Module 1: Mathematical Foundations · Practice · complements Foundations 1–11
One worked example for each Foundation — the calculation in full, and the question a physicist would ask around it.
How to use this page
Every example is set out in three parts. The Physics asks what is happening, before any symbol is written. The Mathematics does the work in full, at the standard the exam expects. The Contemplation asks what the method just revealed about how we describe the world.
Read the problem, close the page, and attempt it. Then compare — not only your answer, but where you started. The first section is the one most students skip, and it is the one that separates a physicist from a calculator.
On this page
- An equation that cannot be right
- Seven kilometres travelled, five kilometres away
- A point in two coordinate systems
- Crossing a flowing river
- A force that does nothing
- The same force, two different torques
- Displacement from a graph, without calculus
- Why circular motion has acceleration
- Twelve metres travelled, four metres moved
- The approximation that makes the pendulum solvable
- Everything at once
1. An equation that cannot be right Foundation 1
Problem. A student, working from memory, writes the displacement of a uniformly accelerating particle as s = ut + at³. Without solving anything, show that this equation cannot describe any physical situation, and say what has gone wrong.
The physics
An equation in physics is not a string of symbols that happens to balance. Each term stands for a physical quantity, and terms that are added together must be quantities of the same kind. You can add a length to a length. You cannot add a length to a time, or to anything that is not a length, whatever the algebra permits. So before checking whether this formula gives the right number, we can ask whether it is capable of giving a number that means anything.
The mathematics
Take the units of each term in turn.
The left side: s is a displacement, measured in m.
First term: ut = (m s−1)(s) = m. Correct.
Second term: at³ = (m s−2)(s³) = m s. Not a length.
The equation therefore reads, in units: metres = metres + metre-seconds. The two terms on the right are not quantities of the same kind, so their sum is meaningless, and no value of u, a or t can repair it.
The correct form is s = ut + ½at², in which the second term gives (m s−2)(s²) = m. The student has mis-remembered the power on t, and the units expose it in three lines without any physical reasoning about motion at all.
The contemplation
Notice that the check required no knowledge of what the particle was doing. The units caught the error before physics was consulted at all — which means units are not a labelling convention applied after the thinking, but a constraint that operates on the thinking itself. This is worth building into a habit. Every equation you write can be tested this way, in seconds, and it catches a whole class of errors that no amount of care with arithmetic would find.
2. Seven kilometres travelled, five kilometres away Foundation 2
Problem. A cyclist rides 3 km due east, then 4 km due north, taking 20 minutes in total. Find the distance travelled, the magnitude of the displacement, the average speed and the magnitude of the average velocity.
The physics
Four quantities are asked for, and they fall into two pairs. Distance and speed are scalars: they accumulate along the path, and they do not care which way the cyclist was pointing. Displacement and velocity are vectors: they depend only on where the journey started and where it ended, and the route between is irrelevant to them. Expect the two pairs to disagree, and expect the vector answers to be smaller — a journey with a turn in it never gets you as far as it takes you.
The mathematics
Distance is the path length, added arithmetically because it is a scalar:
3 + 4 = 7 km
Displacement is the vector sum. The two legs are perpendicular, so:
Δr = 3î + 4ĵ km
|Δr| = √(3² + 4²) = √25 = 5 km, at tan−1(4/3) = 53.1° north of east
With t = 20 min = 1/3 h:
average speed = 7 ÷ (1/3) = 21 km/h
|average velocity| = 5 ÷ (1/3) = 15 km/h
Both first quadrant, so no quadrant correction is needed here — but check the signs before reaching for the arctangent as a habit.
The contemplation
Two speeds have come out of one journey, 21 and 15, and neither is wrong. They are answers to different questions: how hard was the cyclist working, and how much progress was made. Everyday language has one word where physics needs two, and that is the whole source of the confusion — not the mathematics, which is a single application of Pythagoras. When a physical term feels slippery, it is usually because ordinary speech has collapsed a distinction the physics keeps open.
