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Motion with constant acceleration
An object moving in 3 dimensions with constant acceleration
possesses a vector displacement of the form

(62) 
Hence, the object's velocity is given by

(63) 
Note that
, as expected. In the above,
the constant vectors and are the object's displacement and velocity at time
, respectively.
As is easily demonstrated, the vector equivalents of Eqs. (21)(23) are:
These equation fully characterize 3dimensional motion with constant acceleration.
Here,
is the net displacement of the object
between times and .
The quantity
, appearing in Eq. (66), is termed
the scalar product of vectors and , and is defined

(67) 
The above formula has a simple geometric interpretation, which is illustrated
in Fig. 15. If is the magnitude (or length) of vector ,
is the magnitude of vector , and is the angle
subtended between these two vectors, then

(68) 
In other words, the scalar product of vectors and equals
the product of the length of vector times the length of that component of
vector which lies in the same direction as vector .
It immediately follows that if two vectors are mutually perpendicular
(i.e.,
) 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)]:

(69) 
It is also apparent from Eq. (67) that
,
and
,
and
.
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 , and then eliminating .
Next: Projectile motion
Up: Motion in 3 dimensions
Previous: Motion with constant velocity
Richard Fitzpatrick
20060202