9781118230725.pdf

Position and Displacement

One general way of locating a particle (or particle-like object) is with a position
vector , which is a vector that extends from a reference point (usually the
origin) to the particle. In the unit-vector notation of Module 3-2, can be written

(4-1)

wherex,y, and z are the vector components of and the coefficients x,y, and z
are its scalar components.
The coefficients x,y, and zgive the particle’s location along the coordinate
axes and relative to the origin; that is, the particle has the rectangular coordinates
(x,y,z). For instance, Fig. 4-1 shows a particle with position vector

and rectangular coordinates (3 m, 2 m, 5 m). Along the xaxis the particle is
3 m from the origin, in the direction. Along the yaxis it is 2 m from the
origin, in the direction. Along the zaxis it is 5 m from the origin, in the
direction.
As a particle moves, its position vector changes in such a way that the vector
always extends to the particle from the reference point (the origin). If the posi-
tion vector changes—say, from to during a certain time interval—then the
particle’s displacement during that time interval is

(4-2)

Using the unit-vector notation of Eq. 4-1, we can rewrite this displacement as

or as (4-3)

where coordinates (x 1 ,y 1 ,z 1 ) correspond to position vector and coordinates
(x 2 ,y 2 ,z 2 ) correspond to position vector. We can also rewrite the displacement
by substituting xfor (x 2 x 1 ), yfor (y 2 y 1 ), and zfor (z 2 z 1 ):

:rxiˆyjˆzkˆ. (4-4)

     

:r 2

:r 1

:r(x 2 x 1 )iˆ(y 2 y 1 )jˆ(z 2 z 1 )kˆ,

:r(x 2 iˆy 2 jˆz 2 kˆ)(x 1 iˆy 1 jˆz 1 kˆ)

:r:r 2 :r 1.

:r

:r 1 :r 2

jˆ kˆ

iˆ

:r(3 m)iˆ(2 m)jˆ(5 m)kˆ

iˆ jˆ kˆ :r

:rxiˆyjˆzkˆ,

:r

:r

Figure 4-1The position vector for a parti-

cle is the vector sum of its vector compo-

nents.

:r

y

x

z

(–3 m)i

(5 m)k (2 m)j

O

ˆ

ˆ

ˆ

r

To locate the

particle, this

is how far

parallel to z.

This is how far

parallel to y.

This is how far

parallel to x.

position vector. Let’s evaluate those coordinates at the

given time, and then we can use Eq. 3-6 to evaluate the mag-

nitude and orientation of the position vector.

:r

Sample Problem 4.01 Two-dimensional position vector, rabbit run

A rabbit runs across a parking lot on which a set of

coordinate axes has, strangely enough, been drawn. The co-

ordinates (meters) of the rabbit’s position as functions of

timet(seconds) are given by

x0.31t^2 7.2t 28 (4-5)

and y0.22t^2 9.1t30. (4-6)

(a) At t15 s, what is the rabbit’s position vector in unit-

vector notation and in magnitude-angle notation?

KEY IDEA

The xandycoordinates of the rabbit’s position, as given by

Eqs. 4-5 and 4-6, are the scalar components of the rabbit’s

:r

Calculations:We can write

(4-7)

(We write rather than because the components are

functions of t, and thus is also.)

Att15 s, the scalar components are

x(0.31)(15)^2 (7.2)(15) 28 66 m

and y(0.22)(15)^2 (9.1)(15) 30 57 m,

so :r(66 m)iˆ(57 m)jˆ, (Answer)

:r

:r(t) :r

:r(t)x(t)iˆy(t)jˆ.

4-1 POSITION AND DISPLACEMENT 63