3. A point in two coordinate systems Foundation 3
Problem. A point P has Cartesian coordinates (−3, 3√3). Find its polar coordinates with 0 ≤ θ < 2π, and write the polar unit vectors r̂ and θ̂ at P in terms of î and ĵ.
The physics
Before touching a calculator, place the point. The x-coordinate is negative and the y-coordinate positive, so P is up and to the left — second quadrant. That single observation is what will make the answer right, because the arctangent function cannot distinguish the second quadrant from the fourth and will hand back a wrong angle without complaint. Note also that r̂ and θ̂ are not fixed directions in the plane; they are defined at P, and at any other point they would come out differently.
The mathematics
r = √((−3)² + (3√3)²) = √(9 + 27) = √36 = 6
For the angle, take the acute angle first, from the magnitudes only:
α = tan−1|3√3 / −3| = tan−1(√3) = π/3
Second quadrant: θ = π − π/3 = 2π/3 (120°)
So P = (6, 2π/3). A calculator asked for tan−1(−√3) would have returned −π/3, pointing into the fourth quadrant — the diametrically opposite direction.
The unit vectors follow from their standard forms, evaluated at this θ:
r̂ = cos θ î + sin θ ĵ = −½ î + (√3/2) ĵ
θ̂ = −sin θ î + cos θ ĵ = −(√3/2) î − ½ ĵ
Two checks worth doing every time: each has magnitude 1, and their dot product is (−½)(−√3/2) + (√3/2)(−½) = 0, so they are perpendicular as required.
The contemplation
The calculator was not broken. It answered exactly the question it was asked — which number has this tangent — and that question has two answers a half-turn apart. The information distinguishing them was never in the ratio 3√3/(−3); it was in the individual signs, which the division destroyed. This is worth generalising: an operation that combines quantities can lose what distinguished them, and the sketch you drew at the start is holding information the algebra threw away.
4. Crossing a flowing river Foundation 4
Problem. A motorboat is steered due east across a river, with velocity 4 m/s relative to the water. The river flows due north at 3 m/s relative to the ground. Find the boat’s velocity relative to the ground, its speed, and the direction in which it actually travels.
The physics
The boat is doing two things at once, and neither interferes with the other. Its engine drives it east through the water; meanwhile the whole body of water — boat included — is carried north. A passenger watching the shore sees the sum. Note that the boat’s heading and its actual course are different directions, which is the entire content of the problem: pointing east does not mean going east.
The mathematics
Relative velocities compose by addition, with the intermediate reference cancelling:
vboat/ground = vboat/water + vwater/ground
= 4î + 3ĵ m/s
The two components are perpendicular, so the speed follows from Pythagoras:
|v| = √(16 + 9) = 5 m/s
θ = tan−1(3/4) = 36.9° north of east
Both components positive, so the direction is in the first quadrant and the arctangent may be used directly.
The contemplation
Ask which of these three velocities is the real one, and the question dissolves. The boat genuinely moves at 4 m/s through the water and genuinely moves at 5 m/s over the ground, and there is no further fact about which is its true speed. A velocity is always relative to something, and the phrase “the velocity of the boat” is incomplete in the same way that “the torque on the body” is incomplete without a pivot. Physics is full of quantities that appear absolute until you ask what they are measured against.
5. A force that does nothing Foundation 5
Problem. A constant force F = (3î + 4ĵ) N acts on a particle while it undergoes a displacement d = 5î m. Find the work done, the angle between force and displacement, and the contribution of the vertical component of the force.
The physics
The force points up and to the right; the particle moves only to the right. Work measures energy transferred, and only the part of the force lying along the motion can transfer any. The vertical part is pushing in a direction the particle never goes, so it should contribute nothing — and if the arithmetic says otherwise, the arithmetic is wrong. Expect an answer smaller than |F||d| = 25 J.
The mathematics
W = F · d = (3)(5) + (4)(0) = 15 J
The second term is zero because the displacement has no y-component. The vertical force of 4 N contributes exactly nothing, as anticipated.
