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This lesson covers the full AQA A-Level treatment of kinematics (AQA Spec 3.4.1.1–3.4.1.3). We go well beyond simple substitution into SUVAT equations: you will learn to resolve velocity vectors, analyse multi-step problems, and extract detailed information from motion graphs.
Spec mapping (AQA 7408 §3.4.1.1–§3.4.1.3): This lesson develops scalar and vector quantities, the equations of motion for uniform acceleration in a straight line, and the interpretation of displacement-time, velocity-time and acceleration-time graphs. Required Practical 1 (determination of g by free fall, RP1) sits inside this strand and is treated in depth in Lesson 10. The treatment includes vector resolution, multi-stage motion, and instantaneous-versus-average values, which underpin everything in §3.4.1.4–§3.4.1.8 that follows. (Refer to the official AQA specification document for exact wording.)
Synoptic links: Kinematics is the load-bearing topic of A-Level mechanics. (i) Projectile motion (Lesson 2, §3.4.1.3) treats horizontal and vertical SUVAT independently — the techniques here transfer directly to two-dimensional motion under gravity. (ii) Newton's laws (Lesson 3, §3.4.1.4) uses the kinematic acceleration a you compute here as the right-hand side of F = ma; without confident SUVAT, force problems collapse. (iii) Energy conservation (Lesson 5, §3.4.1.7) lets you cross-check SUVAT answers using ½mv² − ½mu² = Fs; many candidates lose marks because they treat SUVAT and energy as separate toolkits rather than two routes to the same answer. The graph-area / graph-gradient duality (area under v–t = displacement; gradient of v–t = acceleration) recurs in simple harmonic motion (Year 13) and in charge-time graphs for capacitors.
| Quantity | Type | SI Unit |
|---|---|---|
| Distance | Scalar | m |
| Displacement | Vector | m |
| Speed | Scalar | m s⁻¹ |
| Velocity | Vector | m s⁻¹ |
| Acceleration | Vector | m s⁻² |
Key Definition: A scalar has magnitude only. A vector has both magnitude and direction.
Any velocity v at angle θ to the horizontal can be resolved into two perpendicular components:
The magnitude of the resultant is found from Pythagoras:
v = √(vₓ² + vᵧ²)
The direction is found from trigonometry:
θ = tan⁻¹(vᵧ / vₓ)
A boat travels at 8.0 m s⁻¹ at 40° north of east. Find the eastward and northward components.
Solution:
Check: √(6.1² + 5.1²) = √(37.2 + 26.0) = √63.2 = 7.95 ≈ 8.0 m s⁻¹ ✓
A swimmer crosses a river swimming due north at 1.5 m s⁻¹ relative to the water. The river flows due east at 0.80 m s⁻¹. Find the swimmer's resultant velocity.
Solution:
Exam Tip: Always draw a vector triangle when adding velocities. Label the directions clearly and use Pythagoras / trigonometry — do not guess.
For motion with constant acceleration in a straight line:
| # | Equation | Missing Variable |
|---|---|---|
| 1 | v = u + at | s |
| 2 | s = ½(u + v)t | a |
| 3 | s = ut + ½at² | v |
| 4 | s = vt − ½at² | u |
| 5 | v² = u² + 2as | t |
Important: These equations are ONLY valid when acceleration is constant. If acceleration changes, you must split the motion into stages or use graphical methods.
From Equation 1: t = (v − u) / a
Substituting into Equation 3:
s = u × (v − u)/a + ½a × [(v − u)/a]²
s = u(v − u)/a + ½(v − u)²/a
s = [2u(v − u) + (v − u)²] / (2a)
s = (v − u)[2u + v − u] / (2a)
s = (v − u)(v + u) / (2a)
2as = v² − u²
Therefore: v² = u² + 2as ✓
Many exam questions involve motion in two or more stages. The key is to identify where the acceleration changes and treat each stage separately.
