   Next: Worked example 3.1: Broken Up: Motion in 3 dimensions Previous: Projectile motion

## Relative velocity

Suppose that, on a windy day, an airplane moves with constant velocity with respect to the air, and that the air moves with constant velocity with respect to the ground. What is the vector velocity of the plane with respect to the ground? In principle, the answer to this question is very simple: (81)

In other words, the velocity of the plane with respect to the ground is the vector sum of the plane's velocity relative to the air and the air's velocity relative to the ground. See Fig. 18. Note that, in general, is parallel to neither nor . Let us now consider how we might implement Eq. (81) in practice. As always, our first task is to set up a suitable Cartesian coordinate system. A convenient system for dealing with 2-dimensional motion parallel to the Earth's surface is illustrated in Fig. 19. The -axis points northward, whereas the -axis points eastward. In this coordinate system, it is conventional to specify a vector in term of its magnitude, , and its compass bearing, . As illustrated in Fig. 20, a compass bearing is the angle subtended between the direction of a vector and the direction to the North pole: i.e., the -direction. By convention, compass bearings run from to . Furthermore, the compass bearings of North, East, South, and West are , , , and , respectively. According to Fig. 20, the components of a general vector , whose magnitude is and whose compass bearing is , are simply (82)

Note that we have suppressed the -component of (which is zero), for ease of notation. Although, strictly speaking, Fig. 20 only justifies the above expression for in the range to , it turns out that this expression is generally valid: i.e., it is valid for in the full range to . As an illustration, suppose that the plane's velocity relative to the air is , at a compass bearing of , and the air's velocity relative to the ground is , at a compass bearing of . It follows that the components of and (measured in units of km/h) are   (83)   (84)

According to Eq. (81), the components of the plane's velocity relative to the ground are simply the algebraic sums of the corresponding components of and . Hence,     (85)

Our final task is to reconstruct the magnitude and compass bearing of vector , given its components . The magnitude of follows from Pythagoras' theorem [see Eq. (35)]:     (86)

In principle, the compass bearing of is given by the following formula: (87)

This follows because and [see Eq. (82)]. Unfortunately, the above expression becomes a little difficult to interpret if is negative. An unambiguous pair of expressions for is given below: (88)

if ; or (89)

if . These expressions can be derived from simple trigonometry. For the case in hand, Eq. (89) is the relevant expression, hence (90)

Thus, the plane's velocity relative to the ground is at a compass bearing of .   Next: Worked example 3.1: Broken Up: Motion in 3 dimensions Previous: Projectile motion
Richard Fitzpatrick 2006-02-02