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) by the time interval. This ratio is called the average velocity over that time interval and is defined as

$${\displaystyle \mathbf {\bar {v}} ={\frac {\Delta \mathbf {r} }{\Delta t}}={\frac {\Delta x}{\Delta t}}{\hat {\mathbf {x} }}+{\frac {\Delta y}{\Delta t}}{\hat {\mathbf {y} }}+{\frac {\Delta z}{\Delta t}}{\hat {\mathbf {z} }}={\bar {v}}_{x}{\hat {\mathbf {x} }}+{\bar {v}}_{y}{\hat {\mathbf {y} }}+{\bar {v}}_{z}{\hat {\mathbf {z} }}\,}$$

where ${\displaystyle \Delta \mathbf {r} }$ is the displacement vector during the time interval ${\displaystyle \Delta t}$. In the limit that the time interval ${\displaystyle \Delta t}$ approaches zero, the average velocity approaches the instantaneous velocity, defined as the time derivative of the position vector,

$${\displaystyle \mathbf {v} =\lim _{\Delta t\to 0}{\frac {\Delta \mathbf {r} }{\Delta t}}={\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}=v_{x}{\hat {\mathbf {x} }}+v_{y}{\hat {\mathbf {y} }}+v_{z}{\hat {\mathbf {z} }}.}$$

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:

$${\displaystyle v=|\mathbf {v} |={\frac {{\text{d}}s}{{\text{d}}t}},}$$

where ${\displaystyle s}$ is the arc-length measured along the trajectory of the particle. This arc-length must always increase as the particle moves. Hence, ${\displaystyle {\text{d}}s/{\text{d}}t}$ 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.

$${\displaystyle \mathbf {\bar {a}} ={\frac {\Delta \mathbf {\bar {v}} }{\Delta t}}={\frac {\Delta {\bar {v}}_{x}}{\Delta t}}{\hat {\mathbf {x} }}+{\frac {\Delta {\bar {v}}_{y}}{\Delta t}}{\hat {\mathbf {y} }}+{\frac {\Delta {\bar {v}}_{z}}{\Delta t}}{\hat {\mathbf {z} }}={\bar {a}}_{x}{\hat {\mathbf {x} }}+{\bar {a}}_{y}{\hat {\mathbf {y} }}+{\bar {a}}_{z}{\hat {\mathbf {z} }}\,}$$

where Δv is the average velocity and Δt 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,

$${\displaystyle \mathbf {a} =\lim _{\Delta t\to 0}{\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {{\text{d}}\mathbf {v} }{{\text{d}}t}}=a_{x}{\hat {\mathbf {x} }}+a_{y}{\hat {\mathbf {y} }}+a_{z}{\hat {\mathbf {z} }}.}$$

Alternatively,

$${\displaystyle \mathbf {a} =\lim _{(\Delta t)^{2}\to 0}{\frac {\Delta \mathbf {r} }{(\Delta t)^{2}}}={\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}=a_{x}{\hat {\mathbf {x} }}+a_{y}{\hat {\mathbf {y} }}+a_{z}{\hat {\mathbf {z} }}.}$$

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:

$${\displaystyle |\mathbf {a} |=|{\dot {\mathbf {v} }}|={\frac {{\text{d}}v}{{\text{d}}t}}.}$$

Relative position 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

${\displaystyle \mathbf {r} _{A/B}=\mathbf {r} _{A}-\mathbf {r} _{B}}$

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

If point A has position components ${\displaystyle \mathbf {r} _{A}=\left(x_{A},y_{A},z_{A}\right)}$

and point B has position components ${\displaystyle \mathbf {r} _{B}=\left(x_{B},y_{B},z_{B}\right)}$

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

Relative velocity

Relative velocities between two particles in classical mechanics.

The velocity of one point relative to another is simply the difference between their velocities

$${\displaystyle \mathbf {v} _{A/B}=\mathbf {v} _{A}-\mathbf {v} _{B}}$$

which is the difference between the components of their velocities.

