3.3.3 Components

You learned in Chapters 1 and 2 to give the kinematic variable vx a positive sign if the velocity vector v ⃗ points toward the positive end of the x-axis, a negative sign if v ⃗ points in the negative x-direction. We need to extend this idea to vectors in general.

Suppose vector A⃗ has been decomposed into component vectors A⃗x and A⃗y parallel to the coordinate axes. We can describe each component vector with a single number called the component. The x-component and y-component of vector A⃗, denoted Ax and Ay , are determined as follows:

TACTICS BOX 3.1 Determining the components of a vector

1 The absolute value |Ax| of the x-component Ax is the magnitude of the component vector A⃗x.

2 The sign of A⃗x is positive if A⃗x points in the positive x-direction (right), negative if A⃗x points in the negative x-direction (left).

3 The y-component Ay is determined similarly.

Workbook Exercises 10–18

In other words, the component Ax tells us two things: how big A⃗x is and, with its sign, which end of the axis A⃗x points toward. FIGURE 3.11 shows three examples of determining the components of a vector.

FIGURE 3.11 Determining the components of a vector.

The figure illustrates components of two displacement vectors labeled vector "A" and vector B using two separate x y coordinate planes.
FIGURE 3.11 Full Alternative Text
Full Alternative Text

NOTE   Beware of the somewhat confusing terminology. A⃗x and A⃗y are called component vectors, whereas Ax and Ay are simply called components. The components Ax and Ay are just numbers (with units), so make sure you do not put arrow symbols over the components.

We will frequently need to decompose a vector into its components. We will also need to “reassemble” a vector from its components. In other words, we need to move back and forth between the geometric and the component representations of a vector. FIGURE 3.12 shows how this is done.

FIGURE 3.12 Moving between the geometric representation and the component representation.

The figure illustrates components of two velocity vectors labeled vector "A" and vector B using two separate x y coordinate planes.
FIGURE 3.12 Full Alternative Text

Each decomposition requires that you pay close attention to the direction in which the vector points and the angles that are defined.

EXAMPLE 3.3 Finding the components of an acceleration vector

Seen from above, a hummingbird’s acceleration is (6.0 m/s2, 30° south of west). Find the x- and y-components of the acceleration vector a⃗.

VISUALIZE It’s important to draw vectors. FIGURE 3.13 establishes a map-like coordinate system with the x-axis pointing east and the y-axis north. Vector a⃗ is then decomposed into components parallel to the axes. Notice that the axes are “acceleration axes” with units of acceleration, not xy-axes, because we’re measuring an acceleration vector.

FIGURE 3.13 Decomposition of a⃗.

The figure illustrates the components of an acceleration vector labeled vector “a” in the third quadrant of the x y coordinate plane.

SOLVE The acceleration vector points to the left (negative x-direction) and down (negative y-direction), so the components ax and ay are both negative:

ax=−a cos 30°=−(6.0 m/s2) cos 30°=−5.2 m/s2ay=−a sin 30°=−(6.0 m/s2) sin 30°=−3.0 m/s2

REVIEW The units of ax and ay are the same as the units of vector a⃗. Notice that we had to insert the minus signs manually by observing that the vector points left and down.

EXAMPLE 3.4 Finding the direction of motion

FIGURE 3.14 shows a car’s velocity vector v ⃗. Determine the car’s speed and direction of motion.

FIGURE 3.14 The velocity vector v ⃗ of Example 3.4.

The figure illustrates a velocity vector labeled v in the second quadrant of the x y coordinate plane.

VISUALIZE FIGURE 3.15 shows the components vx and vy and defines an angle θ with which we can specify the direction of motion.

FIGURE 3.15 Decomposition of v ⃗.

The figure illustrates the components of a velocity vector labeled vector V in the second quadrant of a coordinate plane.

SOLVE We can read the components of v ⃗ directly from the axes: vx=−6.0 m/s and vy=4.0 m/s. Notice that vx is negative. This is enough information to find the car’s speed v, which is the magnitude of v ⃗:

v=vx2+vy2=(−6.0 m/s)2+(4.0 m/s)2=7.2 m/s

From trigonometry, angle θ is

θ=tan−1(vy| vx |)=tan−1(4.0 m/s6.0 m/s)=34°

The absolute value signs are necessary because vx is a negative number. The velocity vector v ⃗ can be written in terms of the speed and the direction of motion as

v ⃗=(7.2 m/s, 34° above the negative x-axis)