A train accelerates from rest at 0.50 m s⁻² for 40 s, then decelerates uniformly to rest over a distance of 300 m. Find (a) the maximum speed, (b) the distance covered while accelerating, (c) the deceleration, and (d) the total journey time.
Stage 1: Acceleration phase
u₁ = 0, a₁ = 0.50 m s⁻², t₁ = 40 s
(a) v₁ = u₁ + a₁t₁ = 0 + 0.50 × 40 = 20 m s⁻¹
(b) s₁ = u₁t₁ + ½a₁t₁² = 0 + ½ × 0.50 × 40² = 400 m
Stage 2: Deceleration phase
u₂ = 20 m s⁻¹, v₂ = 0, s₂ = 300 m
(c) v₂² = u₂² + 2a₂s₂ → 0 = 400 + 2a₂ × 300 → a₂ = −400/600 = −0.67 m s⁻²
The deceleration is 0.67 m s⁻².
(d) t₂ = (v₂ − u₂)/a₂ = (0 − 20)/(−0.67) = 30 s
Total time = t₁ + t₂ = 40 + 30 = 70 s
Total distance = 400 + 300 = 700 m
A ball is thrown vertically upward at 15 m s⁻¹ from a height of 2.0 m above the ground. Taking g = 9.81 m s⁻², find (a) the maximum height above the ground, (b) the total time to reach the ground.
Solution — taking upward as positive:
u = +15 m s⁻¹, a = −9.81 m s⁻²
(a) At the top, v = 0.
v² = u² + 2as → 0 = 15² + 2(−9.81)s → s = 225 / 19.62 = 11.5 m above launch point.
Maximum height above ground = 11.5 + 2.0 = 13.5 m
(b) The ball starts 2.0 m above the ground and must travel to ground level.
Taking the launch point as origin, the ball hits the ground when s = −2.0 m.
s = ut + ½at² → −2.0 = 15t + ½(−9.81)t² → 4.905t² − 15t − 2.0 = 0
Using the quadratic formula:
t = [15 ± √(225 + 39.24)] / 9.81 = [15 ± √264.2] / 9.81 = [15 ± 16.26] / 9.81
Taking the positive root: t = 31.26 / 9.81 = 3.19 s
Common Misconception: Students often forget to account for the initial height above the ground. Always define a clear origin and sign convention at the start.
A velocity–time graph shows the following data for a car:
| Time (s) | Velocity (m s⁻¹) |
|---|---|
| 0 | 0 |
| 5 | 15 |
| 10 | 15 |
| 15 | 0 |
Find (a) the acceleration in the first 5 s, (b) the deceleration in the last 5 s, (c) the total displacement.
Solution:
(a) Acceleration = gradient = (15 − 0) / (5 − 0) = 3.0 m s⁻²
(b) Deceleration = |gradient| = |( 0 − 15) / (15 − 10)| = 15/5 = 3.0 m s⁻²
(c) Total displacement = total area under graph
Total displacement = 37.5 + 75.0 + 37.5 = 150 m
Exam Tip: When calculating area under a v–t graph, split the shape into rectangles and triangles. Always check the signs — regions below the time axis represent negative displacement (motion in the reverse direction).
A cyclist travels 200 m north in 25 s, then 100 m south in 15 s.
Common Misconception: Average speed and average velocity are NOT the same when direction changes. Speed uses total distance (scalar); velocity uses net displacement (vector).
The acceleration due to gravity near the Earth's surface is approximately g = 9.81 m s⁻². In calculations, take g = 9.81 m s⁻² unless told otherwise.
For a freely falling object (no air resistance):
A stone is dropped from the top of a cliff. It takes 3.2 s to reach the water below. Find (a) the height of the cliff and (b) the speed at which the stone hits the water.
Solution: u = 0, a = 9.81 m s⁻², t = 3.2 s (taking downward as positive).
