Pearson Edexcel International Advanced Level Mathematics: Models, vectors and constant acceleration (YMA01)

This guide covers Unit M1: Mechanics 1, sections 1.1, 2.1, 2.2 and 3.1 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). Under the specification, M1 is optional: it can be taken within IAS Mathematics, IAS Further Mathematics, IAL Mathematics and IAL Further Mathematics. It lists P1, P2 and vectors in two dimensions as prerequisites.

Course links: the Edexcel A Level Mathematics hub, the printable checklist and the free diagnostics. After this guide use the revision notes and the practice questions.

What this unit covers

Section	You must be able to
1.1	Use the terms particle, lamina, rigid body, rod, inextensible string, smooth and rough, light smooth pulley, bead, wire and peg, and state the assumptions behind them
2.1	Find the magnitude and direction of a vector, resolve it into two components, find resultants, use i and j
2.2	Apply vectors to displacement, velocity, acceleration and force in a plane
3.1	Use constant acceleration formulae and displacement-time, velocity-time, speed-time and acceleration-time graphs

1.1 Mathematical models in mechanics

A mathematical model replaces a real situation with a simpler one that you can solve. You choose assumptions, solve, then ask whether the answer is reasonable. The specification asks you to know these terms and their assumptions.

Term	Assumption it makes
Particle	All mass at one point; size, shape and rotation ignored
Lamina	Flat object with negligible thickness
Rigid body	Does not bend or deform
Rod	Long thin rigid object, treated as one-dimensional; light if its mass is negligible, uniform if its mass is spread evenly, non-uniform otherwise
Light string	Negligible mass, so tension is the same along it
Inextensible string	Does not stretch, so connected objects move together
Smooth surface	No friction
Rough surface	Friction acts
Light smooth pulley	Negligible mass and no friction, so the string tension is unchanged by passing over it
Bead	A particle threaded on a wire, free to slide
Wire	Thin, rigid, one-dimensional
Peg	A fixed support of negligible thickness

Other common assumptions are no air resistance and constant gravitational acceleration. When you give an assumption in an answer, add its effect, for example: “inextensible, so both particles have the same acceleration”.

2.1 Vectors: magnitude, direction and components

A vector has size and direction. In two dimensions write it as ai + bj, where i and j are unit vectors in perpendicular directions (often east and north, or horizontal and vertical).

magnitude of ai + bj = √(a² + b²)
angle θ with the i direction: tan θ = b/a (use the sign of a and b to place it)
a vector of magnitude R at angle θ to i: R cos θ i + R sin θ j

Worked example 1: magnitude and direction

Find the magnitude of v = (−5i + 12j) m s⁻¹ and the angle between v and the positive j direction.

|v| = √(25 + 144) = 13 m s⁻¹
tan φ = 5/12 (horizontal over vertical, as the angle is from j)
φ = 22.6° to 1 decimal place

The angle is 22.6° from the positive j direction, towards the negative i side.

Resolving a vector

To resolve means to split a vector into two perpendicular parts. Draw a right-angled triangle with the vector as the hypotenuse. The component along the side adjacent to the angle uses cos, the opposite side uses sin.

Worked example 2: resolving

A force of 18 N acts at 35° above the horizontal. With i horizontal and j vertically upwards:

horizontal = 18 cos 35° = 14.7 N
vertical   = 18 sin 35° = 10.3 N
force = (14.7i + 10.3j) N

For a bearing, measure clockwise from north. A velocity of 9 m s⁻¹ on bearing 140°, with i east and j north, has east component 9 sin 140° = 5.79 and north component 9 cos 140° = −6.89. So v = (5.79i − 6.89j) m s⁻¹. The sin and cos swap compared with an angle from the east axis.

Resultants

The resultant of several vectors is their sum. Add the i parts and the j parts separately. A vector diagram places them nose to tail; the resultant joins the start to the end.

Worked example 3: resultant of forces

Forces (6i + 2j) N, (−3i + 7j) N and (4i − j) N act on a particle.

resultant = (6 − 3 + 4)i + (2 + 7 − 1)j = (7i + 8j) N
magnitude = √(49 + 64) = √113 = 10.6 N
angle with i: tan α = 8/7, α = 48.8°

2.2 Vectors for displacement, velocity, acceleration and force

Displacement, velocity, acceleration and force are all vectors in a plane. Their magnitudes are, respectively, the straight-line distance from the start, the speed, the size of the acceleration and the size of the force.

The specification requires these two relationships for constant values:

velocity = (change of displacement)/time
acceleration = (change of velocity)/time

In symbols, for constant velocity v, a particle starting at r₀ has position r = r₀ + vt at time t. For constant acceleration a, a velocity u changes to v = u + at, so a = (v − u)/t.

Worked example 4: constant velocity

A particle is at (−4i + 9j) m at t = 0 and moves with constant velocity (2.5i − 3j) m s⁻¹. Find when it is due north of the origin O, and its position at t = 5.

r = (−4 + 2.5t)i + (9 − 3t)j
due north of O means i component = 0: −4 + 2.5t = 0, so t = 1.6 s
there, j component = 9 − 3(1.6) = 4.2, so it is 4.2 m north of O
at t = 5: r = (−4 + 12.5)i + (9 − 15)j = (8.5i − 6j) m

Speed is the magnitude of velocity: √(2.5² + 3²) = 3.91 m s⁻¹.

