Applications of vectors - relative velocity
Explore relative velocity in Year 11 Specialist Mathematics in Queensland (QCAA), where vectors describe how one moving object appears from another and how a craft moves through a current or a wind. It is the applied heart of the algebra of vectors topic.
You will learn to find the velocity of one object relative to another by subtracting vectors, and the true ground velocity of a boat or plane by adding the current or wind — then read off ground speed, bearings, crossing times and drift for real navigation problems.
Theory
Relative velocity asks how one moving object appears from another, and how a craft’s velocity combines with the medium it moves through. In Year 11 Specialist Mathematics (QCAA, Queensland) you use vector subtraction for the velocity of one object relative to another, and vector addition to find the true (ground) velocity of a boat in a current or a plane in a wind.
The velocity of \(A\) relative to \(B\) is how \(A\) appears to move to an observer travelling with \(B\). It is found by subtracting the velocities: \(\mathbf{v}_{A\text{ rel }B}=\mathbf{v}_A-\mathbf{v}_B\). Reversing the objects reverses the vector: \(\mathbf{v}_{B\text{ rel }A}=-\mathbf{v}_{A\text{ rel }B}\).
A resultant (actual) velocity is what you get when a craft moves through a moving medium. The craft’s velocity relative to the ground equals its velocity relative to the medium plus the velocity of the medium: \(\mathbf{v}_{\text{craft/ground}}=\mathbf{v}_{\text{craft/medium}}+\mathbf{v}_{\text{medium/ground}}\). The current or wind is always added, never subtracted.
The heading is the direction the craft is pointed (its velocity through the water or air); the track or course is the direction it actually travels over the ground. When the two velocities are perpendicular, combine their magnitudes with Pythagoras and find the direction with a tangent ratio.
Two everyday tasks fall out of this: steering a boat upstream so its resultant runs straight across a river, and finding the ground speed and bearing of a plane blown off a heading by wind. Speeds use metric units (m/s or km/h) and directions use compass bearings or “east of north” style descriptions.
Velocity of one object relative to another (subtract the velocity of the observer):
Resultant (ground) velocity of a craft in a moving medium — add the medium’s velocity:
For perpendicular components, the resultant speed and its direction \(\theta\) are:
How to solve a relative-velocity problem
- Set axes and write each velocity as a vector. Take \(\mathbf{i}\) as east and \(\mathbf{j}\) as north (or across and along the bank), and write every speed and direction in component form.
- Choose subtract or add. For “\(A\) relative to \(B\)” compute \(\mathbf{v}_A-\mathbf{v}_B\); for a true/ground velocity add the current or wind to the craft’s own velocity.
- Combine the components. Add or subtract the \(\mathbf{i}\) and \(\mathbf{j}\) parts separately to get the resultant vector.
- Extract speed and direction. Use \(|\mathbf{v}|=\sqrt{v_x^{\,2}+v_y^{\,2}}\) for the speed and a tangent ratio for the angle, then read off the bearing, crossing time or drift the question asks for.
Subtract the observer’s velocity: \(\mathbf{v}_A-\mathbf{v}_B\).
| \(\mathbf{v}_A-\mathbf{v}_B\) | \(=\) | \((6\mathbf{i}+8\mathbf{j})-(2\mathbf{i}+5\mathbf{j})\) |
| \(=\) | \((6-2)\mathbf{i}+(8-5)\mathbf{j}\) | |
| \(=\) | \(4\mathbf{i}+3\mathbf{j}\) |
Take the magnitude of that vector with Pythagoras.
| \(|\mathbf{v}_A-\mathbf{v}_B|\) | \(=\) | \(\sqrt{4^2+3^2}\) |
| \(=\) | \(\sqrt{16+9}\) | |
| \(=\) | \(\sqrt{25}\) | |
| \(=\) | \(5\) |
\(A\) moves at \(4\mathbf{i}+3\mathbf{j}\) m/s relative to \(B\), a speed of \(5\) m/s.
