Velocity Vectors and Moving Objects

Turn a speed and direction into a velocity vector, add velocities, find position at time t, and decide whether two moving objects collide.

In the exam for IGCSE Additional Maths 0606 · 5 real past-paper questions below

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Velocity and speed

Two ferries both travel at 20 km/h. One heads to the island, the other back to the harbour. Same speed, very different journeys. Speed is only a size. Velocity is a size and a direction.

Rule

A velocity is a vector, such as $v = 6 i - 8 j$ m s⁻¹. Each second the object moves 6 m in the i direction and 8 m in the −j direction.

Formula

speed = ∣ v ∣

$v$ — the velocity vector. Its magnitude is the speed, which is never negative.

Worked example

1
A particle has velocity $6 i - 8 j$ m s⁻¹. Find its speed.
2
$∣6 i - 8 j ∣ = 6^{2} + (- 8)^{2}$ — Speed is the magnitude of the velocity.
3
$= 100 = 10 m s^{- 1}$ — Include the units.

In these questions, i usually points east and j points north. The question will say so at the start.

Common mistake

Giving a velocity when the question asks for speed. Speed is one number; velocity is a vector.

Key idea

Velocity tells you where the object is heading. Speed only tells you how fast.

Building a velocity vector

Most questions give a speed and a direction, and you must build the velocity vector. Use the unit vector from the last lesson.

Formula

v = s \times \frac{d}{∣ d ∣}

$s$ — the speed.
$d$ — any vector in the direction of motion.

Worked example

1
A particle moves at 26 m s⁻¹ in the direction of $5 i + 12 j$ . Find its velocity.
2
$∣5 i + 12 j ∣ = 25 + 144 = 13$ — The length of the direction vector.
3
$\frac{26}{13} = 2$ — The velocity must be twice as long.
4
$v = 10 i + 24 j m s^{- 1}$ — Scale both components by 2.

If the direction is a bearing, measure it clockwise from north. With i east and j north, the east part uses sin and the north part uses cos.

Formula

v = s sin θ i + s cos θ j

$s$ — the speed.
$θ$ — the bearing, measured clockwise from north.

Worked example

1
A boat sails at 20 km/h on a bearing of $06 0^{\circ}$ . Find its velocity vector.
2
$20 sin 6 0^{\circ} = 20 \times \frac{3}{2} = 103$ — The east part. A bearing is measured from north, so east uses sin.
3
$20 cos 6 0^{\circ} = 20 \times \frac{1}{2} = 10$ — The north part uses cos.
4
$v = 103 i + 10 j km/h$ — Check: $300 + 100 = 20$ , the speed.

Worked example

1
The same idea works for any bearing. Find the velocity for 8 km/h on a bearing of $30 0^{\circ}$ .
2
$8 sin 30 0^{\circ} = 8 \times (- \frac{3}{2}) = - 43$ — Negative: the boat is moving west.
3
$8 cos 30 0^{\circ} = 8 \times \frac{1}{2} = 4$ — Positive: it is also moving north.
4
$v = - 43 i + 4 j$ — The signs come out of sin and cos by themselves.

Common mistake

Using cos for the east part with a bearing. That rule is for angles from the x-axis. Bearings start from north, so east uses sin.

Rule

No calculator on Paper 1: use exact values such as $sin 6 0^{\circ} = \frac{3}{2}$ and $cos 6 0^{\circ} = \frac{1}{2}$ .

Key idea

Speed times a unit vector gives the velocity. Always check that its magnitude is the speed.

Position at time t

If you know where an object starts and its velocity, you know where it is at any time. Each hour it moves one more velocity vector.

Formula

r = r_{0} + t v

$r_{0}$ — the position vector at the start, when t = 0.
$t$ — the time since the start.
$v$ — the constant velocity.
$r$ — the position vector at time t.

Rule

Learn this. It is not on the 0606 or 4037 list of formulas.

Worked example

1
A boat starts at $2 i + 5 j$ km and moves with velocity $3 i - 4 j$ km/h. Find its position after 2 hours, and at time t.
2
$r = (2 i + 5 j) + 2 (3 i - 4 j)$ — Start position plus 2 lots of the velocity.
3
$= 8 i - 3 j$ — Collect the i parts and the j parts.
4
$r = (2 + 3 t) i + (5 - 4 t) j$ — At time t: the same, with t in place of 2.

Worked example

1
Does the same boat pass through the point $14 i - 11 j$ ?
2
$2 + 3 t = 14 \Rightarrow t = 4$ — Match the i parts.
3
$5 - 4 t = - 11 \Rightarrow t = 4$ — Match the j parts.
4
$same t$ — Both parts give t = 4, so the boat is there after 4 hours.

Key idea

Position = start + time × velocity. Write it with t in it, because later parts of the question use it.

Adding velocities

A boat crossing a river is pushed by the current too. Its actual velocity is the sum of its own velocity through the water and the velocity of the current. This sum is called the resultant velocity.

Formula

v_{actual} = v_{boat} + v_{current}

$v_{boat}$ — how the boat moves through the water.
$v_{current}$ — how the water itself moves (or the wind, for a plane).
$v_{actual}$ — the resultant: how the boat really moves over the ground.