If point A has velocity components ${\displaystyle \mathbf {v} _{A}=\left(v_{A_{x}},v_{A_{y}},v_{A_{z}}\right)}$ and point B has velocity components ${\displaystyle \mathbf {v} _{B}=\left(v_{B_{x}},v_{B_{y}},v_{B_{z}}\right)}$ then the velocity of point A relative to point B is the difference between their components: ${\displaystyle \mathbf {v} _{A/B}=\mathbf {v} _{A}-\mathbf {v} _{B}=\left(v_{A_{x}}-v_{B_{x}},v_{A_{y}}-v_{B_{y}},v_{A_{z}}-v_{B_{z}}\right)}$

Alternatively, this same result could be obtained by computing the time derivative of the relative position vector r_B/A.

Relative acceleration

The acceleration of one point C relative to another point B is simply the difference between their accelerations.

$${\displaystyle \mathbf {a} _{C/B}=\mathbf {a} _{C}-\mathbf {a} _{B}}$$

which is the difference between the components of their accelerations.

If point C has acceleration components ${\displaystyle \mathbf {a} _{C}=\left(a_{C_{x}},a_{C_{y}},a_{C_{z}}\right)}$ and point B has acceleration components ${\displaystyle \mathbf {a} _{B}=\left(a_{B_{x}},a_{B_{y}},a_{B_{z}}\right)}$ then the acceleration of point C relative to point B is the difference between their components: ${\displaystyle \mathbf {a} _{C/B}=\mathbf {a} _{C}-\mathbf {a} _{B}=\left(a_{C_{x}}-a_{B_{x}},a_{C_{y}}-a_{B_{y}},a_{C_{z}}-a_{B_{z}}\right)}$

Alternatively, this same result could be obtained by computing the second time derivative of the relative position vector r_B/A.^[13]

Assuming that the initial conditions of the position, ${\displaystyle \mathbf {r} _{0}}$, and velocity ${\displaystyle \mathbf {v} _{0}}$ at time ${\displaystyle t=0}$ are known, the first integration yields the velocity of the particle as a function of time.^{[lower-alpha 1]}

$${\displaystyle \mathbf {v} (t)=\mathbf {v} _{0}+\int _{0}^{t}\mathbf {a} \,{\text{d}}\tau =\mathbf {v} _{0}+\mathbf {a} t.}$$

A second integration yields its path (trajectory),

$${\displaystyle \mathbf {r} (t)=\mathbf {r} _{0}+\int _{0}^{t}\mathbf {v} (\tau )\,{\text{d}}\tau =\mathbf {r} _{0}+\int _{0}^{t}\left(\mathbf {v} _{0}+\mathbf {a} \tau \right){\text{d}}\tau =\mathbf {r} _{0}+\mathbf {v} _{0}t+{\tfrac {1}{2}}\mathbf {a} t^{2}.}$$

Additional relations between displacement, velocity, acceleration, and time can be derived. Since the acceleration is constant,

$${\displaystyle \mathbf {a} ={\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {\mathbf {v} -\mathbf {v} _{0}}{t}}}$$

can be substituted into the above equation to give:

$${\displaystyle \mathbf {r} (t)=\mathbf {r} _{0}+\left({\frac {\mathbf {v} +\mathbf {v} _{0}}{2}}\right)t.}$$

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

$${\displaystyle t={\frac {\mathbf {v} -\mathbf {v} _{0}}{\mathbf {a} }}}$$

$${\displaystyle \left(\mathbf {r} -\mathbf {r} _{0}\right)\cdot \mathbf {a} =\left(\mathbf {v} -\mathbf {v} _{0}\right)\cdot {\frac {\mathbf {v} +\mathbf {v} _{0}}{2}}\ ,}$$

where ${\displaystyle \cdot }$ denotes the dot product, which is appropriate as the products are scalars rather than vectors.

$${\displaystyle 2\left(\mathbf {r} -\mathbf {r} _{0}\right)\cdot \mathbf {a} =|\mathbf {v} |^{2}-|\mathbf {v} _{0}|^{2}.}$$

