3.2.2 More Vector Mathematics

In addition to adding vectors, we will need to subtract vectors, multiply vectors by scalars, and understand how to interpret the negative of a vector. These operations are illustrated in FIGURE 3.7.

FIGURE 3.7 Working with vectors.

The figure illustrates the multiplication and subtraction of vectors in six parts.
FIGURE 3.7 Full Alternative Text

EXAMPLE 3.2 Velocity and displacement

Carolyn drives her car north at 30 km/h for 1 hour, east at 60 km/h for 2 hours, then north at 50 km/h for 1 hour. What is Carolyn’s net displacement?

SOLVE Chapter 1 defined average velocity as

v ⃗=Δr⃗Δt

so the displacement Δr⃗ during the time interval Δt is Δr⃗=(Δt)v ⃗. This is multiplication of the vector v ⃗ by the scalar Δt. Carolyn’s velocity during the first hour is v ⃗1=(30 km/h, north), so her displacement during this interval is

Δr⃗1=(1 hour)(30 km/h, north)=(30 km, north)

Similarly,

Δr⃗2=(2 hours)(60 km/h, east)=(120 km, east)Δr⃗3=(1 hour)(50 km/h, north)=(50 km, north)

In this case, multiplication by a scalar changes not only the length of the vector but also its units, from km/h to km. The direction, however, is unchanged. Carolyn’s net displacement is

Δr⃗net=Δr⃗1+Δr⃗2+Δr⃗3

This addition of the three vectors is shown in FIGURE 3.8, using the tip-to-tail method. Δr⃗net stretches from Carolyn’s initial position to her final position. The magnitude of her net displacement is found using the Pythagorean theorem:

FIGURE 3.8 The net displacement is the vector sum Δr⃗net=Δr⃗1+Δr⃗2+Δr⃗3 .

The figure illustrates the addition of three displacement vectors using the tip-to-tail method. The three displacement vector arrows are labeled delta vector r subscript 1, delta vector r subscript 2, and delta vector r subscript 3.
rnet=(120 km)2+(80 km)2=144 km

The direction of Δr⃗net is described by angle θ, which is

θ=tan−1(80 km120 km)=34°

Thus Carolyn’s net displacement is Δr⃗net=(144 km, 34° northof east).