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IGCSE · Practice questions · Velocity vectors: position, time and collisions

Start plus t times velocity.

Six original Cambridge-style questions on velocity vectors: speed and bearing, building a velocity from a direction, position at time \(t\), the least distance between two particles, collisions and a boat crossing a river.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Write \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\), then compare components.

01
Recall
[2 marks]

The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point due east and due north. A boat has velocity \((-7\mathbf{i} + 24\mathbf{j})\) km/h. Find its speed and the bearing on which it travels.

  • Speed \(= \sqrt{(-7)^{2} + 24^{2}} = \sqrt{625} = 25\) km/h. ✓
  • It moves 24 north and 7 west each hour, so it heads \(\tan^{-1}\dfrac{7}{24} = 16.3^{\circ}\) west of north. ✓
  • Bearing \(= 360^{\circ} - 16.3^{\circ} = 343.7^{\circ}\). ✓
02
Calculation
[3 marks]

A particle starts at the point with position vector \((4\mathbf{i} + 9\mathbf{j})\) m and moves with constant speed 39 m/s in the direction of \(12\mathbf{i} - 5\mathbf{j}\).

(a) Find its velocity.

(b) Find its position vector after 2 seconds.

  • (a) \(|12\mathbf{i} - 5\mathbf{j}| = \sqrt{144 + 25} = 13\), so \(\mathbf{v} = \dfrac{39}{13}(12\mathbf{i} - 5\mathbf{j}) = (36\mathbf{i} - 15\mathbf{j})\) m/s. ✓
  • (b) \(\mathbf{r} = (4\mathbf{i} + 9\mathbf{j}) + 2(36\mathbf{i} - 15\mathbf{j})\). ✓
  • \(\mathbf{r} = (76\mathbf{i} - 21\mathbf{j})\) m. ✓
03
Calculation
[4 marks]

The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) point due east and due north. At 09:00 a ship is at \((-4\mathbf{i} + 10\mathbf{j})\) km and it moves with velocity \((6\mathbf{i} - 2\mathbf{j})\) km/h. A lighthouse \(L\) is at \((8\mathbf{i} + 2\mathbf{j})\) km.

(a) Find the time at which the ship is due north of \(L\), and how far it is from \(L\) then.

(b) Find the time at which the ship is due east of \(L\), and how far it is from \(L\) then.

  • \(\mathbf{r} = (-4 + 6t)\mathbf{i} + (10 - 2t)\mathbf{j}\), \(t\) in hours after 09:00. ✓
  • (a) Equal \(\mathbf{i}\) parts: \(-4 + 6t = 8\), so \(t = 2\), at 11:00. Then \(\mathbf{r} = 8\mathbf{i} + 6\mathbf{j}\), which is \(6 - 2 = 4\) km north of \(L\). ✓
  • (b) Equal \(\mathbf{j}\) parts: \(10 - 2t = 2\), so \(t = 4\), at 13:00. ✓
  • Then \(\mathbf{r} = 20\mathbf{i} + 2\mathbf{j}\), which is \(20 - 8 = 12\) km east of \(L\). ✓
04
Analysis
[5 marks]

At time \(t\) hours after noon, boat \(A\) has position \(\mathbf{r}_A = (8\mathbf{i} + 3\mathbf{j}) + t(-\mathbf{i} + 2\mathbf{j})\) km and boat \(B\) has position \(\mathbf{r}_B = (\mathbf{i} + 2\mathbf{j}) + t(2\mathbf{i} + 6\mathbf{j})\) km.

(a) Show that the square of the distance between the boats is \(25t^{2} - 50t + 50\).

(b) Find the least distance between the boats and the time at which it occurs.

(c) Find the position vectors of the boats at that time.

