Editing Kinematics (section)

== Kinematics of a particle trajectory in a non-rotating frame of reference ==
[[File:Kinematics.svg|thumb|300px|Kinematic quantities of a classical particle: mass ''m'', position '''r''', velocity '''v''', acceleration '''a'''.]]
{{multiple image
|align = vertical
|width1 = 100
|image1 = Position vector plane polar coords.svg
|caption1 = Position vector '''r''', always points radially from the origin.
|width2 = 150
|image2 = Velocity vector plane polar coords.svg
|caption2 = Velocity vector '''v''', always tangent to the path of motion.
|width3 = 200
|image3 = Acceleration vector plane polar coords.svg
|caption3 = Acceleration vector '''a''', not parallel to the radial motion but offset by the angular and Coriolis accelerations, nor tangent to the path but offset by the centripetal and radial accelerations.
|footer = Kinematic vectors in plane polar coordinates. Notice the setup is not restricted to 2-d space, but a plane in any higher dimension.}}

Particle kinematics is the study of the trajectory of particles. The position of a particle is defined as the coordinate vector from the origin of a coordinate frame to the particle. For example, consider a tower 50&nbsp;m south from your home, where the coordinate frame is centered at your home, such that east is in the direction of the ''x''-axis and north is in the direction of the ''y''-axis, then the coordinate vector to the base of the tower is '''r''' = (0&nbsp;m, −50&nbsp;m, 0&nbsp;m). If the tower is 50&nbsp;m high, and this height is measured along the ''z''-axis, then the coordinate vector to the top of the tower is '''r''' = (0&nbsp;m, −50&nbsp;m, 50&nbsp;m).

In the most general case, a three-dimensional coordinate system is used to define the position of a particle. However, if the particle is constrained to move within a plane, a two-dimensional coordinate system is sufficient. All observations in physics are incomplete without being described with respect to a reference frame.

The position vector of a particle is a [[Euclidean vector|vector]] drawn from the origin of the [[reference frame]] to the particle. It expresses both the distance of the point from the origin and its direction from the origin. In three dimensions, the position vector <math>{\bf r}</math> can be expressed as
<math display="block">\mathbf r = (x,y,z) = x\hat\mathbf x + y\hat\mathbf y + z\hat\mathbf z,</math>
where <math>x</math>, <math>y</math>, and <math>z</math> are the [[Cartesian coordinates]] and <math> \hat\mathbf x</math>, <math>\hat\mathbf y</math> and <math>\hat\mathbf z</math> are the [[unit vectors]] along the <math>x</math>, <math>y</math>, and <math>z</math> coordinate axes, respectively. The magnitude of the position vector <math>\left|\mathbf r\right|</math> gives the distance between the point <math>\mathbf r</math> and the origin.
<math display="block">|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}.</math>
The [[direction cosine]]s of the position vector provide a quantitative measure of direction. In general, an object's position vector will depend on the frame of reference; different frames will lead to different values for the position vector.

The ''trajectory'' of a particle is a vector function of time, <math>\mathbf{r}(t)</math>, which defines the curve traced by the moving particle, given by
<math display="block"> \mathbf r(t) = x(t)\hat\mathbf x + y(t) \hat\mathbf y +z(t) \hat\mathbf z,</math>
where <math>x(t)</math>, <math>y(t)</math>, and <math>z(t)</math> describe each coordinate of the particle's position as a function of time.

[[File:Distancedisplacement.svg|thumb|300px|right|The distance travelled is always greater than or equal to the displacement.]]