(a) s = ut + ½at² = 0 + ½ × 9.81 × 3.2² = 4.905 × 10.24 = 50.2 m
(b) v = u + at = 0 + 9.81 × 3.2 = 31.4 m s⁻¹
Check: v² = u² + 2as = 0 + 2 × 9.81 × 50.2 = 985.9 → v = 31.4 m s⁻¹ ✓
Two cars are 800 m apart on a straight road, travelling toward each other. Car A starts from rest and accelerates uniformly at 2.0 m s⁻². Car B starts from rest at the same instant and accelerates uniformly at 1.5 m s⁻² toward A. After what time do the cars meet, and what distance has each travelled?
Solution: Let car A start at x = 0, car B start at x = 800 m. Take rightward as positive for car A.
Position of A at time t: x_A = ½ × 2.0 × t² = t².
Position of B at time t (moving left): x_B = 800 − ½ × 1.5 × t² = 800 − 0.75t².
The cars meet when x_A = x_B: t² = 800 − 0.75t² → 1.75t² = 800 → t² = 457.1 → t = 21.4 s.
Distance travelled by A: x_A = (21.4)² = 458 m.
Distance travelled by B: 800 − 458 = 342 m.
Check: B accelerates at 1.5 m s⁻² for 21.4 s, so distance = ½ × 1.5 × (21.4)² = ½ × 1.5 × 458 = 343 m ✓ (within rounding).
This style of "two objects, when do they meet" problem is common in extended-question kinematics and is solved efficiently by writing the position of each object as a function of time, then equating.
A passenger on a train notices a station signpost moving past the window and starts a stopwatch. The stopwatch reads 1.4 s when a second signpost, known to be 35.0 m further down the track, passes the window. The train is decelerating uniformly. At the first signpost the train's speed was 28.0 m s⁻¹. Find (a) the train's speed as the second signpost passes, (b) the deceleration, and (c) the additional distance the train travels before coming to rest, assuming the deceleration is maintained.
Solution:
Set up: u = 28.0 m s⁻¹, s = 35.0 m, t = 1.4 s.
(a) Use s = ½(u + v)t → 35.0 = ½(28.0 + v) × 1.4 → 35.0/0.70 = 28.0 + v → 50.0 = 28.0 + v → v = 22.0 m s⁻¹.
(b) Use v = u + at → 22.0 = 28.0 + a × 1.4 → a = (22.0 − 28.0) / 1.4 = −6.0/1.4 = −4.29 m s⁻² (deceleration of 4.29 m s⁻²).
(c) From the second signpost, u = 22.0 m s⁻¹, v = 0, a = −4.29 m s⁻². Use v² = u² + 2as → 0 = 22.0² + 2(−4.29)s → s = 484/8.58 = 56.4 m.
This worked example shows the value of choosing the SUVAT equation that uses three known quantities to find the fourth, without needing intermediate calculations. Equation 2 (s = ½(u + v)t) is the most often overlooked of the five and is the fastest route here.
Specimen question modelled on the AQA paper format — not a verbatim past-paper item.
A small steel ball is released from rest from a height of 1.80 m above a soft landing mat. Air resistance is negligible. Take g = 9.81 m s⁻².
(a) Calculate the time the ball takes to reach the mat. [3 marks]
(b) Calculate the speed of the ball just before it strikes the mat. [2 marks]
(c) The ball is now thrown vertically upward at 4.5 m s⁻¹ from the same release point. Calculate the total time before the ball lands on the mat. [4 marks]
The 9 marks are distributed across the standard AO bracket descriptions used at A-Level. AO1 marks (around 3 of 9) reward correct recall of the SUVAT equations and the identification of the missing variable in each part. AO2 marks (around 4 of 9) reward correct substitution with consistent units and a defensible sign convention. AO3 marks (around 2 of 9) reward the integrated reasoning in part (c) — recognising that the upward and downward phases can be handled in one quadratic if you commit to a single positive direction at the outset rather than splitting the motion into two stages. The AO descriptions used here are generic bracket descriptions; the assessor's own wording on any specific mark scheme is not reproduced.
