Home / IGCSE Add Maths 0606 / Vectors in two dimensions / Velocity vectors: position, time and collisions
Cambridge IGCSE · Topic 13.4
IGCSE Add Maths 0606 / Topic 13 / IGCSE

Where it is, where it is going.

A velocity vector carries two facts at once: its length is the speed and its direction is the way the particle moves. Add \(t\) lots of it to a starting position and you know where the particle is at any time, which answers 'when is it due north?', 'how close do they get?' and 'do they collide?'.

Topic 13All vectors in two dimensions lessons →
The key idea

Speed is the magnitude of velocity: \(|\mathbf{v}|\). A velocity of speed \(s\) in the direction of \(\mathbf{d}\) is \(s \times \dfrac{\mathbf{d}}{|\mathbf{d}|}\). A particle that starts at \(\mathbf{r}_0\) with constant velocity \(\mathbf{v}\) is at \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\) after time \(t\). Two particles collide when \(\mathbf{r}_A = \mathbf{r}_B\) for the SAME value of \(t\) in both components. Velocities add: a boat's velocity over the ground is its velocity through the water plus the velocity of the current.

x (km) y (km) 2 4 6 8 10 2 4 6 O A (t = 0) B (t = 0) vA = 3i + 2j vB = −i + 2j t = 1 t = 1 collide at (7, 5) when t = 2
Fig. 1: A starts at (1, 1) with velocity 3i + 2j km/h and B starts at (9, 1) with velocity -i + 2j km/h. The open circles show each particle after 1 hour. Both reach (7, 5) when t = 2, so they collide there
Section 01

Speed, direction and velocity.

In this lesson \(\mathbf{i}\) is a unit vector due east and \(\mathbf{j}\) a unit vector due north, unless a question says otherwise. A velocity \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) means the particle moves \(a\) units east and \(b\) units north every unit of time. Its speed is the magnitude, and its direction is the direction of the vector. Going the other way, to build a velocity from a speed and a direction, turn the direction into a unit vector and multiply by the speed.

\[ \text{speed} = |\mathbf{v}| = \sqrt{a^{2} + b^{2}}, \qquad \mathbf{v} = \text{speed} \times \frac{\mathbf{d}}{|\mathbf{d}|} \]

Worked example From velocity to speed and bearing
A ship has velocity \((6\mathbf{i} - 8\mathbf{j})\) km/h. Find its speed and the bearing on which it is sailing.
  1. Speed \(= \sqrt{6^{2} + (-8)^{2}} = \sqrt{100} = 10\) km/h.
  2. The ship moves 6 east and 8 south each hour, so it heads below the \(\mathbf{i}\) direction at \(\tan^{-1}\dfrac{8}{6} = 53.1^{\circ}\).
  3. East is a bearing of \(090^{\circ}\); turning a further \(53.1^{\circ}\) clockwise towards south gives \(090^{\circ} + 53.1^{\circ}\).
Answer: \(10\) km/h on a bearing of \(143.1^{\circ}\)
Worked example From speed and direction to velocity
Find the velocity of (a) a particle moving at 26 m/s in the direction of \(5\mathbf{i} - 12\mathbf{j}\) (b) a boat moving at 12 km/h on a bearing of \(030^{\circ}\).
  1. (a) \(|5\mathbf{i} - 12\mathbf{j}| = \sqrt{25 + 144} = 13\), so the unit vector is \(\dfrac{1}{13}(5\mathbf{i} - 12\mathbf{j})\).
  2. \(\mathbf{v} = 26 \times \dfrac{1}{13}(5\mathbf{i} - 12\mathbf{j}) = 2(5\mathbf{i} - 12\mathbf{j})\).
  3. (b) East component \(12\sin 30^{\circ} = 6\); north component \(12\cos 30^{\circ} = 6\sqrt{3} \approx 10.4\).
Answer: (a) \((10\mathbf{i} - 24\mathbf{j})\) m/s   (b) \((6\mathbf{i} + 6\sqrt{3}\,\mathbf{j})\) km/h
Your turn
A particle moves at 25 m/s in the direction of \(-6\mathbf{i} + 8\mathbf{j}\). Find its velocity.
Examiner trap

The direction vector is not the velocity. Multiplying the speed straight onto \(5\mathbf{i} - 12\mathbf{j}\) gives \(130\mathbf{i} - 312\mathbf{j}\), with speed 338 instead of 26. Divide by \(|\mathbf{d}|\) first, then multiply by the speed, and check that \(|\mathbf{v}|\) comes back to the speed.

Section 02

Position at time t.