For the angle, use the two definitions of the dot product together. With |F| = √(9 + 16) = 5 N and |d| = 5 m:
cos θ = 15 / (5 × 5) = 0.6, so θ = 53.1°
No quadrant correction is possible or needed: cos−1 returns a value between 0° and 180°, which is exactly the range an angle between two vectors can occupy.
The contemplation
A force of 4 N was applied throughout and accomplished nothing whatever. It was not too weak, and it did not cancel against anything — it simply pointed somewhere the particle was not going. This is why work is a dot product rather than a product: nature is not asking how large the force was, but how much of it was aligned with the motion. The mathematics did not impose that; it was chosen because the world behaves that way.
6. The same force, two different torques Foundation 6
Problem. A force F = 5ĵ N acts at the point Q = (2, 0, 0) m. Find the torque about the origin O, and about the point P = (−1, 0, 0) m. Comment on the comparison.
The physics
One force, one point of application, two questions. Torque measures turning effect about a chosen axis, so before computing anything we must accept that the answer depends on a choice we make. The lever arm from P is longer than from O, so the torque about P should come out larger. Both rotations are anticlockwise seen from above, so both torques should point along +k̂.
The mathematics
About the origin. The position vector from O to Q is r = 2î m.
τO = (2î) × (5ĵ) = 10(î × ĵ) = 10k̂ N m
About P. The position vector now runs from P to Q:
rQ/P = rQ − rP = 2î − (−î) = 3î m
τP = (3î) × (5ĵ) = 15k̂ N m
Both point along +k̂, as expected, and the second is larger in the ratio 3:2 — exactly the ratio of the two lever arms. Reading the magnitudes as force times perpendicular distance gives the same answers directly: 5 × 2 = 10 and 5 × 3 = 15.
The contemplation
Two different answers, both correct, for one unchanged force. Nothing about the physical situation differed between the calculations — only where we chose to stand. So “the torque on this body” names nothing until a pivot is supplied, and a problem that omits one has not asked a complete question. Compare the boat in example 4: velocity needs a reference frame, torque needs a pivot. Neither incompleteness is a flaw in the physics. It is the physics telling you that some quantities are relations rather than properties.
7. Displacement from a graph, without calculus Foundation 7
Problem. A body moves in a straight line. Its velocity increases uniformly from 2 m/s at t = 0 to 10 m/s at t = 4 s. Find the displacement over that interval using the geometry of the velocity–time graph, and state the acceleration.
The physics
Velocity is not constant, so displacement is not simply velocity times time. But the increase is uniform, which means the graph is a straight line, and the area under a straight line is a shape school geometry can handle. Before computing: the velocity averages somewhere near 6 m/s over 4 s, so expect roughly 24 m. That estimate is worth making, because it will catch any slip in the arithmetic.
The mathematics
The region under the line is a trapezium with parallel sides v1 and v2 and width Δt:
Δx = ½(v1 + v2) Δt = ½(2 + 10)(4)
= 24 m
Equivalently, split the region: a rectangle 2 × 4 = 8 m under the initial velocity, plus a triangle ½ × 4 × 8 = 16 m for the gain. Same 24 m, and the split shows where each part of the motion came from.
The acceleration is the gradient of the line:
a = (10 − 2)/4 = 2 m/s²
The contemplation
One graph answered two entirely different questions. Its gradient gave a rate; its area gave an accumulation. Those are the only two questions calculus exists to answer, and here both were settled with a ruler. The straight line is what made it possible — and when the line curves, the questions do not change, only the technique for answering them. Foundations 8 and 9 are that technique. This example is what they are generalising.
8. Why circular motion has acceleration Foundation 8
Problem. A particle moves in a circle of radius R at constant angular speed ω. One student argues that since the speed Rω never changes, the acceleration must be zero. Another argues that the acceleration cannot be zero because the direction of motion is always changing. Settle the disagreement by differentiating the position vector in polar coordinates.