Grade C response (part c):
"Taking up as positive, u = +4.5, a = −9.81, s = −1.80 when the ball lands. Use s = ut + ½at². So −1.80 = 4.5t − 4.905t². Rearranging gives 4.905t² − 4.5t − 1.80 = 0. Using the quadratic formula, t = [4.5 ± √(20.25 + 35.32)] / 9.81 = [4.5 ± 7.45] / 9.81. The positive root gives t = 11.95 / 9.81 = 1.22 s. The negative root is not physical so I reject it. The total time is 1.22 s."
Examiner commentary: M1 for committing to a sign convention; M2 for the correct quadratic in t; M3 for the positive root and a sensible rejection of the unphysical root. A Grade C response often stops after producing the quadratic and substituting, without explaining the choice of sign or commenting on the rejected root.
Grade B response (part c):
"Taking upward as positive: u = +4.5 m s⁻¹, a = −9.81 m s⁻². The mat is 1.80 m below the launch point so the final displacement is s = −1.80 m. Apply s = ut + ½at²: −1.80 = 4.5t − 4.905t². Rearranged: 4.905t² − 4.5t − 1.80 = 0. Quadratic formula: t = [4.5 ± √(4.5² + 4 × 4.905 × 1.80)] / (2 × 4.905) = [4.5 ± √55.57] / 9.81 = [4.5 ± 7.454] / 9.81. The positive root gives t = 1.22 s. As a check, the ball rises 4.5²/(2 × 9.81) = 1.03 m above the launch point and then falls 1.03 + 1.80 = 2.83 m, which takes √(2 × 2.83/9.81) = 0.76 s after a rise time of 4.5/9.81 = 0.46 s, total 1.22 s ✓."
Examiner commentary: M1, M2 and M3 as above, plus the consistency check at the end which would attract the AO3 mark in most mark schemes. The response is missing the explicit statement of why the negative root is unphysical — that is the move that takes it from B to A*. The check is welcome and shows synoptic understanding, but the framing could be tighter; the candidate is doing the right work but not yet packaging it for full credit.
Grade A* response (part c):
"Define upward as positive throughout, with the origin at the release point. Then u = +4.5 m s⁻¹ and a = −9.81 m s⁻² for the entire flight (the only force is weight). The ball lands at displacement s = −1.80 m. Substituting into s = ut + ½at² gives 4.905t² − 4.5t − 1.80 = 0. Using the quadratic formula, t = (4.5 ± √(20.25 + 35.316))/9.81 = (4.5 ± 7.454)/9.81. Two roots: t₁ = 1.218 s and t₂ = −0.301 s. The negative root corresponds to the past trajectory along the same parabola — it is mathematically valid but physically irrelevant because the ball was at rest in the thrower's hand before t = 0. The positive root, t = 1.22 s (3 s.f.), is the time at which the ball reaches the mat. Sanity check: peak occurs at t = u/g = 0.459 s at height u²/(2g) = 1.03 m above release, giving total fall distance 2.83 m. Time to fall from rest through 2.83 m is √(2 × 2.83/9.81) = 0.760 s. Total time 0.459 + 0.760 = 1.219 s ✓."
Examiner commentary: All three procedural marks (M1 commitment to sign convention; M2 correct quadratic; M3 positive root) plus the AO3 mark for the explicit physical interpretation of the rejected root and the independent two-stage check. The interpretive sentence about the negative root is the A* discriminator — Grade B candidates compute the same numbers but do not articulate why they reject one of them. Note also the appropriate significant-figure quotation throughout (3 s.f., matching the data).
Sign conventions are the single largest source of avoidable marks lost in SUVAT questions. A common pattern is for candidates to decide that up is positive while writing the working but then to use g = +9.81 m s⁻² because that is what they have memorised; the answer comes out reversed. Always declare the convention in writing before substituting, and let the negative sign on g appear naturally.
A second category of error is the unit slip: writing 1.5 mm as 1.5 × 10⁻³ m correctly during measurement but then substituting 1.5 anyway, or mixing centimetres and metres in a single equation. Force yourself to put every quantity into SI base units before substitution, even when the question gives you values in mixed units.
