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Next: Motion with constant velocity Up: Motion in 3 dimensions Previous: Diagonals of a parallelogram

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

\begin{displaymath}
{\bf r}(t) = [x(t), y(t), z(t)].
\end{displaymath} (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,
\begin{displaymath}
{\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}.
\end{displaymath} (44)

When written in component form, the above definition yields
$\displaystyle v_x$ $\textstyle =$ $\displaystyle \frac{dx}{dt},$ (45)
$\displaystyle v_y$ $\textstyle =$ $\displaystyle \frac{dy}{dt},$ (46)
$\displaystyle v_z$ $\textstyle =$ $\displaystyle \frac{dz}{dt}.$ (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,

\begin{displaymath}
{\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}.
\end{displaymath} (48)

When written in component form, the above definition yields
$\displaystyle a_x$ $\textstyle =$ $\displaystyle \frac{dv_x}{dt}=\frac{d^2 x}{dt^2},$ (49)
$\displaystyle a_y$ $\textstyle =$ $\displaystyle \frac{dv_y}{dt}=\frac{d^2 y}{dt^2},$ (50)
$\displaystyle a_z$ $\textstyle =$ $\displaystyle \frac{dv_z}{dt}=\frac{d^2 z}{dt^2}.$ (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

$\displaystyle x$ $\textstyle =$ $\displaystyle \sin t,$ (52)
$\displaystyle y$ $\textstyle =$ $\displaystyle \cos t,$ (53)
$\displaystyle z$ $\textstyle =$ $\displaystyle 3 t.$ (54)

The corresponding components of the body's velocity are then simply
$\displaystyle v_x$ $\textstyle =$ $\displaystyle \frac{dx}{dt}=\cos t,$ (55)
$\displaystyle v_y$ $\textstyle =$ $\displaystyle \frac{dy}{dt}= - \sin t,$ (56)
$\displaystyle v_z$ $\textstyle =$ $\displaystyle \frac{dz}{dt}=3,$ (57)

whilst the components of the body's acceleration are given by
$\displaystyle a_x$ $\textstyle =$ $\displaystyle \frac{dv_x}{dt}=-\sin t,$ (58)
$\displaystyle a_y$ $\textstyle =$ $\displaystyle \frac{dv_y}{dt}=- \cos t,$ (59)
$\displaystyle a_z$ $\textstyle =$ $\displaystyle \frac{dv_z}{dt}=0.$ (60)


next up previous
Next: Motion with constant velocity Up: Motion in 3 dimensions Previous: Diagonals of a parallelogram
Richard Fitzpatrick 2006-02-02