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

These revision notes condense 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. For full explanations and worked examples use the study guide, then test yourself with the practice questions.

Course links: the Edexcel A Level Mathematics hub, the printable checklist and the free diagnostics.

1.1 Models: terms and assumptions

Term	What you assume	Effect to quote
Particle	Mass at a point	Size and rotation ignored
Lamina	Flat, negligible thickness	Treated as two-dimensional
Rigid body	No bending or deforming	Shape stays fixed
Rod	Thin and rigid	One-dimensional
Light	Negligible mass	Tension equal along a string or rod
Uniform / non-uniform	Mass spread evenly / unevenly	Even spread puts the weight at the middle
Inextensible string	Does not stretch	Connected objects have the same speed and acceleration magnitude
Smooth	No friction	No friction force
Rough	Friction can act	Friction opposes motion
Light smooth pulley	No mass, no friction	Tension unchanged over the pulley
Bead	Particle on a wire	Slides freely (unless the wire is rough)
Wire	Thin and rigid	One-dimensional path
Peg	Fixed support	Object can rest or turn on it

Also state “air resistance ignored” and “g constant” where relevant. Always pair an assumption with its effect.

2.1 Vectors: key results

Task	Method
Magnitude of ai + bj	√(a² + b²)
Angle with i	tan θ = b/a, then place by signs
Angle with j	tan φ = a/b
Resolve R at angle θ above horizontal	R cos θ i + R sin θ j
Bearing β, speed v (i east, j north)	v sin β i + v cos β j
Resultant	Add i parts, add j parts

Method in steps: resolve a vector

1. Sketch the vector and mark the angle with a named reference line.
2. Draw the right-angled triangle with the vector as hypotenuse.
3. Adjacent side: R cos. Opposite side: R sin.
4. Give each component its sign from the diagram.

Small reminder

(6i + 2j) + (−3i + 7j) + (4i − j) = 7i + 8j, magnitude √113 = 10.6, angle with i of 48.8°.

2.2 Vectors in motion

Quantity	Relationship (constant values)
Velocity	change of displacement ÷ time
Acceleration	change of velocity ÷ time
Position with constant velocity	r = r₀ + vt
Velocity with constant acceleration	v = u + at, so a = (v − u)/t
Speed	Magnitude of velocity
Distance between two particles	Magnitude of the difference of position vectors

Method in steps: position vector problems

1. Write the position as r = r₀ + vt, giving an i part and a j part.
2. “Due north of O”: i part is zero. “Due east of O”: j part is zero.
3. Solve for t, then substitute to get the other component.
4. For two particles meeting, equate their position vectors and solve for t using both components. The same t must satisfy both.

Must-know distinctions

Pair	Difference
Speed and velocity	Speed is a size only; velocity has direction
Distance and displacement	Distance is the path length; displacement is the straight-line change of position with direction
Average speed and average velocity	Distance ÷ time versus displacement ÷ time

3.1 Constant acceleration

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

The specification lists these as formulae you are expected to know and says they will not be in the formulae booklet. They apply only when acceleration is constant.

Method in steps: constant acceleration

1. Draw the motion. Choose a positive direction.
2. Note which three of s, u, v, a and t are given and which one is asked for.
3. Choose the formula that leaves out the quantity you neither have nor want.
4. For a stop, v = 0. For rest start, u = 0. At the highest point, v = 0.
5. If acceleration changes, split the motion into stages and use the final velocity of one stage as the initial velocity of the next.

Small reminder

Car from 28 to 12 m s⁻¹ in 5 s: a = −3.2 m s⁻². Distance = ½(28 + 12)(5) = 100 m. Further stopping distance = 12²/(2 × 3.2) = 22.5 m.

Vertical motion under gravity

Take a = −9.8 m s⁻² if up is positive, and a = +9.8 if down is positive. Keep to one choice. At the top v = 0, and the time up equals the time down when the start and end are at the same level. When the end point is below the start, the displacement is negative and you solve a quadratic in t, rejecting the negative root.