Worked example 5: acceleration from velocities

A particle’s velocity changes from (3i + 2j) m s⁻¹ to (11i − 6j) m s⁻¹ in 4 s, with constant acceleration.

a = ((11 − 3)i + (−6 − 2)j)/4 = (2i − 2j) m s⁻²
magnitude = √8 = 2√2 = 2.83 m s⁻²
direction: 45° below the positive i direction

Always take final minus initial. To find the distance between two moving particles at time t, subtract their position vectors and take the magnitude.

3.1 Motion in a straight line with constant acceleration

Constant acceleration in a straight line means the acceleration does not change. Use s for displacement, u for starting velocity, v for final velocity, a for acceleration and t for time. The specification lists these formulae as ones you are expected to know, and says they will not appear in the formulae booklet.

Formula	Use it when you have no value for
v = u + at	s
s = ut + ½at²	v
s = vt − ½at²	u
v² = u² + 2as	t
s = ½(u + v)t	a

Method

1. Choose a positive direction and keep it for every quantity.
2. Write down which of s, u, v, a and t you have been given, and which one you need.
3. Pick the formula that does not contain the fifth quantity.
4. Substitute, solve, and check that the answer has a sensible sign and size.

For a body falling or thrown vertically, a = ±9.8 m s⁻² with the sign set by your positive direction, and v = 0 at the highest point.

Worked example 6: braking

A car slows uniformly from 28 m s⁻¹ to 12 m s⁻¹ in 5 s. Find its acceleration and the distance in that time, and the further distance to stop.

a = (v − u)/t = (12 − 28)/5 = −3.2 m s⁻²
s = ½(u + v)t = ½(28 + 12)(5) = 100 m
to stop from 12 m s⁻¹: 0 = 12² + 2(−3.2)s, so s = 144/6.4 = 22.5 m

The two distances add to 122.5 m, which matches stopping from 28 m s⁻¹ directly: 28²/(2 × 3.2).

Worked example 7: vertical motion

A stone is thrown upwards at 25 m s⁻¹ from the top of a cliff 30 m above the sea. Find the time to reach the sea and its speed on arrival.

take up as positive: s = −30, u = 25, a = −9.8
−30 = 25t − 4.9t²  →  4.9t² − 25t − 30 = 0
t = (25 + √1213)/9.8 = 6.10 s  (the negative root is rejected)
v² = 25² + 2(9.8)(30) = 1213, so speed = 34.8 m s⁻¹

The velocity is −34.8 m s⁻¹, downwards; speed is the magnitude, 34.8 m s⁻¹.

Motion graphs

Type of graph	What its slope gives	What the enclosed area gives
Displacement against time	Velocity	Nothing useful
Velocity against time	Acceleration	Displacement
Speed against time	Size of the acceleration	Distance
Acceleration against time	Nothing needed here	Change in velocity

For a constant acceleration, a velocity-time graph is a straight line. A horizontal line means constant velocity. Where a velocity-time graph crosses the axis, the object turns round. Areas above the axis are positive displacement and areas below are negative. For total distance, add all areas as positive. A speed-time graph is the velocity-time graph with any part below the axis reflected above it.

A curve on a displacement-time graph means changing velocity. A horizontal section means the object is at rest.

Worked example 8: a journey in stages

A train accelerates uniformly from rest at 0.4 m s⁻² for 50 s, then travels at constant speed, then decelerates uniformly at 0.5 m s⁻² to rest. The total distance is 2500 m. Find the total time.

top speed = 0.4 × 50 = 20 m s⁻¹
braking time = 20/0.5 = 40 s
distance while accelerating = ½(20)(50) = 500 m
distance while braking = ½(20)(40) = 400 m
constant stage: 20T = 2500 − 900, so T = 80 s
total time = 50 + 80 + 40 = 170 s

Worked example 9: reading a velocity-time graph

A particle’s velocity falls uniformly from 8 m s⁻¹ at t = 0 to −4 m s⁻¹ at t = 6.

acceleration = (−4 − 8)/6 = −2 m s⁻²
at rest when 8 − 2t = 0, so t = 4
area 0 to 4 = ½(4)(8) = 16 (forwards)
area 4 to 6 = ½(2)(4) = 4 (backwards)
displacement = 16 − 4 = 12 m, distance = 16 + 4 = 20 m

The final speed is 4 m s⁻¹, but the velocity is −4 m s⁻¹.

Common errors

• Giving the assumption but not its consequence for the motion.
• Using tan θ = b/a for an angle measured from j, when it should be a/b.
• Swapping sin and cos when converting a bearing to components.
• Treating deceleration of 3 as a = +3 in a formula, when the acceleration is −3.
• Using a formula with a different positive direction from the one you set.
• Using one formula for a journey with more than one acceleration, instead of splitting it into stages.
• Confusing speed with velocity, or distance with displacement.
• Reading the gradient of a velocity-time graph as velocity.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. This guide covers Unit M1: Mechanics 1, sections 1.1, 2.1, 2.2 and 3.1.