3 MOTION IN 3 DIMENSIONS 3.12 Relative velocity

v

v

g y

As an illustration, suppose that the plane’s velocity relative to the air is 300 km/h,
at a compass bearing of 120 ◦, and the air’s velocity relative to the ground is

85 km/h, at a compass bearing of 225 ◦. It follows that the components of va
and u (measured in units of km/h) are

va = (300 cos 120 ◦, 300 sin 120 ◦) = (−1.500 × 102 , 2.598 × 102 ), (3.54)

u = (85 cos 225 ◦, 85 sin 225 ◦) = (−6.010 × 101 , −6.010 × 101 ). (3.55)

According to Eq. (3.52), the components of the plane’s velocity vg relative to the

ground are simply the algebraic sums of the corresponding components of va and

u. Hence,

vg = (−1.500 × 102 − 6.010 × 101 , 2.598 × 102 − 6.010 × 101 )

= (−2.101 × 102 , 1.997 × 102 ). (3.56)

Our final task is to reconstruct the magnitude and compass bearing of vector

vg, given its components (vg (^) x, vg (^) y). The magnitude of vg follows from Pythagoras’
theorem [see Eq. (3.6)]:
vg =
q
(vg (^) x)^2 + (vg (^) y)^2

q
(−2.10 1 × 102 )^2 + (1.99 7 × 102 )^2 = 289. 9 km/h. (3.57)
In principle, the compass bearing of vg is given by the following formula:
φ = tan−^1
vg y

. (3.58)
g x

This follows because vg (^) x = vg cos φ and vg (^) y = vg sin φ [see Eq. (3.53)]. Un-
fortunately, the above expression becomes a little difficult to interpret if vg (^) x is
negative. An unambiguous pair of expressions for φ is given below:
φ = tan−^1
vg y
, (3.59)
g x
if vg (^) x ≥ 0 ; or
φ = 180 ◦ − tan−^1
v
|vg (^) x|
, (3.60)