Mechanics Overview – A Level Maths Year 1

Mechanics is the mathematics of how things move and why. In the first year of A Level it covers three DfE content areas: P · Quantities and units in mechanics, Q · Kinematics, and R · Forces and Newton's laws. Everything happens in a straight line, or with forces given as $\mathbf{i}$, $\mathbf{j}$ vectors.

Mechanics has two halves that meet in the middle. Kinematics describes motion — position, velocity, acceleration, time. Dynamics explains it — forces cause acceleration. The bridge between them is acceleration: Newton's second law gives you $a$ from the forces, and kinematics turns $a$ into speeds, distances and times.

The chain that solves most problems

forces $\xrightarrow{\;F = ma\;}$ acceleration $\xrightarrow{\;\text{suvat or calculus}\;}$ velocity, displacement, time

Worked Example — Forces then motion

A sledge of mass $20$ kg, initially at rest, is pulled across horizontal snow by a horizontal rope with tension $50$ N against a resistance of $14$ N. Find its speed and the distance travelled after $5$ s.

①Model the sledge as a particle; the resistance is constant.

②Newton's second law: $50 - 14 = 20a \Rightarrow a = 1.8$ m s$^{-2}$.

③Constant acceleration, so use suvat: $u = 0$, $a = 1.8$, $t = 5$.

④$v = 0 + 1.8(5) = 9$ m s$^{-1}$; $s = \tfrac12(1.8)(25) = 22.5$ m.

Calculators are allowed. Use $g = 9.8$ m s$^{-2}$ unless told otherwise, and give final answers to 2 or 3 significant figures, since $g$ is only given to 2 s.f. Always draw a diagram first — most lost marks in mechanics come from a missing or wrong force, not from the algebra.

Word	Meaning
Displacement $s$	Position relative to a fixed origin, with direction (vector).
Velocity $v$	Rate of change of displacement (vector). Speed is its magnitude.
Acceleration $a$	Rate of change of velocity (vector).
Particle	An object whose size can be ignored.
Light	Of negligible mass (strings, rods, pulleys).
Inextensible	Does not stretch: connected particles share the same acceleration.
Smooth	No friction.
Equilibrium	Resultant force zero: at rest or moving at constant velocity.
Newton (N)	Unit of force; $1$ N $= 1$ kg m s$^{-2}$.

Mass is not weight. Mass is measured in kg and is the same everywhere. Weight is a force, $W = mg$, measured in newtons.

1 — Units and weight

SI units m, s, kg; km h$^{-1}$ $\div 3.6$ $=$ m s$^{-1}$; $W = mg$

2 — Motion graphs

gradient of $s$–$t$ $=$ velocity; gradient of $v$–$t$ $=$ acceleration; area under $v$–$t$ $=$ displacement

3 — Constant acceleration

$v = u + at$, $s = \tfrac12(u+v)t$, $s = ut + \tfrac12at^2$, $s = vt - \tfrac12at^2$, $v^2 = u^2 + 2as$

4 — Variable acceleration

$v = \dfrac{\mathrm{d}s}{\mathrm{d}t}$, $a = \dfrac{\mathrm{d}v}{\mathrm{d}t}$, $v = \displaystyle\int a\,\mathrm{d}t$, $s = \displaystyle\int v\,\mathrm{d}t$

5 — Newton's second law

resultant force $= ma$ (in vector form, $\mathbf{F} = m\mathbf{a}$)

Which kinematics tool? If the acceleration is constant (and especially if it comes from constant forces), use suvat. If $s$, $v$ or $a$ is given as a function of $t$, use calculus.

The DfE content areas for Year 1 mechanics, and the page that covers each part.

PQuantities, Units and ModellingSI and derived units, scalars and vectors, $g = 9.8$ m s$^{-2}$, and the modelling words: particle, light, inextensible, smooth, rigid.
QMotion GraphsDisplacement–time and velocity–time graphs; what gradients and areas mean.
QConstant Acceleration (suvat)The five equations, their derivation, vertical motion under gravity and multi-stage problems.
QVariable AccelerationUsing differentiation and integration in one-dimensional kinematics.
RForces and Newton's LawsForce diagrams, Newton's three laws, $F = ma$ with vectors, equilibrium, lifts, towbars and pulleys.

Mixing units. Convert grams to kg, km to m and km h$^{-1}$ to m s$^{-1}$ before substituting.

Changing the positive direction part-way. In vertical motion, if up is positive then $a = -9.8$ for the whole flight — on the way up, at the top and on the way down.

Using suvat with variable acceleration. If $a$ depends on $t$, the suvat equations do not apply.

Leaving out the weight, or putting $ma$ on the force diagram. $ma$ is not a force: it is what the resultant force equals.

Confusing displacement with distance. When a particle turns round, the distance travelled is more than the change in displacement.

Units

m, s, kg; N $=$ kg m s$^{-2}$.

Weight

$W = mg$, $g = 9.8$ m s$^{-2}$.