The dot product can be replaced by the cosine of the angle α between the vectors (see Geometric interpretation of the dot product for more details) and the vectors by their magnitudes, in which case:

$${\displaystyle 2\left|\mathbf {r} -\mathbf {r} _{0}\right|\left|\mathbf {a} \right|\cos \alpha =|\mathbf {v} |^{2}-|\mathbf {v} _{0}|^{2}.}$$

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 (α) is 0, so ${\displaystyle \cos 0=1}$, and

$${\displaystyle |\mathbf {v} |^{2}=|\mathbf {v} _{0}|^{2}+2\left|\mathbf {a} \right|\left|\mathbf {r} -\mathbf {r} _{0}\right|.}$$

This can be simplified using the notation for the magnitudes of the vectors ${\displaystyle |\mathbf {a} |=a,|\mathbf {v} |=v,|\mathbf {r} -\mathbf {r} _{0}|=\Delta r}$ where ${\displaystyle \Delta r}$ can be any curvaceous path taken as the constant tangential acceleration is applied along that path, so

$${\displaystyle v^{2}=v_{0}^{2}+2a\Delta r.}$$

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 ${\textstyle \Delta r=\int v\,{\text{d}}t}$ or ${\displaystyle \Delta r}$ is the area under a velocity–time graph.^[15]

Velocity Time physics graph

We can take ${\displaystyle \Delta r}$ by adding the top area and the bottom area. The bottom area is a rectangle, and the area of a rectangle is the ${\displaystyle A\cdot B}$ where ${\displaystyle A}$ is the width and ${\displaystyle B}$ is the height.^[16] In this case ${\displaystyle A=t}$ and ${\displaystyle B=v_{0}}$ (the ${\displaystyle A}$ here is different from the acceleration ${\displaystyle a}$). This means that the bottom area is ${\displaystyle tv_{0}}$. Now let's find the top area (a triangle). The area of a triangle is ${\textstyle {\frac {1}{2}}BH}$ where ${\displaystyle B}$ is the base and ${\displaystyle H}$ is the height.^[17] In this case, ${\displaystyle B=t}$ and ${\displaystyle H=at}$ or ${\textstyle A={\frac {1}{2}}BH={\frac {1}{2}}att={\frac {1}{2}}at^{2}={\frac {at^{2}}{2}}}$. Adding ${\displaystyle v_{0}t}$ and ${\textstyle {\frac {at^{2}}{2}}}$ results in the equation ${\displaystyle \Delta r}$ results in the equation ${\textstyle \Delta r=v_{0}t+{\frac {at^{2}}{2}}}$.^[18] This equation is applicable when the final velocity v is unknown.

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_θ that increases the rate of rotation: dω/dt = |a_θ|/R.

Constant radius

If the trajectory of the particle is constrained to lie on a cylinder, then the radius r is constant and the velocity and acceleration vectors simplify. The velocity of v_P is the time derivative of the trajectory r(t),

$${\displaystyle \mathbf {v} _{P}={\frac {\text{d}}{{\text{d}}t}}\left(r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }}\right)=r\omega {\hat {\mathbf {\theta } }}+v_{z}{\hat {\mathbf {z} }}=v{\hat {\mathbf {\theta } }}+v_{z}{\hat {\mathbf {z} }}.}$$

Planar circular trajectories

Each particle on the wheel travels in a planar circular trajectory (Kinematics of Machinery, 1876).^[19]

A special case of a particle trajectory on a circular cylinder occurs when there is no movement along the z axis:

$${\displaystyle \mathbf {r} (t)=r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }},}$$

where r and z₀ are constants. In this case, the velocity v_P is given by:

$${\displaystyle \mathbf {v} _{P}={\frac {\text{d}}{{\text{d}}t}}\left(r{\hat {\mathbf {r} }}+z{\hat {\mathbf {z} }}\right)=r\omega {\hat {\mathbf {\theta } }}=v{\hat {\mathbf {\theta } }},}$$

where ${\displaystyle \omega }$ is the angular velocity of the unit vector θ^{^} around the z axis of the cylinder.