Worked example

1
A boat moves through the water with velocity $4 i + 3 j$ m s⁻¹. The current is $2 i - j$ m s⁻¹. Find the boat's actual velocity and speed.
2
$(4 i + 3 j) + (2 i - j) = 6 i + 2 j$ — Add the two velocities part by part.
3
$∣6 i + 2 j ∣ = 40 = 4 \times 10 = 210$ — The actual speed, as an exact surd.

Worked example

1
With the same current, what velocity must the boat steer so that it actually moves at $5 i$ m s⁻¹, due east?
2
$v_{boat} + (2 i - j) = 5 i$ — Write the vector equation.
3
$v_{boat} = 3 i + j$ — Subtract the current. The boat aims a little north to cancel the current pushing south.

Common mistake

Adding the speeds instead of the velocities. Speeds in different directions do not simply add; add the vectors, then find the magnitude.

Key idea

Velocities add like vectors. Add first, then find the speed.

Do they collide?

Two objects collide if they are at the same place at the same time. Write each position with t in it, then match both parts.

Steps

1
1. Write $r$ for each object at time t.
2. Make the i parts equal and solve for t.
3. Make the j parts equal and solve for t.
4. Same t from both: they collide, and the position tells you where. Different t: they do not collide.

Worked example

1
P starts at $2 i + 9 j$ with velocity $4 i - j$ . Q starts at $10 i + j$ with velocity $2 i + j$ , at the same moment. Do they collide?
2
$r_{P} = (2 + 4 t) i + (9 - t) j$ — Position of P at time t.
3
$r_{Q} = (10 + 2 t) i + (1 + t) j$ — Position of Q at time t.
4
$2 + 4 t = 10 + 2 t \Rightarrow t = 4$ — Match the i parts.
5
$9 - t = 1 + t \Rightarrow t = 4$ — Match the j parts. The same time.
6
$r = 18 i + 5 j$ — They collide at t = 4, at this point.

If one object starts later, its clock starts later. If B sets off 1 hour after A, then B has been moving for only $t - 1$ hours.

Worked example

1
A starts at the origin at noon with velocity $6 i + 8 j$ km/h. B starts at $10 i + 4 j$ at 1 pm with velocity $4 i + 10 j$ km/h. Show that they collide.
2
$r_{A} = 6 t i + 8 t j$ — t is the time in hours after noon.
3
$r_{B} = (10 + 4 (t - 1)) i + (4 + 10 (t - 1)) j$ — B moves for t − 1 hours.
4
$6 t = 4 t + 6 \Rightarrow t = 3$ — Match the i parts.
5
$8 t = 10 t - 6 \Rightarrow t = 3$ — Match the j parts. The same time, so they collide.
6
$r = 18 i + 24 j at 3 pm$ — State where and when.

Common mistake

Finding t from the i parts only. You must show the j parts give the same t, or the paths may cross at different times.

Key idea

Same place, same time. One t must fit both components.

Distance between two objects

The vector from P to Q at time t is $r_{Q} - r_{P}$ . Its magnitude is the distance between them.

Formula

d = ∣ r_{Q} - r_{P} ∣

$r_{P}$ — the position of P at time t.
$r_{Q}$ — the position of Q at the same time t.

Worked example

1
$r_{P} = (1 + 3 t) i + (2 + t) j$ and $r_{Q} = (7 + t) i + (- 2 + 3 t) j$ . Find $d^{2}$ in terms of t, and show P and Q never collide.
2
$r_{Q} - r_{P} = (6 - 2 t) i + (2 t - 4) j$ — Subtract part by part.
3
$d^{2} = (6 - 2 t)^{2} + (2 t - 4)^{2}$ — Magnitude squared.
4
$= 8 t^{2} - 40 t + 52$ — Expand and collect.
5
$= 8 (t - \frac{5}{2})^{2} + 2$ — Complete the square.
6
$d^{2} \geq 2 > 0$ — The distance is never 0, so they never collide. The closest they get is $2$ , when t = 2.5.

At a single time, put the number in first. At t = 1 the vector is $4 i - 2 j$ , so $d = 20 = 25$ .

Common mistake

Finding the distance between the start points. Both objects have moved, so use both positions at the same time t.

Key idea

Distance apart = magnitude of the difference of the two position vectors, at the same moment.

Which method? Read the question

Velocity questions come in a chain: build the velocity, then the position, then compare. Read each part for its clue.

You see	You do
Speed and a direction vector	Unit vector × speed
Speed and a bearing	$s sin θ i + s cos θ j$ (i east, j north)
“Find the speed”	Magnitude of the velocity
“Position at time t”	$r = r_{0} + t v$
Starts later, at time T	Use $t - T$ in place of t
Current or wind	Add the velocities, then find the magnitude
“Do they collide?”	Equate both components; same t means yes
“Distance between them”	$∣ r_{Q} - r_{P} ∣$ at the same time

Common mistake

Mixing up a position vector and a velocity vector. Position is where it is; velocity is how it moves each unit of time.

Key idea

Build v, then r. Almost every later part of the question uses the r you wrote with t in it.

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Real past-paper questions on velocity vectors and moving objects

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