)+t\mathbf{v}(t_0)+\frac{1}{2}{t}^2\mathbf{a}\end{array}$$

Moreover:

$$\begin{array}{rl}& {\mathbf{v}}_{\text{avg}}=\frac{\mathbf{v}(t_0)+\mathbf{v}(t)}{2}\\ & \mathbf{r}(t)=\mathbf{r}(t_0)+(t-t_0){\mathbf{v}}_{\text{avg}}\\ & ||\mathbf{v}(t)|{|}^2=||\mathbf{v}(t_0)|{|}^2+2\mathbf{a}\cdot (\mathbf{r}(t)-\mathbf{r}(t_0))\end{array}$$

PROOF

We need to prove five things:

• (1) $\mathbf{v}(t)=\mathbf{v}(t_0)+t\mathbf{a}$
• (2) $\mathbf{r}(t)=\mathbf{r}(t_0)+t\mathbf{v}(t_0)+\frac{1}{2}{t}^2\mathbf{a}$
• (3) ${\mathbf{v}}_{\text{avg}}=\frac{\mathbf{v}(t_0)+\mathbf{v}(t)}{2}$
• (4) $\mathbf{r}(t)=\mathbf{r}(t_0)+(t-t_0){\mathbf{v}}_{\text{avg}}$
• (5) $||\mathbf{v}(t)|{|}^2=||\mathbf{v}(t_0)|{|}^2+2\mathbf{a}\cdot (\mathbf{r}(t)-\mathbf{r}(t_0))$

Proof of (1): TODO

Proof of (2): TODO

Proof of (3): TODO

Proof of (4): TODO

Proof of (5): TODO