The northward and eastward velocities are perpendicular, so add head to tail and use Pythagoras.
| \(|\mathbf{v}|\) | \(=\) | \(\sqrt{12^2+9^2}\) |
| \(=\) | \(\sqrt{144+81}\) | |
| \(=\) | \(\sqrt{225}\) | |
| \(=\) | \(15\) |
The track angle east of north uses \(\tan\theta=\dfrac{\text{east}}{\text{north}}\).
| \(\tan\theta\) | \(=\) | \(\dfrac{9}{12}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\!\left(\dfrac{9}{12}\right)\) |
| \(=\) | \(36.87^\circ\) | |
| \(\approx\) | \(37^\circ\) |
The ferry travels at \(15\) m/s along a track about \(37^\circ\) east of north.
Point the boat upstream so its along-bank part cancels the current: \(10\sin\theta=6\).
| \(\sin\theta\) | \(=\) | \(\dfrac{6}{10}\) |
| \(\theta\) | \(=\) | \(\sin^{-1}(0.6)\) |
| \(\approx\) | \(37^\circ \text{ upstream}\) |
The across-speed is the remaining side of the right triangle.
| \(v_{\text{across}}\) | \(=\) | \(\sqrt{10^2-6^2}\) |
| \(=\) | \(\sqrt{100-36}\) | |
| \(=\) | \(\sqrt{64}\) | |
| \(=\) | \(8\) |
Crossing time uses only the across-speed.
| \(t\) | \(=\) | \(\dfrac{\text{width}}{v_{\text{across}}}\) |
| \(=\) | \(\dfrac{240}{8}\) | |
| \(=\) | \(30\) |
The boat crosses at \(8\) m/s and takes \(30\) s (steering about \(37^\circ\) upstream).
Add the eastward airspeed and the northward wind to get the ground velocity.
| \(\mathbf{v}_g\) | \(=\) | \((160\mathbf{i}+0\mathbf{j})+(0\mathbf{i}+120\mathbf{j})\) |
| \(=\) | \(160\mathbf{i}+120\mathbf{j}\) |
Ground speed is the magnitude of that vector.
| \(|\mathbf{v}_g|\) | \(=\) | \(\sqrt{160^2+120^2}\) |
| \(=\) | \(\sqrt{25600+14400}\) | |
| \(=\) | \(\sqrt{40000}\) | |
| \(=\) | \(200\) |
East is \(090^\circ\); the wind turns the track north of east by \(\alpha\).
| \(\alpha\) | \(=\) | \(\tan^{-1}\!\left(\dfrac{120}{160}\right)\) |
| \(=\) | \(36.87^\circ\) | |
| \(\text{bearing}\) | \(=\) | \(090^\circ-37^\circ\) |
| \(=\) | \(053^\circ\) |
The ground velocity is \(160\mathbf{i}+120\mathbf{j}\) km/h, a ground speed of \(200\) km/h on a bearing of about \(053^\circ\).
Common pitfalls
Frequently asked questions
How do you find the velocity of A relative to B?
Subtract the velocities: \(\mathbf{v}_{A\text{ rel }B}=\mathbf{v}_A-\mathbf{v}_B\). It is how \(A\) appears to move to an observer travelling with \(B\).
When do you add and when do you subtract velocities?
Add to find a true (ground) velocity of a craft in a current or wind: craft-in-medium plus medium. Subtract to find how one object looks from another (relative velocity).
How do you find a boat’s actual speed across a river?
When the boat’s velocity and the current are perpendicular, combine their speeds with Pythagoras: \(|\mathbf{v}|=\sqrt{v_x^{\,2}+v_y^{\,2}}\).
How do you steer a boat so it goes straight across a current?
Point upstream at an angle \(\theta\) so the upstream part cancels the current, using \(\sin\theta=\dfrac{\text{current}}{\text{boat speed}}\); the across-speed is then \(\sqrt{\text{boat}^2-\text{current}^2}\).
What is the difference between heading and track?
The heading is the direction the craft is pointed (its velocity through the water or air); the track is the direction it actually travels over the ground once the current or wind is added.
How do you find the crossing time and drift?
The crossing time is the river width divided by the across-component of the velocity; the downstream drift is the current speed multiplied by that time.