===Velocity and speed===
The [[velocity]] of a particle is a vector quantity that describes the ''direction'' as well as the magnitude of motion of the particle. More mathematically, the rate of change of the position vector of a point with respect to time is the velocity of the point. Consider the ratio formed by dividing the difference of two positions of a particle ([[Displacement (geometry)|displacement]]) by the time interval. This ratio is called the [[average velocity]] over that time interval and is defined as<math display="block"> \mathbf\bar v = \frac{\Delta \mathbf r}{\Delta t} = \frac{\Delta x}{\Delta t}\hat\mathbf x + \frac{\Delta y}{\Delta t}\hat\mathbf y + \frac{\Delta z}{\Delta t}\hat\mathbf z =\bar v_x\hat\mathbf x + \bar v_y\hat\mathbf y + \bar v_z \hat\mathbf z \,</math>where <math>\Delta \mathbf{r}</math> is the displacement vector during the time interval <math>\Delta t</math>. In the limit that the time interval <math>\Delta t</math> approaches zero, the average velocity approaches the instantaneous velocity, defined as the time derivative of the position vector,
<math display="block"> \mathbf v
= \lim_{\Delta t\to 0}\frac{\Delta\mathbf{r}}{\Delta t}
= \frac{\text{d}\mathbf r}{\text{d}t}
= v_x\hat\mathbf x + v_y\hat\mathbf y + v_z \hat\mathbf z .</math>
Thus, a particle's velocity is the time rate of change of its position. Furthermore, this velocity is [[tangent]] to the particle's trajectory at every position along its path. In a non-rotating frame of reference, the derivatives of the coordinate directions are not considered as their directions and magnitudes are constants.

The [[speed]] of an object is the magnitude of its velocity. It is a scalar quantity:
<math display="block"> v=|\mathbf{v}|= \frac {\text{d}s}{\text{d}t},</math>
where <math>s</math> is the arc-length measured along the trajectory of the particle. This arc-length must always increase as the particle moves. Hence, <math>\frac{\text{d}s}{\text{d}t}</math> is non-negative, which implies that speed is also non-negative.

===Acceleration===
The velocity vector can change in magnitude and in direction or both at once. Hence, the acceleration accounts for both the rate of change of the magnitude of the velocity vector and the rate of change of direction of that vector. The same reasoning used with respect to the position of a particle to define velocity, can be applied to the velocity to define acceleration. The [[acceleration]] of a particle is the vector defined by the rate of change of the velocity vector. The [[average acceleration]] of a particle over a time interval is defined as the ratio.
<math display="block"> \mathbf\bar a = \frac{\Delta \mathbf\bar v}{\Delta t} = \frac{\Delta \bar v_x}{\Delta t}\hat\mathbf x + \frac{\Delta \bar v_y}{\Delta t}\hat\mathbf y + \frac{\Delta \bar v_z}{\Delta t}\hat\mathbf z =\bar a_x\hat\mathbf x + \bar a_y\hat\mathbf y + \bar a_z \hat\mathbf z \,</math>
where Δ'''v''' is the average velocity and Δ''t'' is the time interval.

The acceleration of the particle is the limit of the average acceleration as the time interval approaches zero, which is the time derivative,
<math display="block"> \mathbf a
= \lim_{\Delta t\to 0}\frac{\Delta\mathbf{v}}{\Delta t}
=\frac{\text{d}\mathbf v}{\text{d}t}
= a_x\hat\mathbf x + a_y\hat\mathbf y + a_z \hat\mathbf z  . </math>

Alternatively,
<math display="block"> \mathbf a
= \lim_{(\Delta t)^2 \to 0}\frac{\Delta\mathbf{r}}{(\Delta t)^2}
= \frac{\text{d}^2\mathbf r}{\text{d}t^2}
= a_x\hat\mathbf x + a_y\hat\mathbf y + a_z \hat\mathbf z . </math>

Thus, acceleration is the first derivative of the velocity vector and the second derivative of the position vector of that particle. In a non-rotating frame of reference, the derivatives of the coordinate directions are not considered as their directions and magnitudes are constants.

The magnitude of the [[acceleration]] of an object is the magnitude |'''a'''| of its acceleration vector. It is a scalar quantity:
<math display="block"> |\mathbf{a}| = |\dot{\mathbf{v}} | = \frac{\text{d}v}{\text{d}t}.</math>

===Relative position vector===
[[Displacement (vector)|A relative position vector]] is a vector that defines the position of one point relative to another. It is the difference in position of the two points.
The position of one point ''A'' relative to another point ''B'' is simply the difference between their positions

:<math>\mathbf{r}_{A/B} = \mathbf{r}_{A} - \mathbf{r}_{B} </math>

which is the difference between the components of their position vectors.