Small reminder

Thrown up at 25 m s⁻¹ from a cliff top 30 m above the sea: 4.9t² − 25t − 30 = 0, so t = 6.10 s and the speed on arrival is √1213 = 34.8 m s⁻¹.

Motion graphs

Type of graph	Slope gives	Area gives
Displacement against time	Velocity	Nothing useful
Velocity against time	Acceleration	Displacement
Speed against time	Size of acceleration	Distance
Acceleration against time	Nothing needed here	Change in velocity

• A horizontal velocity-time line is constant velocity, and a straight sloping line is constant acceleration.
• Crossing the time axis on a velocity-time graph means a change of direction.
• Area below the axis is negative displacement. Count it as positive for distance.
• A horizontal displacement-time line means at rest.
• Use trapezium area ½(a + b)h for a journey graph, with the parallel sides vertical.

Small reminder

Velocity from 8 to −4 in 6 s: a = −2 m s⁻², rest at t = 4, displacement 16 − 4 = 12 m, distance 20 m.

Quick self-test

1. Find the magnitude of 9i − 12j.
2. A tow rope exerts 20 N, inclined 30° up from the horizontal. Resolve this into horizontal and vertical parts.
3. Find the resultant of (2i + 5j) and (−6i + j).
4. A particle moves from (i + 2j) m to (13i − 10j) m in 4 s at constant velocity. Find its velocity and speed (3 significant figures).
5. A body has u = 5 m s⁻¹, a = 2 m s⁻² and t = 6 s. Find s and v.
6. A body starts from rest with acceleration 1.5 m s⁻² and travels 48 m. Find its final speed.
7. A body finishes at 14 m s⁻¹ having accelerated at 2 m s⁻² for 5 s. Find the distance travelled using s = vt − ½at².
8. A car at 20 m s⁻¹ brakes uniformly at 4 m s⁻². Find the time and distance to stop.
9. A velocity-time graph rises from 0 to 8 m s⁻¹ in 4 s and then stays at 8 m s⁻¹ for 6 s. Find the distance travelled.
10. A ball is dropped from rest and falls for 3 s (g = 9.8 m s⁻²). Find the distance fallen and speed reached.
11. A boat travels at 10 m s⁻¹ on bearing 060°. Write the velocity in terms of i (east) and j (north), to 3 significant figures.
12. State what “light” means for a string and what it implies.

Answers

1. √(81 + 144) = 15.
2. 20 cos 30° = 17.3 N horizontal and 20 sin 30° = 10 N vertical.
3. −4i + 6j.
4. v = ((13 − 1)i + (−10 − 2)j)/4 = (3i − 3j) m s⁻¹, speed = √18 = 4.24 m s⁻¹.
5. s = 5(6) + ½(2)(36) = 66 m; v = 5 + 12 = 17 m s⁻¹.
6. v² = 0 + 2(1.5)(48) = 144, so v = 12 m s⁻¹.
7. s = 14(5) − ½(2)(25) = 70 − 25 = 45 m.
8. t = 20/4 = 5 s; s = 400/(2 × 4) = 50 m.
9. ½(8)(4) + 8(6) = 16 + 48 = 64 m.
10. s = ½(9.8)(9) = 44.1 m; v = 9.8 × 3 = 29.4 m s⁻¹.
11. 10 sin 60° = 8.66 east; 10 cos 60° = 5.00 north; (8.66i + 5.00j) m s⁻¹.
12. It has negligible mass, so the tension is the same throughout it (provided it passes only over smooth surfaces or pulleys).

Where marks are usually lost

• Writing an assumption such as “inextensible” without saying that it makes the accelerations equal.
• Using the wrong trigonometric ratio in a component because there was no sketch.
• Quoting an angle without naming the line it is measured from.
• Reversing the order of subtraction when finding a change in velocity or displacement.
• Using an acceleration of +3 for a deceleration of 3 m s⁻².
• Applying one constant acceleration formula across a change of acceleration.
• Mixing positive directions, such as taking up as positive for u but down for a.
• Giving displacement when the question asks for distance, or velocity when it asks for speed.
• Accepting both roots of a time equation when one is negative or earlier than the start of motion.

Official syllabus

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