$v$–$t$ graph

Gradient $= a$; area $=$ displacement.

suvat

Constant $a$ only. Pick the equation without the unwanted letter.

Greatest height

$v = 0$.

Calculus

Differentiate $s \to v \to a$; integrate back, using initial conditions.

Newton 2

Resultant force $= ma$.

Equilibrium

$\sum\mathbf{F} = \mathbf{0}$.

Connected particles

Whole system for $a$; one part for the internal force.

Ten questions drawing on the whole of Year 1 mechanics. Take $g = 9.8$ m s$^{-2}$.

Question 1

Convert $54$ km h$^{-1}$ to m s$^{-1}$, and find the weight of a $750$ g parcel.

▶ Show solution

$54 \div 3.6 = 15$ m s$^{-1}$.

$750$ g $= 0.75$ kg, so $W = 0.75 \times 9.8 = 7.35$ N.

Question 2

State what is meant by modelling an object as a particle, and give one consequence.

▶ Show solution

The object's dimensions are negligible, so its mass is treated as concentrated at a single point.

Consequence: it cannot rotate, and all the forces act at that one point (air resistance is often also ignored).

Question 3

A car travelling at $15$ m s$^{-1}$ brakes uniformly and stops in $3$ s. Find the deceleration and the stopping distance.

▶ Show solution

$0 = 15 + 3a \Rightarrow a = -5$: deceleration $5$ m s$^{-2}$.

$s = \tfrac12(15 + 0)(3) = 22.5$ m.

Question 4

A cyclist accelerates uniformly from rest to $8$ m s$^{-1}$ in $4$ s, then rides at $8$ m s$^{-1}$ for $20$ s. Find the total distance using a velocity–time graph.

▶ Show solution

Area $= \tfrac12(4)(8) + 20(8) = 16 + 160 = 176$ m.

Question 5

A ball is thrown vertically upwards at $19.6$ m s$^{-1}$. Find the greatest height and the time taken to reach it.

▶ Show solution

$v = 0$ at the top: $0 = 19.6 - 9.8t \Rightarrow t = 2$ s.

$0 = 19.6^2 - 2(9.8)s \Rightarrow s = \dfrac{384.16}{19.6} = 19.6$ m.

Question 6

A particle's displacement is $s = t^3 - 3t^2$ m at time $t$ s. Find its velocity and acceleration when $t = 4$.

▶ Show solution

$v = 3t^2 - 6t$, so $v(4) = 48 - 24 = 24$ m s$^{-1}$.

$a = 6t - 6$, so $a(4) = 18$ m s$^{-2}$.

Question 7

A car of mass $1500$ kg accelerates uniformly from rest to $20$ m s$^{-1}$ in $8$ s against a constant resistance of $600$ N. Find the driving force.

▶ Show solution

$a = \dfrac{20}{8} = 2.5$ m s$^{-2}$.

$D - 600 = 1500(2.5) \Rightarrow D = 3750 + 600 = 4350$ N.

Question 8

Forces $(4\mathbf{i} - \mathbf{j})$ N and $(2\mathbf{i} + 9\mathbf{j})$ N act on a particle of mass $2$ kg. Find the magnitude of its acceleration.

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Resultant $= (6\mathbf{i} + 8\mathbf{j})$ N; $\mathbf{a} = (3\mathbf{i} + 4\mathbf{j})$ m s$^{-2}$.

$|\mathbf{a}| = \sqrt{9 + 16} = 5$ m s$^{-2}$.

Question 9

Particles of mass $5$ kg and $3$ kg are connected by a light inextensible string passing over a smooth pulley, and released from rest. Find the acceleration and the tension.

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$5g - T = 5a$ and $T - 3g = 3a$. Adding: $2g = 8a \Rightarrow a = 2.45$ m s$^{-2}$.

$T = 3(9.8 + 2.45) = 36.75 \approx 36.8$ N.

Question 10

A lift starts from rest and rises with acceleration $0.8$ m s$^{-2}$ for $3$ s, then moves at constant speed for $10$ s, then decelerates uniformly to rest in $2$ s. A crate of mass $50$ kg stands on the lift floor.

(a) Sketch the velocity–time graph and find the total height risen. (b) Find the normal reaction on the crate during each of the three stages.

▶ Show solution

(a) Maximum speed $= 0.8 \times 3 = 2.4$ m s$^{-1}$. Trapezium with parallel sides $10$ and $15$, height $2.4$:

Height $= \tfrac12(10 + 15)(2.4) = 30$ m. (Check: $3.6 + 24 + 2.4 = 30$ ✓)

(b) Up positive, $R - 50g = 50a$.

Stage 1, $a = 0.8$: $R = 50(9.8 + 0.8) = 530$ N.

Stage 2, $a = 0$: $R = 490$ N.

Stage 3, $a = -\dfrac{2.4}{2} = -1.2$: $R = 50(9.8 - 1.2) = 430$ N.