  • (a) \(\mathbf{r}_B - \mathbf{r}_A = (1 + 2t - 8 + t)\mathbf{i} + (2 + 6t - 3 - 2t)\mathbf{j} = (3t - 7)\mathbf{i} + (4t - 1)\mathbf{j}\). ✓
  • \(d^{2} = (3t - 7)^{2} + (4t - 1)^{2} = 9t^{2} - 42t + 49 + 16t^{2} - 8t + 1 = 25t^{2} - 50t + 50\). ✓
  • (b) \(25t^{2} - 50t + 50 = 25(t - 1)^{2} + 25\), least value 25 when \(t = 1\). ✓
  • Least distance \(\sqrt{25} = 5\) km, at 13:00. ✓
  • (c) \(\mathbf{r}_A = (7\mathbf{i} + 5\mathbf{j})\) km and \(\mathbf{r}_B = (3\mathbf{i} + 8\mathbf{j})\) km. Check: \(\sqrt{4^{2} + 3^{2}} = 5\). ✓
05
Analysis
[5 marks]

Particle \(P\) starts at \((-3\mathbf{i} + 4\mathbf{j})\) m with velocity \((5\mathbf{i} + 2\mathbf{j})\) m/s. At the same moment particle \(Q\) starts at \((9\mathbf{i} + 16\mathbf{j})\) m with velocity \((\mathbf{i} + q\mathbf{j})\) m/s, where \(q\) is a constant.

(a) Given that \(P\) and \(Q\) collide, find \(q\), the time of the collision and the position vector of the point where they collide.

(b) If instead \(q = -1\), find the distance between the particles at the time found in (a).

  • (a) \(\mathbf{r}_P = (-3 + 5t)\mathbf{i} + (4 + 2t)\mathbf{j}\) and \(\mathbf{r}_Q = (9 + t)\mathbf{i} + (16 + qt)\mathbf{j}\). ✓
  • \(\mathbf{i}\): \(-3 + 5t = 9 + t\), so \(t = 3\). ✓
  • \(\mathbf{j}\) at \(t = 3\): \(4 + 6 = 16 + 3q\), so \(q = -2\). ✓
  • They collide after 3 s at \((12\mathbf{i} + 10\mathbf{j})\) m. ✓
  • (b) With \(q = -1\), at \(t = 3\): \(\mathbf{r}_Q = 12\mathbf{i} + 13\mathbf{j}\) while \(\mathbf{r}_P = 12\mathbf{i} + 10\mathbf{j}\), so they are \(3\) m apart (\(Q\) due north of \(P\)). ✓
06
Analysis
[6 marks]

A river is 80 m wide with straight parallel banks. The water flows at 1.2 m/s parallel to the banks. A boat can travel at 2 m/s in still water. Take \(\mathbf{i}\) as a unit vector downstream and \(\mathbf{j}\) as a unit vector straight across the river.

(a) The boat is steered at right angles to the bank. Find its speed over the ground, the time it takes to cross and how far downstream it lands.

(b) The boat is instead steered so that it lands directly opposite its starting point. Find the angle between the bank and the direction in which the boat is steered, and the time it takes to cross.

  • (a) Ground velocity \(= 2\mathbf{j} + 1.2\mathbf{i}\); speed \(= \sqrt{1.2^{2} + 2^{2}} = \sqrt{5.44} \approx 2.33\) m/s. ✓
  • Only the \(\mathbf{j}\) part crosses the river: time \(= \dfrac{80}{2} = 40\) s. ✓
  • Downstream distance \(= 1.2 \times 40 = 48\) m. ✓
  • (b) Steer with velocity \(-1.2\mathbf{i} + b\mathbf{j}\) through the water, so the ground velocity is \(b\mathbf{j}\): \(1.2^{2} + b^{2} = 2^{2}\), so \(b = \sqrt{2.56} = 1.6\). ✓
  • The steering direction makes \(\tan^{-1}\dfrac{1.6}{1.2} = 53.1^{\circ}\) with the upstream bank. ✓
  • Time \(= \dfrac{80}{1.6} = 50\) s. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.