A third is conflating average and instantaneous quantities. When a question says "average velocity over the first 4 seconds" it wants Δs / Δt; when it says "the velocity at t = 4 s" it wants the gradient of the displacement-time tangent at that instant or the SUVAT result v = u + at. The distinction matters for any motion where acceleration is non-zero.
A fourth — and the discriminator between B and A* in extended problems — is failing to verify the answer with a different equation. SUVAT contains five equations for a reason: at least two of them can usually deliver any specific quantity, so a five-second cross-check eliminates almost all numerical slips.
Marion and Thornton's Classical Dynamics of Particles and Systems (Chapter 2) treats the same equations of motion you have learned here, but using the calculus framework: acceleration is dv/dt, velocity is ds/dt, and SUVAT becomes the integrated form of the equation a = constant. Once you are comfortable with that, you can solve problems where acceleration is a known function of time, displacement, or velocity (drag, springs, variable rocket thrust) — none of which yield to the five A-Level SUVAT equations. For physics-admissions interviews and Physics Olympiad preparation, Irodov's Problems in General Physics (Section 1.1) and the MIT 8.01 OpenCourseWare lecture notes are the standard further-reading set. Students preparing for the engineering side will also encounter the same kinematic framework as the foundation for vehicle dynamics and projectile-tracking radar systems.
Three subtle errors distinguish A from A* candidates on kinematics. First, sign on the rejected quadratic root: when projectile motion or vertical-throw problems produce a quadratic in t, the negative root is mathematically valid — it is the time on the same parabola at which the trajectory would have passed the same point in the past. A* candidates state this explicitly; A candidates simply discard it. Second, the difference between average velocity and the instantaneous-velocity midpoint: for uniformly accelerated motion, ⟨v⟩ = (u + v)/2 = v(t/2). This is a special property of constant acceleration and fails the moment acceleration varies — many A-grade candidates use it as if it were a general identity. Third, graphical reading of an acceleration-time graph: the area under an a-t graph gives the change in velocity, not the velocity itself. A candidate who writes "v = area under graph" loses the AO1 mark for the definition and the AO2 mark that follows it.
Spend the first 30 seconds of any SUVAT question writing down the five quantities (s, u, v, a, t), with a question mark next to the one you want and a "?" next to anything not given. This eliminates the most common error in the topic, which is to grab the wrong equation. Always state your sign convention before substituting — write the words "taking upward as positive" or similar at the top of the working. If you have committed to a convention, an examiner can mark backwards through your working even if your arithmetic slips; if you have not, a sign error becomes invisible and unrecoverable.
For multi-stage problems (acceleration, then deceleration; or upward, then downward), draw a quick v-t sketch even when not asked. The sketch is almost always the fastest route to the answer for total-distance, total-time and average-velocity questions, and it eliminates the temptation to apply SUVAT across a discontinuity in acceleration.
When the question gives a quadratic, write the discriminant explicitly before substituting into the quadratic formula. This catches sign errors in b² and 4ac before they propagate. Quote answers to the same number of significant figures as the data — three sig figs is normal at A-Level — and check the order of magnitude before circling the answer; a free-fall problem cannot deliver a time of 0.01 s for a metre-scale drop, and noticing the absurdity is worth one mark in itself.
| Quantity | Formula | Unit |
|---|---|---|
| Displacement | s (vector) | m |
| Velocity | v = Δs/Δt | m s⁻¹ |
| Acceleration | a = Δv/Δt | m s⁻² |
| SUVAT 1 | v = u + at | — |
| SUVAT 2 | s = ½(u + v)t | — |
| SUVAT 3 | s = ut + ½at² | — |
| SUVAT 4 | s = vt − ½at² | — |
| SUVAT 5 | v² = u² + 2as | — |
| Resolving | vₓ = v cos θ, vᵧ = v sin θ | — |
AQA Specification Reference: Section 3.4.1.1 (Scalars and vectors), 3.4.1.2 (Equations of motion for constant acceleration), 3.4.1.3 (Motion graphs).