The acceleration a_P of the particle P is now given by:

$${\displaystyle \mathbf {a} _{P}={\frac {{\text{d}}(v{\hat {\mathbf {\theta } }})}{{\text{d}}t}}=a{\hat {\mathbf {\theta } }}-v\theta {\hat {\mathbf {r} }}.}$$

The components

$${\displaystyle a_{r}=-v\theta ,\quad a_{\theta }=a,}$$

are called, respectively, the radial and tangential components of acceleration.

The notation for angular velocity and angular acceleration is often defined as

$${\displaystyle \omega ={\dot {\theta }},\quad \alpha ={\ddot {\theta }},}$$

so the radial and tangential acceleration components for circular trajectories are also written as

$${\displaystyle a_{r}=-r\omega ^{2},\quad a_{\theta }=r\alpha .}$$

Displacements and motion

The movement of each of the components of the Boulton & Watt Steam Engine (1784) is modeled by a continuous set of rigid displacements.

The position of one component of a mechanical system relative to another is defined by introducing a reference frame, say M, on one that moves relative to a fixed frame, F, on the other. The rigid transformation, or displacement, of M relative to F defines the relative position of the two components. A displacement consists of the combination of a rotation and a translation.

The set of all displacements of M relative to F is called the configuration space of M. A smooth curve from one position to another in this configuration space is a continuous set of displacements, called the motion of M relative to F. The motion of a body consists of a continuous set of rotations and translations.

Matrix representation

The combination of a rotation and translation in the plane R² can be represented by a certain type of 3×3 matrix known as a homogeneous transform. The 3×3 homogeneous transform is constructed from a 2×2 rotation matrix A(φ) and the 2×1 translation vector d = (d_x, d_y), as:

$${\displaystyle [T(\phi ,\mathbf {d} )]={\begin{bmatrix}A(\phi )&\mathbf {d} \\\mathbf {0} &1\end{bmatrix}}={\begin{bmatrix}\cos \phi &-\sin \phi &d_{x}\\\sin \phi &\cos \phi &d_{y}\\0&0&1\end{bmatrix}}.}$$

These homogeneous transforms perform rigid transformations on the points in the plane z = 1, that is, on points with coordinates r = (x, y, 1).

In particular, let r define the coordinates of points in a reference frame M coincident with a fixed frame F. Then, when the origin of M is displaced by the translation vector d relative to the origin of F and rotated by the angle φ relative to the x-axis of F, the new coordinates in F of points in M are given by:

$${\displaystyle \mathbf {P} =[T(\phi ,\mathbf {d} )]\mathbf {r} ={\begin{bmatrix}\cos \phi &-\sin \phi &d_{x}\\\sin \phi &\cos \phi &d_{y}\\0&0&1\end{bmatrix}}{\begin{bmatrix}x\\y\\1\end{bmatrix}}.}$$

Homogeneous transforms represent affine transformations. This formulation is necessary because a translation is not a linear transformation of R². However, using projective geometry, so that R² is considered a subset of R³, translations become affine linear transformations.^[21]

Position

This allows the description of a rotation as the angular position of a planar reference frame M relative to a fixed F about this shared z-axis. Coordinates p = (x, y) in M are related to coordinates P = (X, Y) in F by the matrix equation:

$${\displaystyle \mathbf {P} (t)=[A(t)]\mathbf {p} ,}$$

where

$${\displaystyle [A(t)]={\begin{bmatrix}\cos(\theta (t))&-\sin(\theta (t))\\\sin(\theta (t))&\cos(\theta (t))\end{bmatrix}},}$$

is the rotation matrix that defines the angular position of M relative to F as a function of time.