If point ''A'' has position components <math>\mathbf{r}_{A} = \left( x_{A}, y_{A}, z_{A} \right) </math>

and point ''B'' has position components <math>\mathbf{r}_{B} = \left( x_{B}, y_{B}, z_{B} \right) </math>

then the position of point ''A'' relative to point ''B'' is the difference between their components: <math>\mathbf{r}_{A/B} = \mathbf{r}_{A} - \mathbf{r}_{B} = \left( x_{A} - x_{B}, y_{A} - y_{B}, z_{A} - z_{B} \right) </math>

===Relative velocity===
{{main|Relative velocity}}
[[File:Relative velocity.svg|300px|thumb|Relative velocities between two particles in classical mechanics.]]

The velocity of one point relative to another is simply the difference between their velocities
<math display="block">\mathbf{v}_{A/B} = \mathbf{v}_{A} - \mathbf{v}_{B} </math>
which is the difference between the components of their velocities.

If point ''A'' has velocity components <math>\mathbf{v}_{A} = \left( v_{A_x}, v_{A_y}, v_{A_z} \right) </math> and point ''B'' has velocity components <math>\mathbf{v}_{B} = \left( v_{B_x}, v_{B_y}, v_{B_z} \right) </math> then the velocity of point ''A'' relative to point ''B'' is the difference between their components:
<math>\mathbf{v}_{A/B} = \mathbf{v}_{A} - \mathbf{v}_{B} = \left( v_{A_x} - v_{B_x}, v_{A_y} - v_{B_{y}}, v_{A_z} - v_{B_z} \right) </math>

Alternatively, this same result could be obtained by computing the time derivative of the relative position vector '''r'''<sub>B/A</sub>.

===Relative acceleration===

The acceleration of one point ''C'' relative to another point ''B'' is simply the difference between their accelerations.
<math display="block">\mathbf{a}_{C/B} = \mathbf{a}_{C} - \mathbf{a}_{B} </math>
which is the difference between the components of their accelerations.

If point ''C'' has acceleration components <math>\mathbf{a}_{C} = \left( a_{C_x}, a_{C_y}, a_{C_z} \right) </math>
and point ''B'' has acceleration components <math>\mathbf{a}_{B} = \left( a_{B_x}, a_{B_y}, a_{B_z} \right) </math>
then the acceleration of point ''C'' relative to point ''B'' is the difference between their components: <math>\mathbf{a}_{C/B} = \mathbf{a}_{C} - \mathbf{a}_{B} = \left( a_{C_x} - a_{B_x} , a_{C_y} - a_{B_y} , a_{C_z} - a_{B_z} \right) </math>

Assuming that the initial conditions of the position, <math>\mathbf{r}_0</math>, and velocity <math>\mathbf{v}_0</math> at time <math>t = 0</math> are known, the first integration yields the velocity of the particle as a function of time.<ref>{{Citation | title=2.4 Integration | date=2 June 2017 | url=https://www.youtube.com/watch?v=H7xmTMQ265s | archive-url=https://ghostarchive.org/varchive/youtube/20211113/H7xmTMQ265s| archive-date=2021-11-13 | url-status=live| publisher=MIT | language=en | access-date=2021-07-04}}{{cbignore}}</ref>
<math display="block">\mathbf{v}(t) = \mathbf{v}_0 + \int_0^t \mathbf{a}(\tau) \, \text{d}\tau</math>

Additional relations between displacement, velocity, acceleration, and time can be derived. If the acceleration is constant,
<math display="block">\mathbf{a} = \frac{\Delta\mathbf{v}}{\Delta t} = \frac{\mathbf{v}-\mathbf{v}_0}{ t } </math> can be substituted into the above equation to give:
<math display="block">\mathbf{r}(t) = \mathbf{r}_0 + \left(\frac{\mathbf{v} + \mathbf{v}_0}{2}\right) t .</math>

A relationship between velocity, position and acceleration without explicit time dependence can be obtained by solving the average acceleration for time and substituting and simplifying