A particle with constant velocity \(\mathbf{v}\) moves \(\mathbf{v}\) every unit of time, so after \(t\) units it has moved \(t\mathbf{v}\) from its start. Its position vector is the start plus that displacement. Each component is then a simple function of \(t\), and the questions 'when is it due north of a point?' or 'how far apart are two particles?' become equations in \(t\).

\[ \mathbf{r} = \mathbf{r}_0 + t\mathbf{v} \quad\Longleftrightarrow\quad \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} + t\begin{pmatrix} v_x \\ v_y \end{pmatrix} \]

Worked example Due north of a lighthouse
At noon a ship is at the point with position vector \((-6\mathbf{i} + 2\mathbf{j})\) km and it moves with velocity \((3\mathbf{i} + 4\mathbf{j})\) km/h. A lighthouse is at \((3\mathbf{i} + \mathbf{j})\) km. Find (a) the position of the ship at 13:30 (b) the time when the ship is due north of the lighthouse, and its distance from the lighthouse then.
  1. \(\mathbf{r} = (-6\mathbf{i} + 2\mathbf{j}) + t(3\mathbf{i} + 4\mathbf{j}) = (-6 + 3t)\mathbf{i} + (2 + 4t)\mathbf{j}\), with \(t\) in hours after noon.
  2. (a) 13:30 is \(t = 1.5\): \(\mathbf{r} = (-6 + 4.5)\mathbf{i} + (2 + 6)\mathbf{j} = -1.5\mathbf{i} + 8\mathbf{j}\).
  3. (b) Due north: the \(\mathbf{i}\) components match, \(-6 + 3t = 3\), so \(t = 3\), which is 15:00.
  4. Then \(\mathbf{r} = 3\mathbf{i} + 14\mathbf{j}\), which is \(14 - 1 = 13\) km north of the lighthouse.
Answer: (a) \((-1.5\mathbf{i} + 8\mathbf{j})\) km   (b) 15:00, 13 km north of the lighthouse
Worked example How close do they get?
Particle \(A\) starts at \(O\) with velocity \((3\mathbf{i} + 4\mathbf{j})\) m/s. At the same moment particle \(B\) starts at \(10\mathbf{i}\) m with velocity \((\mathbf{i} + 2\mathbf{j})\) m/s. Find the least distance between them and the time when it occurs.
  1. \(\mathbf{r}_A = 3t\mathbf{i} + 4t\mathbf{j}\) and \(\mathbf{r}_B = (10 + t)\mathbf{i} + 2t\mathbf{j}\).
  2. \(\mathbf{r}_A - \mathbf{r}_B = (2t - 10)\mathbf{i} + 2t\mathbf{j}\).
  3. \[ d^{2} = (2t - 10)^{2} + (2t)^{2} = 8t^{2} - 40t + 100 = 8(t - 2.5)^{2} + 50 \]
  4. The square is never negative, so \(d^{2}\) is least, 50, when \(t = 2.5\).
Answer: Least distance \(\sqrt{50} = 5\sqrt{2} \approx 7.07\) m, after \(2.5\) s
Your turn
A particle has position \(\mathbf{r} = (5\mathbf{i} - 2\mathbf{j}) + t(-2\mathbf{i} + 3\mathbf{j})\) m at time \(t\) s. Find its position when \(t = 4\), and the time when it is due north of \(O\).
Examiner trap

'Due north of the lighthouse' means the EAST components are equal, not the north ones; setting \(2 + 4t = 1\) would find when the ship is due WEST of it. And for the distance, subtract the two positions before squaring: \(|\mathbf{r}_A| - |\mathbf{r}_B|\) is not the distance between the particles.

Stage 1 · Learn

Check your understanding

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Quick check+10 XP

What is the speed of a particle with velocity \((5\mathbf{i} - 12\mathbf{j})\) m/s?

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A particle moves at 20 m/s in the direction of \(3\mathbf{i} + 4\mathbf{j}\). What is its velocity?

Quick check+10 XP

A particle starts at \((2\mathbf{i} - \mathbf{j})\) m and moves with velocity \((3\mathbf{i} + 4\mathbf{j})\) m/s. Where is it after 3 seconds?

Quick check+10 XP

Which condition shows that particles \(A\) and \(B\) collide?

Section 03

Collisions, currents and winds.

Two particles collide only if they are at the same place at the same time. Crossing paths is not enough: in Fig. 1 a particle could pass through \((7, 5)\) after the other has gone. Write both position vectors, use one component to find \(t\), then check the other component with that same \(t\). Velocities also add like any vectors: a boat's velocity over the ground is its velocity through the water plus the current, and a plane's ground velocity is its velocity in still air plus the wind.

\[ \text{collision} \iff \mathbf{r}_{A0} + t\mathbf{v}_A = \mathbf{r}_{B0} + t\mathbf{v}_B \ \text{ for one value of } t \text{ in BOTH components} \]

\[ \mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{boat in still water}} + \mathbf{v}_{\text{current}} \]