The physics
Both students have noticed something true. The speed really is constant, and the direction really does change. The disagreement is about whether acceleration responds to speed or to velocity — and velocity is a vector, so anything that changes its direction changes it. The place to look is the basis vectors: in polar coordinates they are attached to the particle and turn with it, so differentiating cannot treat them as constants.
The mathematics
The position is simply r = R r̂, with R constant. Write r̂ in Cartesian components, where the basis really is fixed, and differentiate using the chain rule with θ̇ = ω:
r̂ = cos θ î + sin θ ĵ
dr̂/dt = (−sin θ î + cos θ ĵ)θ̇ = ω θ̂
So the velocity is
v = R dr̂/dt = Rω θ̂
Its magnitude is Rω, constant — the first student’s observation, confirmed. But θ̂ is itself turning:
θ̂ = −sin θ î + cos θ ĵ
dθ̂/dt = (−cos θ î − sin θ ĵ)θ̇ = −ω r̂
Differentiating the velocity therefore gives
a = Rω dθ̂/dt = −Rω² r̂
Magnitude Rω², directed inward along −r̂. The second student is right, and the first student’s premise was true but did not support the conclusion.
The contemplation
Centripetal acceleration was never inserted into this calculation. Nobody wrote down a formula for it or invoked a rule about circular motion; it emerged because the basis vectors turned, and their turning had to be differentiated along with everything else. That is the strongest possible answer to a student who suspects centripetal acceleration is a bookkeeping fiction. It is what you get when you take seriously the fact that “outward” means something different at every point on the circle.
9. Twelve metres travelled, four metres moved Foundation 9
Problem. A particle moves along the x-axis with acceleration a(t) = (6t − 12) m/s², starting with v(0) = 9 m/s at x(0) = 5 m. Find the times at which it is momentarily at rest, and the total distance it travels between t = 0 and t = 4 s.
The physics
The acceleration starts negative and later turns positive, so the particle will slow, stop, reverse, and eventually turn again. That means the journey has direction changes in it, and the moment a journey reverses, distance and displacement part company. The question asks for distance, so the answer cannot simply be the difference between the final and initial positions — the reversals have to be located first.
The mathematics
Velocity. Integrate the acceleration, fixing the constant from v(0) = 9:
v(t) = 3t² − 12t + 9
Setting v = 0 gives 3(t − 1)(t − 3) = 0, so the particle is at rest at t = 1 s and t = 3 s.
Position. Integrate again, with x(0) = 5:
x(t) = t³ − 6t² + 9t + 5
Evaluate at the four moments that matter — the endpoints and the two reversals:
x(0) = 5 x(1) = 9 x(3) = 5 x(4) = 9 (all in m)
Now add the legs separately, taking each magnitude:
S = |9 − 5| + |5 − 9| + |9 − 5| = 4 + 4 + 4 = 12 m
Displacement = x(4) − x(0) = 4 m
The contemplation
A student who integrated velocity straight from 0 to 4 would get 4 m and stop, with a plausible number and no warning. The integral is not wrong — it faithfully returns the displacement it was asked for. It is the question that was wrong. This is the recurring danger of a fluent technique: it produces an answer to whatever it was actually asked, and rarely signals that you asked for the wrong thing. Locating the reversals first is not extra care; it is the physics deciding how the mathematics should be set up.
10. The approximation that makes the pendulum solvable Foundation 10
Problem. A simple pendulum of length L obeys d²θ/dt² + (g/L) sin θ = 0. Show how the small-angle approximation converts this into simple harmonic motion, obtain the period, and estimate the error at an amplitude of 10°.
The physics
The equation as written is exact and unsolvable in elementary functions — the sine of the unknown is what blocks it. The obstacle is not the physics but the shape of the restoring force: it is not proportional to the displacement, only nearly so when the displacement is small. If we restrict ourselves to small swings, we can replace the true restoring force with the straight-line one that matches it near zero, and the equation becomes one we recognise. What we lose is the large-swing behaviour, and we should say by how much.
