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Another vector quantity which crops up frequently in applied mathematics is velocity. In everyday English, the words ‘speed? and ‘velocity? mean much the same as each other, but in scientific parlance there is a significant difference between them.

Velocity and speed

The velocity v of an object is its rate of change of position. This is a (vector) measure of how fast it is moving and of its direction of motion.
The speed |v| of an object is a (scalar) measure of how fast it is moving, irrespective of its direction of motion. As the notation indicates, the speed is the magnitude of the velocity vector. The SI unit for speed is metres per second (ms⁻¹).

Here we consider only constant velocities, which have both constant speed and fixed direction. The direction of a velocity can be expressed in several ways, but it is often given as a bearing, as for displacements. Indeed, constant velocities are closely related to displacements. If an object travels with constant velocity v ms⁻¹ (metres per second) for t seconds, then its displacement d metres over that time is given by d = t v. So each constant velocity is a scalar multiple of a displacement (v = (1/t)d), and vice versa. Provided t > 0, the vectors v and d have the same direction.

One particular example is wind velocity. Note that when the direction of a wind is specified in English (as in ‘a westerly wind?), this describes the direction from which the wind blows, and not the direction to which it moves.

Activity 7 Velocity vector for wind

Let i be 1 ms⁻¹ East and let j be 1 ms⁻¹ North. Give in component form the velocity vector for a south-westerly wind of speed 2 ms⁻¹.

Answer

Solution

The wind blows from the South-West, and hence moves towards the North-East. Hence the wind velocity vector v has direction θ = 45° (measured anticlockwise from the i-direction). Its component form is

The sorts of problems that arise for velocities are similar to those that you have met already for displacements, as the following example illustrates.

Example 4 Crossing the river

A river flows due East at a speed of 0.3 ms⁻¹. A boy in a rowing boat, who can row at 0.5 ms⁻¹ in still water, starts from a point on the South bank and steers at right-angles to the bank. The boat is also blown by a wind with speed 0.4 ms⁻¹ from a N 20° E direction.

Find the resultant velocity of the boat, in terms of its speed (correct to two decimal places) and a bearing (to the nearest degree).
Suppose that the river has constant width 10 metres. How long does it take the boy to cross the river, and how far upstream or downstream has he then travelled?

Answer

Solution

Figure 14 Three velocity vectorsLong description

Let i be 1 ms⁻¹ East and let j be 1 ms⁻¹ North. Assume that the boy rows throughout at the maximum speed of which he is capable. Suppose that v _b is the velocity of the boat in still water, v _r is the velocity of the river and v _w is the velocity of the wind. These vectors are shown in Figure 14.

From the information given, v _b has magnitude |v _b| = 0.5 and direction θ _b = 90°, while v _r has magnitude |v _r| = 0.3 and direction θ _r = 0.

The wind comes from N 20° E and hence blows towards S 20° W, for which the direction is −(90° + 20°). Hence the vector v _w has magnitude |v _w| = 0.4 and direction θ _w = −110°.

The component forms of the vectors are therefore
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The first two component forms can in this case be written down directly from Figure 14. The resultant velocity is
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The strategy from Section 1.1 can now be applied. The resultant speed of the boat is

Since the components of v are v ₁ ≈ 0.1632, v ₂ ≈ 0.1241, we have

Also, (0.1632, 0.1241) lies in the first quadrant, so the direction of v is θ = φ ≈ 37°. This corresponds to the bearing N 53° E.

Thus the resultant velocity is 0.21 ms⁻¹ at N 53° E.
The i-component of v indicates how fast the boat travels downstream (or upstream, if v ₁ < 0), while the j-component, v ₂, is the rate of progress across the river. Since v ₂ ≈ 0.1241 (in ms⁻¹), the 10-metre width of the river will be crossed in 10/0.1241 ≈ 81 seconds. (Here we ignore the length of the boat!)

In that time, the boat will have travelled 81 × 0.1632 ≈ 13 metres downstream.

Comment

It is natural to draw the arrows for velocity vectors in Figure 14 with their tails at the same point, since the boat can be thought of as moving under the simultaneous effect of the three velocities. When adding displacements, on the other hand, it is natural to think of the vectors taking effect consecutively and hence to place the arrows nose to tail. This is simply a choice of visualisation, and does not affect the mathematics involved. However, placing the arrows nose to tail does give some indication of the magnitude and direction of the resultant vector, which provides a rough check on the calculation.

Activity 8 Ship in a current

A ship has a speed in still water of 5 ms⁻¹ and is pointed in the direction S 50° W, but there is a current of speed 2 ms⁻¹ flowing towards the direction N 40° W. Find the resultant velocity of the ship, in terms of its speed (correct to one decimal place) and a bearing (to the nearest degree).

Answer

Solution

Let i be 1 ms⁻¹ East and let j be 1 ms⁻¹ North, Suppose that v _s is the velocity of the ship in still water and v _c is the velocity of the current. These vectors are shown in Figure 15.

Figure 15 Long description

From the information given, v _s has magnitude |v _s| = 5 and direction θ _s = −(90° + 50°) = −140°, while v _c has magnitude |v _c| = 2 and direction θ _c = 90° + 40° = 130°.