Worked example Do they collide?
Ships \(A\) and \(B\) leave at the same time from \((\mathbf{i} + \mathbf{j})\) km and \((9\mathbf{i} + \mathbf{j})\) km with velocities \((3\mathbf{i} + 2\mathbf{j})\) km/h and \((-\mathbf{i} + 2\mathbf{j})\) km/h. (a) Show that they collide, and find where. (b) If \(B\) had started at \((9\mathbf{i} + 2\mathbf{j})\) km instead, would they collide?
  1. \(\mathbf{r}_A = \begin{pmatrix} 1 + 3t \\ 1 + 2t \end{pmatrix}\) and \(\mathbf{r}_B = \begin{pmatrix} 9 - t \\ 1 + 2t \end{pmatrix}\).
  2. (a) \(\mathbf{i}\): \(1 + 3t = 9 - t\), so \(t = 2\). \(\mathbf{j}\) at \(t = 2\): \(A\) has \(5\) and \(B\) has \(5\). Both match, so they collide at \(t = 2\), at \((7\mathbf{i} + 5\mathbf{j})\) km (Fig. 1).
  3. (b) Now \(\mathbf{r}_B = \begin{pmatrix} 9 - t \\ 2 + 2t \end{pmatrix}\). The \(\mathbf{i}\) components still match only at \(t = 2\), but then \(B\) has \(\mathbf{j}\) component \(6\), not \(5\).
  4. They are never in the same place at the same time; at \(t = 2\), \(B\) is 1 km due north of \(A\).
Answer: (a) They collide after 2 hours at \((7\mathbf{i} + 5\mathbf{j})\) km   (b) No
Worked example Straight across a river
A river 78 m wide has straight parallel banks and flows at 1 m/s. A boat can move at 2.6 m/s in still water. Find the time to cross and the distance carried downstream if (a) the boat is steered at right angles to the bank (b) the boat is steered upstream so that it travels straight across.
  1. Take \(\mathbf{i}\) downstream and \(\mathbf{j}\) straight across. The current is \(1\mathbf{i}\) m/s.
  2. (a) Ground velocity \(= 2.6\mathbf{j} + 1\mathbf{i}\). Only the \(\mathbf{j}\) part crosses the river: time \(= \dfrac{78}{2.6} = 30\) s; downstream drift \(= 1 \times 30 = 30\) m.
  3. (b) Steer with velocity \(-1\mathbf{i} + b\mathbf{j}\) so the current is cancelled: \(1^{2} + b^{2} = 2.6^{2}\), so \(b = \sqrt{5.76} = 2.4\).
  4. Time \(= \dfrac{78}{2.4} = 32.5\) s. The boat points upstream at \(\cos^{-1}\dfrac{1}{2.6} = 67.4^{\circ}\) to the bank.
Answer: (a) 30 s, 30 m downstream   (b) 32.5 s, steering at \(67.4^{\circ}\) to the upstream bank
Your turn
A plane flies with velocity \((160\mathbf{i} + 120\mathbf{j})\) km/h in still air. The wind blows with velocity \((40\mathbf{i} - 120\mathbf{j})\) km/h. Find the plane's velocity over the ground and its speed.
Examiner trap

Finding \(t\) from the \(\mathbf{i}\) components and stopping there proves nothing: the paths always cross somewhere unless they are parallel. You must show the \(\mathbf{j}\) components agree at the SAME \(t\). In river questions, steering straight across is not the same as travelling straight across; only the component of the ground velocity across the river sets the crossing time.

Stage 2 · Exam

Exam-style questions

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A ship's position at time \(t\) hours is \(\mathbf{r} = (3\mathbf{i} + 2\mathbf{j}) + t(4\mathbf{i} - 2\mathbf{j})\) km, where \(\mathbf{i}\) is due east and \(\mathbf{j}\) due north. Find the position of the ship when it is due east of the origin \(O\).

Exam style+20 XP

At time \(t\) hours, \(\mathbf{r}_B - \mathbf{r}_A = (6 - 3t)\mathbf{i} + (4 - t)\mathbf{j}\) km. Find the least distance between \(A\) and \(B\).

Exam style+20 XP

A river 60 m wide flows at 1.5 m/s. A boat can move at 2.5 m/s in still water and is steered so that it travels straight across. How long does the crossing take?

Exam style+20 XP

Particle \(A\) has position \((2\mathbf{i} + 7\mathbf{j}) + t(3\mathbf{i} - \mathbf{j})\) and particle \(B\) has position \((8\mathbf{i} + \mathbf{j}) + t(\mathbf{i} + p\mathbf{j})\). Given that they collide, find \(p\).

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Six original Cambridge-style questions
Speed and bearing from a velocity, velocity from a speed and direction, due north and due east of a point, the closest approach of two boats, a collision with an unknown constant, and crossing a river.