Velocity

If the point p does not move in M, its velocity in F is given by

$${\displaystyle \mathbf {v} _{P}={\dot {\mathbf {P} }}=[{\dot {A}}(t)]\mathbf {p} .}$$

It is convenient to eliminate the coordinates p and write this as an operation on the trajectory P(t),

$${\displaystyle \mathbf {v} _{P}=[{\dot {A}}(t)][A(t)^{-1}]\mathbf {P} =[\Omega ]\mathbf {P} ,}$$

where the matrix

$${\displaystyle [\Omega ]={\begin{bmatrix}0&-\omega \\\omega &0\end{bmatrix}},}$$

is known as the angular velocity matrix of M relative to F. The parameter ω is the time derivative of the angle θ, that is:

$${\displaystyle \omega ={\frac {{\text{d}}\theta }{{\text{d}}t}}.}$$

Acceleration

The acceleration of P(t) in F is obtained as the time derivative of the velocity,

$${\displaystyle \mathbf {A} _{P}={\ddot {P}}(t)=[{\dot {\Omega }}]\mathbf {P} +[\Omega ]{\dot {\mathbf {P} }},}$$

which becomes

$${\displaystyle \mathbf {A} _{P}=[{\dot {\Omega }}]\mathbf {P} +[\Omega ][\Omega ]\mathbf {P} ,}$$

where

$${\displaystyle [{\dot {\Omega }}]={\begin{bmatrix}0&-\alpha \\\alpha &0\end{bmatrix}},}$$

is the angular acceleration matrix of M on F, and

$${\displaystyle \alpha ={\frac {{\text{d}}^{2}\theta }{{\text{d}}t^{2}}}.}$$

The description of rotation then involves these three quantities:

• Angular position: the oriented distance from a selected origin on the rotational axis to a point of an object is a vector r(t) locating the point. The vector r(t) has some projection (or, equivalently, some component) r_⊥(t) on a plane perpendicular to the axis of rotation. Then the angular position of that point is the angle θ from a reference axis (typically the positive x-axis) to the vector r_⊥(t) in a known rotation sense (typically given by the right-hand rule).
• Angular velocity: the angular velocity ω is the rate at which the angular position θ changes with respect to time t:
  $${\displaystyle \omega ={\frac {{\text{d}}\theta }{{\text{d}}t}}}$$

  
  The angular velocity is represented in Figure 1 by a vector Ω pointing along the axis of rotation with magnitude ω and sense determined by the direction of rotation as given by the right-hand rule.
• Angular acceleration: the magnitude of the angular acceleration α is the rate at which the angular velocity ω changes with respect to time t:
  $${\displaystyle \alpha ={\frac {{\text{d}}\omega }{{\text{d}}t}}}$$

  

The equations of translational kinematics can easily be extended to planar rotational kinematics for constant angular acceleration with simple variable exchanges:

$${\displaystyle \omega _{\mathrm {f} }=\omega _{\mathrm {i} }+\alpha t\!}$$

$${\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }=\omega _{\mathrm {i} }t+{\tfrac {1}{2}}\alpha t^{2}}$$

$${\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }={\tfrac {1}{2}}(\omega _{\mathrm {f} }+\omega _{\mathrm {i} })t}$$

$${\displaystyle \omega _{\mathrm {f} }^{2}=\omega _{\mathrm {i} }^{2}+2\alpha (\theta _{\mathrm {f} }-\theta _{\mathrm {i} }).}$$

Here θ_i and θ_f are, respectively, the initial and final angular positions, ω_i and ω_f are, respectively, the initial and final angular velocities, and α is the constant angular acceleration. Although position in space and velocity in space are both true vectors (in terms of their properties under rotation), as is angular velocity, angle itself is not a true vector.