The mathematics
Near θ = 0 the tangent line to sin θ has gradient cos 0 = 1 and passes through the origin, so the local straight-line approximation is sin θ ≈ θ, with θ in radians. Substituting:
d²θ/dt² + (g/L)θ = 0
Compare with the standard form d²x/dt² = −ω²x. The match is exact, with
ω² = g/L, so ω = √(g/L)
T = 2π/ω = 2π√(L/g)
The error. At 10° = 0.1745 rad, sin θ = 0.1736. The approximation overstates the restoring force by about 0.5%, and the resulting period is short by roughly 0.2% — well inside the precision of a stopwatch. At 30° the discrepancy in sin θ is 4.5%; at 90° the true period is about 18% longer than the formula gives.
The contemplation
The most famous formula in elementary mechanics is not a law. It is what one equation becomes after an approximation is applied, and it is false for a pendulum swinging widely. Notice too what the approximation bought — not merely an easier calculation, but a solvable one where none existed. Physics does this constantly, and the honest version of the result always carries its condition alongside it. A student who knows T = 2π√(L/g) has a formula; a student who knows it holds for small swings has the physics.
11. Everything at once Foundation 11
Problem. A particle of mass 1 kg moves along the x-axis with acceleration a(t) = (6t − 18) m/s². At t = 0 it is at x = 2 m moving at 24 m/s. For 0 ≤ t ≤ 6 s, find: the velocity and position functions; all turning points and their nature; the displacement and the total distance; and the net work done on the particle.
The physics
This one problem needs four tools. Integration to rebuild the motion from the acceleration, since only the acceleration is given. Differentiation, or rather its inverse read backwards, to find where the particle turns. Graph reasoning to classify those turning points. And the work–energy theorem to connect the beginning to the end without following the middle. Note that the acceleration is negative until t = 3 s and positive after, so the particle will slow, reverse, and reverse again — expect distance and displacement to differ.
The mathematics
Rebuilding the motion. Integrate twice, fixing each constant from the initial conditions:
v(t) = 3t² − 18t + C1, and v(0) = 24 gives C1 = 24
v(t) = 3t² − 18t + 24
x(t) = t³ − 9t² + 24t + 2
Turning points. These occur where v = 0:
3(t² − 6t + 8) = 3(t − 2)(t − 4) = 0, so t = 2 s and t = 4 s
To classify them, use the second derivative of position, which is the acceleration. At t = 2, a = 6(2) − 18 = −6 m/s², negative, so x has a local maximum. At t = 4, a = +6 m/s², so x has a local minimum.
Positions at the four key moments:
x(0) = 2 x(2) = 22 x(4) = 18 x(6) = 38 (all in m)
Displacement = 38 − 2 = 36 m
Distance = |22−2| + |18−22| + |38−18| = 20 + 4 + 20 = 44 m
Net work. By the work–energy theorem, only the endpoint speeds matter. Since v(6) = 3(36) − 18(6) + 24 = 108 − 108 + 24 = 24 m/s:
W = ½(1)(24²) − ½(1)(24²) = 0 J
The contemplation
The particle covered forty-four metres, reversed twice, and had work done on it totalling zero. Nothing is wrong: a force acted throughout, doing negative work while the particle slowed and positive work while it sped up, and the two happened to balance by t = 6 s. What is worth noticing is that the last calculation ignored everything in between. Four tools were needed to describe the journey, and none of them was needed to answer the final question — the work–energy theorem reached the endpoint directly. Knowing which tool a question actually requires is not a lesser skill than knowing how to use them all.
What to take from these eleven
Look back at where the difficulty actually sat. In example 3 the calculator was correct and the question was underspecified. In example 9 the integral was correct and the question was the wrong one. In example 6 there was no single right answer until a pivot was chosen. Not one of those was a failure of technique.
The mathematics in this module is not difficult, and with practice it becomes fast. What takes longer to build is the habit of asking, before the first line, what kind of question is being asked — and afterwards, whether the number that emerged is answering it. That habit is what the first and third sections of each example were for.
Complements: Foundations 1–11 · Type: Worked examples · Reading: 35 min · Best used: after reading the Foundation, before attempting problems