Vector velocity and vector acceleration

Consider a body moving in 3 dimensions. Suppose that we know the Cartesian coordinates, $x$, $y$, and $z$, of this body as time, $t$, progresses. Let us consider how we can use this information to determine the body's instantaneous velocity and acceleration as functions of time.

The vector displacement of the body is given by

$${\bf r}(t) = [x(t), y(t), z(t)].$$ (43)

By analogy with the 1-dimensional equation (13), the body's vector velocity ${\bf v}=(v_x,v_y,v_z)$ is simply the derivative of ${\bf r}$ with respect to $t$. In other words,

$${\bf v}(t) = \lim_{{\mit\Delta} t\rightarrow 0}\frac{{\bf r}... ...\Delta} t)-{\bf r}(t)}{ {\mit\Delta} t} = \frac{d{\bf r}}{dt}.$$ (44)

When written in component form, the above definition yields

$$v_x = \frac{dx}{dt},$$ (45)

$$v_y = \frac{dy}{dt},$$ (46)

$$v_z = \frac{dz}{dt}.$$ (47)

Thus, the $x$-component of velocity is simply the time derivative of the $x$-coordinate, and so on.

By analogy with the 1-dimensional equation (16), the body's vector acceleration ${\bf a}=(a_x,a_y,a_z)$ is simply the derivative of ${\bf v}$ with respect to $t$. In other words,

$${\bf a}(t) = \lim_{{\mit\Delta} t\rightarrow 0}\frac{{\bf v}... ...{\mit\Delta} t} = \frac{d{\bf v}}{dt}=\frac{d^2{\bf r}}{dt^2}.$$ (48)

When written in component form, the above definition yields

$$a_x = \frac{dv_x}{dt}=\frac{d^2 x}{dt^2},$$ (49)

$$a_y = \frac{dv_y}{dt}=\frac{d^2 y}{dt^2},$$ (50)

$$a_z = \frac{dv_z}{dt}=\frac{d^2 z}{dt^2}.$$ (51)

Thus, the $x$-component of acceleration is simply the time derivative of the $x$-component of velocity, and so on.

As an example, suppose that the coordinates of the body are given by

$$x = \sin t,$$ (52)

$$y = \cos t,$$ (53)

$$z = 3 t.$$ (54)

The corresponding components of the body's velocity are then simply

$$v_x = \frac{dx}{dt}=\cos t,$$ (55)

$$v_y = \frac{dy}{dt}= - \sin t,$$ (56)

$$v_z = \frac{dz}{dt}=3,$$ (57)

whilst the components of the body's acceleration are given by

$$a_x = \frac{dv_x}{dt}=-\sin t,$$ (58)

$$a_y = \frac{dv_y}{dt}=- \cos t,$$ (59)

$$a_z = \frac{dv_z}{dt}=0.$$ (60)