<math display="block"> t = \frac{\mathbf{v}-\mathbf{v}_0}{ \mathbf{a} } </math>

<math display="block"> \left(\mathbf{r} - \mathbf{r}_0\right) \cdot \mathbf{a} = \left( \mathbf{v} - \mathbf{v}_0 \right) \cdot \frac{\mathbf{v} + \mathbf{v}_0}{2} \ , </math>
where <math> \cdot </math> denotes the [[dot product]], which is appropriate as the products are scalars rather than vectors.
<math display="block">2 \left(\mathbf{r} - \mathbf{r}_0\right) \cdot \mathbf{a} = |\mathbf{v}|^2 - |\mathbf{v}_0|^2.</math>

The dot product can be replaced by the cosine of the angle {{mvar|α}} between the vectors (see [[Dot product#Geometric definition|Geometric interpretation of the dot product]] for more details) and the vectors by their magnitudes, in which case:
<math display="block">2 \left|\mathbf{r} - \mathbf{r}_0\right| \left|\mathbf{a}\right| \cos \alpha = |\mathbf{v}|^2 - |\mathbf{v}_0|^2.</math>

In the case of acceleration always in the direction of the motion and the direction of motion should be in positive or negative, the angle between the vectors ({{mvar|α}}) is 0, so <math>\cos 0 = 1</math>, and
<math display="block"> |\mathbf{v}|^2= |\mathbf{v}_0|^2 + 2 \left|\mathbf{a}\right| \left|\mathbf{r}-\mathbf{r}_0\right|.</math>
This can be simplified using the notation for the magnitudes of the vectors <math>|\mathbf{a}|=a, |\mathbf{v}|=v, |\mathbf{r}-\mathbf{r}_0| = \Delta r </math>{{citation needed|date=April 2018}} where <math>\Delta r</math> can be any curvaceous path taken as the constant tangential acceleration is applied along that path{{citation needed|date=April 2018}}, so
<math display="block"> v^2= v_0^2 + 2a \Delta r.</math>
This reduces the parametric equations of motion of the particle to a Cartesian relationship of speed versus position. This relation is useful when time is unknown. We also know that <math display="inline">\Delta r = \int v \, \text{d}t</math> or <math>\Delta r</math> is the area under a velocity–time graph.<ref>https://www.youtube.com/watch?v=jLJLXka2wEM Crash course physics integrals</ref> [[File:Velocity Time physics graph.svg|thumb|Velocity Time physics graph]] We can take <math>\Delta r</math> by adding the top area and the bottom area. The bottom area is a rectangle, and the area of a rectangle is the <math>A \cdot B</math> where <math>A</math> is the width and <math>B</math> is the height. In this case <math>A = t</math> and <math>B = v_0</math> (the <math>A</math> here is different from the acceleration <math>a</math>). This means that the bottom area is <math>tv_0</math>. Now let's find the top area (a triangle). The area of a triangle is <math display="inline">\frac{1}{2} BH</math> where <math>B</math> is the base and <math>H</math> is the height.<ref>https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html Area of Triangles Without Right Angles</ref> In this case, <math>B = t</math> and <math>H = at</math> or <math display="inline">A = \frac{1}{2} BH = \frac{1}{2} att = \frac{1}{2} at^2 = \frac{at^2}{2}</math>. Adding <math>v_0 t</math> and <math display="inline">\frac{at^2}{2}</math> results in the equation <math>\Delta r</math> results in the equation <math display="inline">\Delta r = v_0 t + \frac{at^2}{2}</math>.<ref>{{Cite AV media |url=https://www4.uwsp.edu/physastr/kmenning/Phys203/eqs/kinematics.gif |title=kinematics.gif (508×368) |type=Image |language=en |access-date=3 November 2023}}</ref> This equation is applicable when the final velocity {{mvar|v}} is unknown.

[[File:Nonuniform circular motion.svg|thumb|250px|Figure 2: Velocity and acceleration for nonuniform circular motion: the velocity vector is tangential to the orbit, but the acceleration vector is not radially inward because of its tangential component '''a'''<sub>''θ''</sub> that increases the rate of rotation: d''ω''/d''t'' = <nowiki>|</nowiki>'''a'''<sub>''θ''</sub><nowiki>|</nowiki>/''R''.]]