Position

In order to define these formulas, the movement of a component B of a mechanical system is defined by the set of rotations [A(t)] and translations d(t) assembled into the homogeneous transformation [T(t)]=[A(t), d(t)]. If p is the coordinates of a point P in B measured in the moving reference frame M, then the trajectory of this point traced in F is given by:

$${\displaystyle \mathbf {P} (t)=[T(t)]\mathbf {p} ={\begin{bmatrix}\mathbf {P} \\1\end{bmatrix}}={\begin{bmatrix}A(t)&\mathbf {d} (t)\\0&1\end{bmatrix}}{\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}.}$$

This notation does not distinguish between P = (X, Y, Z, 1), and P = (X, Y, Z), which is hopefully clear in context.

This equation for the trajectory of P can be inverted to compute the coordinate vector p in M as:

$${\displaystyle \mathbf {p} =[T(t)]^{-1}\mathbf {P} (t)={\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}={\begin{bmatrix}A(t)^{\text{T}}&-A(t)^{\text{T}}\mathbf {d} (t)\\0&1\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}.}$$

This expression uses the fact that the transpose of a rotation matrix is also its inverse, that is:

$${\displaystyle [A(t)]^{\text{T}}[A(t)]=I.\!}$$

Velocity

The velocity of the point P along its trajectory P(t) is obtained as the time derivative of this position vector,

$${\displaystyle \mathbf {v} _{P}=[{\dot {T}}(t)]\mathbf {p} ={\begin{bmatrix}\mathbf {v} _{P}\\0\end{bmatrix}}=\left({\frac {d}{dt}}{\begin{bmatrix}A(t)&\mathbf {d} (t)\\0&1\end{bmatrix}}\right){\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}={\begin{bmatrix}{\dot {A}}(t)&{\dot {\mathbf {d} }}(t)\\0&0\end{bmatrix}}{\begin{bmatrix}\mathbf {p} \\1\end{bmatrix}}.}$$

The dot denotes the derivative with respect to time; because p is constant, its derivative is zero.

This formula can be modified to obtain the velocity of P by operating on its trajectory P(t) measured in the fixed frame F. Substituting the inverse transform for p into the velocity equation yields:

$${\displaystyle {\begin{aligned}\mathbf {v} _{P}&=[{\dot {T}}(t)][T(t)]^{-1}\mathbf {P} (t)\\[4pt]&={\begin{bmatrix}\mathbf {v} _{P}\\0\end{bmatrix}}={\begin{bmatrix}{\dot {A}}&{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}{\begin{bmatrix}A&\mathbf {d} \\0&1\end{bmatrix}}^{-1}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[4pt]&={\begin{bmatrix}{\dot {A}}&{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}A^{-1}{\begin{bmatrix}1&-\mathbf {d} \\0&A\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[4pt]&={\begin{bmatrix}{\dot {A}}A^{-1}&-{\dot {A}}A^{-1}\mathbf {d} +{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[4pt]&={\begin{bmatrix}{\dot {A}}A^{\text{T}}&-{\dot {A}}A^{\text{T}}\mathbf {d} +{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}{\begin{bmatrix}\mathbf {P} (t)\\1\end{bmatrix}}\\[6pt]\mathbf {v} _{P}&=[S]\mathbf {P} .\end{aligned}}}$$

The matrix [S] is given by:

$${\displaystyle [S]={\begin{bmatrix}\Omega &-\Omega \mathbf {d} +{\dot {\mathbf {d} }}\\0&0\end{bmatrix}}}$$

where

$${\displaystyle [\Omega ]={\dot {A}}A^{\text{T}},}$$

is the angular velocity matrix.

Multiplying by the operator [S], the formula for the velocity v_P takes the form:

$${\displaystyle \mathbf {v} _{P}=[\Omega ](\mathbf {P} -\mathbf {d} )+{\dot {\mathbf {d} }}=\omega \times \mathbf {R} _{P/O}+\mathbf {v} _{O},}$$

where the vector ω is the angular velocity vector obtained from the components of the matrix [Ω]; the vector

$${\displaystyle \mathbf {R} _{P/O}=\mathbf {P} -\mathbf {d} ,}$$

is the position of P relative to the origin O of the moving frame M; and

$${\displaystyle \mathbf {v} _{O}={\dot {\mathbf {d} }},}$$

is the velocity of the origin O.

