Motion with constant acceleration

An object moving in 3 dimensions with constant acceleration ${\bf a}$ possesses a vector displacement of the form

$${\bf r}(t) = {\bf r}_0 + {\bf v}_0 t + \frac{1}{2} {\bf a} t^2.$$ (62)

Hence, the object's velocity is given by

$${\bf v}(t) = \frac{d{\bf r}}{dt}= {\bf v}_0 + {\bf a} t.$$ (63)

Note that $d{\bf v}/dt ={\bf a}$, as expected. In the above, the constant vectors ${\bf r}_0$ and ${\bf v}_0$ are the object's displacement and velocity at time $t=0$, respectively.

As is easily demonstrated, the vector equivalents of Eqs. (21)-(23) are:

$${\bf s} = {\bf v}_0 t + \frac{1}{2} {\bf a} t^2,$$ (64)

$${\bf v} = {\bf v}_0 + {\bf a} t,$$ (65)

$$v^2 = v_0^{ 2} + 2 {\bf a}\!\cdot\!{\bf s}.$$ (66)

These equation fully characterize 3-dimensional motion with constant acceleration. Here, ${\bf s} = {\bf r}-{\bf r}_0$ is the net displacement of the object between times $t=0$ and $t$.

The quantity ${\bf a}\!\cdot\!{\bf s}$, appearing in Eq. (66), is termed the scalar product of vectors ${\bf a}$ and ${\bf s}$, and is defined

$${\bf a}\!\cdot\!{\bf s} = a_x s_x+ a_y s_y+a_z s_z.$$ (67)

The above formula has a simple geometric interpretation, which is illustrated in Fig. 15. If $\vert{\bf a}\vert$ is the magnitude (or length) of vector ${\bf a}$, $\vert{\bf s}\vert$ is the magnitude of vector ${\bf s}$, and $\theta$ is the angle subtended between these two vectors, then

$${\bf a} \! \cdot \! {\bf s} = \vert{\bf a}\vert \vert{\bf s}\vert \cos\theta.$$ (68)

In other words, the scalar product of vectors ${\bf a}$ and ${\bf s}$ equals the product of the length of vector ${\bf a}$ times the length of that component of vector ${\bf s}$ which lies in the same direction as vector ${\bf a}$. It immediately follows that if two vectors are mutually perpendicular (i.e., $\theta =90^\circ$) then their scalar product is zero. Furthermore, the scalar product of a vector with itself is simply the magnitude squared of that vector [this is immediately apparent from Eq. (67)]:

$${\bf a} \!\cdot \!{\bf a} = \vert{\bf a}\vert^2 = a^2.$$ (69)

It is also apparent from Eq. (67) that ${\bf a}\!\cdot{\bf s}= {\bf s}\!\cdot{\bf a}$, and ${\bf a}\!\cdot\!({\bf b} +{\bf c}) = {\bf a}\!\cdot\!{\bf b}+{\bf a}\!\cdot\!{\bf c}$, and ${\bf a}\!\cdot\!(\lambda {\bf s}) = \lambda({\bf a}\!\cdot{\bf s})$.

Figure 15: The scalar product

Incidentally, Eq. (66) is obtained by taking the scalar product of Eq. (65) with itself, taking the scalar product of Eq. (64) with ${\bf a}$, and then eliminating $t$.