Acceleration

Let a particle travel a distance $s(t)$ as a function of time $t$ (here, $s$ can be thought of as the Arc Length of the curve traced out by the particle). The Speed (the Scalar Norm of the Vector Velocity) is then given by

$${ds\over dt}=\sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2+\left({dz\over dt}\right)^2}.$$ (1)

The acceleration is defined as the time Derivative of the Velocity, so the Scalar acceleration is given by

$$a \equiv {dv\over dt}$$ (2)

$$= {d^2s\over dt^2}$$ (3)

$$= {{dx\over dt}{d^2x\over dt^2}+{dy\over dt}{d^2y\over dt^2}+{dz\ov... ...{dx\over dt}\right)^2+\left({dy\over dt}\right)^2+\left({dz\over dt}\right)^2}}$$ (4)

$$= {dx\over ds}{d^2x\over dt^2}+{dy\over ds}{d^2y\over dt^2}+{dz\over ds}{d^2z\over dt^2}$$ (5)

$$= {d{\bf r}\over ds}\cdot{d^2{\bf r}\over dt^2}.$$ (6)

The Vector acceleration is given by

$${\bf a} \equiv {d{\bf v}\over dt} = {{d^2 {\bf r}}\over {dt^... ... {\bf T} + \kappa \left({{ds}\over{dt}}\right)^2 \hat {\bf N},$$ (7)

where $\hat {\bf T}$ is the Unit Tangent Vector, $\kappa$ the Curvature, $s$ the Arc Length, and $\hat {\bf N}$ the Unit Normal Vector.

Let a particle move along a straight Line so that the positions at times $t_1$, $t_2$, and $t_3$ are $s_1$, $s_2$, and $s_3$, respectively. Then the particle is uniformly accelerated with acceleration $a$ Iff

$$a\equiv 2\left[{(s_2-s_3)t_1+(s_3-s_1)t_2+(s_1-s_2)t_3\over (t_1-t_2)(t_2-t_3)(t_3-t_1)}\right]$$ (8)

is a constant (Klamkin 1995, 1996).

Consider the measurement of acceleration in a rotating reference frame. Apply the Rotation Operator

$$\tilde R \equiv \left({d \over {dt}}\right)_{\rm body}+\boldsymbol{\omega}\times$$ (9)

twice to the Radius Vector r and suppress the body notation,

$${\bf a}_{\rm space} = {\tilde R}^2 {\bf r} = \left({{d \over dt}+ \boldsymbol{\omega}\times}\right)^2 {\bf r}$$

$$= \left({{d \over dt}+\boldsymbol{\omega}\times}\right)\left({{d {\bf r} \over dt}+ \boldsymbol{\omega}\times {\bf r}}\right)$$

$$= {{d^2 \bf r} \over {dt^2}} + {d \over {dt}} ({\boldsymbol{\omega}... ...bf r} \over dt} + \boldsymbol{\omega}\times (\boldsymbol{\omega}\times {\bf r})$$

$$= {{d^2 \bf r} \over {dt^2}}+ \boldsymbol{\omega}\times {{d \bf r}\... ... \boldsymbol{\omega}}\over dt} + \boldsymbol{\omega}\times {{d \bf r} \over dt}$$

$$\phantom{=} + \boldsymbol{\omega}\times (\boldsymbol{\omega}\times {\bf r}).$$ (10)

Grouping terms and using the definitions of the Velocity ${\bf v} \equiv d {\bf r}/ {dt}$ and Angular Velocity $\boldsymbol{\alpha}\equiv {d\boldsymbol{\omega}/dt}$ give the expression

$${\bf a}_{\rm space} = {{d^2 \bf r} \over {dt^2}}+ 2\boldsymb... ...l{\omega}\times {\bf r}) + {\bf r} \times \boldsymbol{\alpha}.$$ (11)

Now, we can identify the expression as consisting of three terms

$${\bf a}_{\rm body} \equiv {{d^2 \bf r} \over {dt^2}},$$ (12)

$${\bf a}_{\rm Coriolis} \equiv 2\boldsymbol{\omega}\times {\bf v},$$ (13)

$${\bf a}_{\rm centrifugal} \equiv \boldsymbol{\omega}\times (\boldsymbol{\omega}\times {\bf r}),$$ (14)

a ``body'' acceleration, centrifugal acceleration, and Coriolis acceleration. Using these definitions finally gives

$${\bf a}_{\rm space} = {\bf a}_{\rm body} + {\bf a}_{\rm Cori... ...\bf a}_{\rm centrifugal} + {\bf r} \times \boldsymbol{\alpha},$$ (15)

where the fourth term will vanish in a uniformly rotating frame of reference (i.e., $\boldsymbol{\alpha}= {\bf0}$). The centrifugal acceleration is familiar to riders of merry-go-rounds, and the Coriolis acceleration is responsible for the motions of hurricanes on Earth and necessitates large trajectory corrections for intercontinental ballistic missiles.

See also Angular Acceleration, Arc Length, Jerk, Velocity

References

Klamkin, M. S. ``Problem 1481.'' Math. Mag. 68, 307, 1995.

Klamkin, M. S. ``A Characteristic of Constant Acceleration.'' Solution to Problem 1481. Math. Mag. 69, 308, 1996.