Acceleration

The acceleration of a point P in a moving body B is obtained as the time derivative of its velocity vector:

$${\displaystyle \mathbf {A} _{P}={\frac {d}{dt}}\mathbf {v} _{P}={\frac {d}{dt}}\left([S]\mathbf {P} \right)=[{\dot {S}}]\mathbf {P} +[S]{\dot {\mathbf {P} }}=[{\dot {S}}]\mathbf {P} +[S][S]\mathbf {P} .}$$

This equation can be expanded firstly by computing

$${\displaystyle [{\dot {S}}]={\begin{bmatrix}{\dot {\Omega }}&-{\dot {\Omega }}\mathbf {d} -\Omega {\dot {\mathbf {d} }}+{\ddot {\mathbf {d} }}\\0&0\end{bmatrix}}={\begin{bmatrix}{\dot {\Omega }}&-{\dot {\Omega }}\mathbf {d} -\Omega \mathbf {v} _{O}+\mathbf {A} _{O}\\0&0\end{bmatrix}}}$$

and

$${\displaystyle [S]^{2}={\begin{bmatrix}\Omega &-\Omega \mathbf {d} +\mathbf {v} _{O}\\0&0\end{bmatrix}}^{2}={\begin{bmatrix}\Omega ^{2}&-\Omega ^{2}\mathbf {d} +\Omega \mathbf {v} _{O}\\0&0\end{bmatrix}}.}$$

The formula for the acceleration A_P can now be obtained as:

$${\displaystyle \mathbf {A} _{P}={\dot {\Omega }}(\mathbf {P} -\mathbf {d} )+\mathbf {A} _{O}+\Omega ^{2}(\mathbf {P} -\mathbf {d} ),}$$

or

$${\displaystyle \mathbf {A} _{P}=\alpha \times \mathbf {R} _{P/O}+\omega \times \omega \times \mathbf {R} _{P/O}+\mathbf {A} _{O},}$$

where α is the angular acceleration vector obtained from the derivative of the angular velocity matrix;

$${\displaystyle \mathbf {R} _{P/O}=\mathbf {P} -\mathbf {d} ,}$$

is the relative position vector (the position of P relative to the origin O of the moving frame M); and

$${\displaystyle \mathbf {A} _{O}={\ddot {\mathbf {d} }}}$$

is the acceleration of the origin of the moving frame M.

Kinematic coupling

A kinematic coupling exactly constrains all 6 degrees of freedom.

Rolling without slipping

An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass:

$${\displaystyle {\boldsymbol {v}}_{G}(t)={\boldsymbol {\Omega }}\times {\boldsymbol {r}}_{G/O}.}$$

For the case of an object that does not tip or turn, this reduces to ${\displaystyle v=r\omega }$.

Inextensible cord

This is the case where bodies are connected by an idealized cord that remains in tension and cannot change length. The constraint is that the sum of lengths of all segments of the cord is the total length, and accordingly the time derivative of this sum is zero.^[23][24][25] A dynamic problem of this type is the pendulum. Another example is a drum turned by the pull of gravity upon a falling weight attached to the rim by the inextensible cord.^[26] An equilibrium problem (i.e. not kinematic) of this type is the catenary.^[27]

Kinematic pairs

Reuleaux called the ideal connections between components that form a machine kinematic pairs. He distinguished between higher pairs which were said to have line contact between the two links and lower pairs that have area contact between the links. J. Phillips shows that there are many ways to construct pairs that do not fit this simple classification.^[28]

Kinematic chains

Illustration of a four-bar linkage from Kinematics of Machinery, 1876

Rigid bodies ("links") connected by kinematic pairs ("joints") are known as kinematic chains. Mechanisms and robots are examples of kinematic chains. The degree of freedom of a kinematic chain is computed from the number of links and the number and type of joints using the mobility formula. This formula can also be used to enumerate the topologies of kinematic chains that have a given degree of freedom, which is known as type synthesis in machine design.