Modelling displacements and velocities

1.4.2 Displacements and bearings

The displacement from a point P to a point Q is the change of position between the two points, as described by the displacement vector

If P and Q represent places on the ground, then it is natural to use a bearing to describe the direction of Q from P. It is straightforward to move between bearings and vector directions (relative to Cartesian axes), once the Cartesian unit vectors have been specified in terms of compass directions.

Activity 4 Converting to and from bearings

Suppose that i is 1 km East and j is 1 km North.

1. Find the direction θ that corresponds to each of the following bearings.

   1. N 40° E

   2. S 50° W

2. Find the bearing that corresponds to each of the following directions.

   1. θ = −70°

   2. θ = 120°

Answer

Solution

1. The directions are shown in Figure 10.

   1. The direction is θ = 90° − 40° = 50°.

   2. The direction is θ = −(90° + 50°) = −140°.

2. The directions are shown in Figure 11.

   1. The bearing is S 20° E.

   2. The bearing is N 30° W.

The next activity asks you to apply to displacements the strategies from Section 1 for converting between geometric and component form.

Activity 5 Displacements in geometric and component form

Suppose that i is 1 km East and j is 1 km North.

1. The displacement from Bristol to Leeds is given by the vector d = 77i + 286j. Verify that d may also be described by 296 km at N 15° E.

2. The displacement from Exeter to Belfast is 465 km at N 21° W.

   1. How far East or West of Exeter is Belfast?

   2. How far North or South of Exeter is Belfast?

3. Specify the displacement vector from Belfast to Exeter:

   1. in component form;

   2. in terms of direct distance and a bearing.

Answer

Solution

1. The strategy from Section 1.1 may be applied. The magnitude of the vector d = 77i + 286j is

   

   Since the components are d ₁ = 77, d ₂ = 286, we have

   

   Also, (77, 286) lies in the first quadrant, so the direction of d is θ =  φ ≈ 75°. This direction corresponds to the bearing N 15° E. Thus d describes the displacement 296 km at N 15° E, as claimed.

2. If e is the displacement vector from Exeter to Belfast, then |e| = 465. The direction of e is θ = 90° + 21° = 111°, Hence the component form

   

   (Note that we have extended use of the ≈ symbol to an approximate vector equation.)

   1. The i-component, −167, gives in kilometres the distance by which Belfast is East of Exeter. However, the minus sign indicates that Belfast is 167 km West of Exeter.

   2. The j-component shows that Belfast is 434 km North of Exeter.

3. The displacement from Belfast to Exeter is the opposite of that from Exeter to Belfast. Hence it is described by the vector −e, where e was given in part (b).

   1. In component form, the displacement vector from Belfast to Exeter is

      

   2. The bearing opposite to N 21° W is S 21° E. Hence, in terms of direct distance and a bearing, the displacement from Belfast to Exeter is 465 km at S 21° E.

The approach outlined in Activity 5 can be applied to two or more displacements, in order to find their resultant. This is demonstrated in the following example.

Example 3 Finding the resultant displacement

A surveyor walks 200 metres due North. She then turns clockwise through 120° and walks 100 metres, after which she walks 300 metres due West. Find her resultant displacement, relative to her starting position, in terms of direct distance and a bearing. Give your answers correct to one decimal place.

Answer

Solution

Let i be 1 m East and let j be 1 m North. Suppose that the displacement vectors for the three phases of the walk are denoted by a, b and c. The path of the surveyor is sketched in Figure 12.

Note that, here and in subsequent vector diagrams, the Cartesian unit vectors i and j are not drawn to the scale indicated for other vectors.

The vector a has magnitude |a|  =  200 and direction 90°. The vector b has magnitude |b| = 100 and direction 90°− 120° =  −30°. The vector c has magnitude |c| = 300 and direction 180°. Hence the respective component forms are

Since the magnitudes of i and j are each 1 m, the magnitudes of a, b and c are measured in metres. This convention is adopted in all similar problems in this chapter.

The resultant displacement is

Note the convenient extension of the use of the symbol ≈ to vector contexts.

The strategy from Section 1.1 can now be applied. The vector d has magnitude

Since the components of d are d ₁ ≈ −213.3975, d ₂ = 150, we have

Also, (−213.3975, 150) lies in the second quadrant, so the direction of d is θ = 180° − φ ≈ 144.9°. This corresponds to the bearing N 54.9° W.

Thus the surveyor's resultant displacement, relative to her starting point, is 260.8 m at N 54.9° W. (An arrow representing the resultant vector d should have its tail at the tail of the arrow representing a and its tip at the tip of the arrow representing c. You might like to check the calculation of d by drawing, to scale, a suitable arrow.)

Activity 6 Finding the resultant displacement

The displacement from Exeter to Belfast is 465 km at N 21° W (as given in Activity 5(b)). The displacement from Belfast to Edinburgh is 230 km at N 48° E. Find, in terms of direct distance and a bearing, the displacement from Exeter to Edinburgh. Give your answers to the nearest kilometre or degree.

Answer

Solution

Let i be 1 km East and let j be 1 km North. Denote the displacement from Exeter to Belfast by the vector a, and the displacement from Belfast to Edinburgh by the vector b. These vectors are sketched in Figure 13.

You showed in Activity 5(b) that

Similarly, the vector b has magnitude |b| = 230 and direction 90° − 48° = 42°, so its component form is

The resultant is

The strategy from Section 1.1 may now be applied. The vector c has magnitude

Since the components are c ₁ ≈ 4, c ₂ ≈ 588, we have

Also, (4, 588) lies in the first quadrant, so the direction of c is θ =  φ ≈ 89.6°, which is 90° to the nearest degree. This corresponds to the bearing due North.

Hence the displacement from Exeter to Edinburgh is 588 km due North.

Except for third party materials and otherwise stated (see terms and conditions), this content is made available under a Creative Commons Attribution-NonCommercial-ShareAlike 2.0 Licence