Kinematics and SUVAT Equations in Depth

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.

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1 Scalars and Vectors — Quick Recap

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.

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2 Resolving Velocity Vectors

Any velocity v at angle θ to the horizontal can be resolved into two perpendicular components:

• Horizontal component: vₓ = v cos θ
• Vertical component: vᵧ = v sin θ

The magnitude of the resultant is found from Pythagoras:

v = √(vₓ² + vᵧ²)

The direction is found from trigonometry:

θ = tan⁻¹(vᵧ / vₓ)

Worked Example 1 — Resolving a Velocity

A boat travels at 8.0 m s⁻¹ at 40° north of east. Find the eastward and northward components.

Solution:

• Eastward component: vₓ = 8.0 cos 40° = 8.0 × 0.766 = 6.1 m s⁻¹
• Northward component: vᵧ = 8.0 sin 40° = 8.0 × 0.643 = 5.1 m s⁻¹

Check: √(6.1² + 5.1²) = √(37.2 + 26.0) = √63.2 = 7.95 ≈ 8.0 m s⁻¹ ✓

Worked Example 2 — Adding Two Velocities

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:

• Resultant speed = √(1.5² + 0.80²) = √(2.25 + 0.64) = √2.89 = 1.7 m s⁻¹
• Direction: θ = tan⁻¹(0.80 / 1.5) = tan⁻¹(0.533) = 28° east of north

Exam Tip: Always draw a vector triangle when adding velocities. Label the directions clearly and use Pythagoras / trigonometry — do not guess.

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3 The Five SUVAT Equations

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.

Deriving Equation 5 from Equations 1 and 3

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 ✓

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4 Multi-Step SUVAT Problems

Many exam questions involve motion in two or more stages. The key is to identify where the acceleration changes and treat each stage separately.

Worked Example 3 — Two-Stage Motion

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

Worked Example 4 — Vertical Multi-Step (Ball Thrown Upward)

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.

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5 Motion Graphs — Detailed Analysis

Displacement–Time Graphs

• Gradient = velocity (at any instant for a curve, use a tangent)
• Straight line → constant velocity
• Curve → changing velocity (acceleration)
• Horizontal line → stationary
• Negative gradient → moving in the negative direction

Velocity–Time Graphs

• Gradient = acceleration
• Area under graph = displacement (signed — area below the t-axis is negative displacement)
• Straight line → constant acceleration
• Curve → changing acceleration
• Horizontal line → constant velocity

Worked Example 5 — Analysing a v–t Graph

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

• Phase 1 (0–5 s): Triangle = ½ × 5 × 15 = 37.5 m
• Phase 2 (5–10 s): Rectangle = 5 × 15 = 75.0 m
• Phase 3 (10–15 s): Triangle = ½ × 5 × 15 = 37.5 m

Total displacement = 37.5 + 75.0 + 37.5 = 150 m

Acceleration–Time Graphs

• Area under graph = change in velocity
• A horizontal line at a = g represents free fall
• A horizontal line at a = 0 represents constant velocity

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).

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6 Instantaneous vs Average Values

• Average velocity = total displacement / total time
• Instantaneous velocity = gradient of displacement–time graph at a point (draw a tangent)
• Average acceleration = change in velocity / time interval
• Instantaneous acceleration = gradient of velocity–time graph at a point

Worked Example 6 — Average vs Instantaneous

A cyclist travels 200 m north in 25 s, then 100 m south in 15 s.

• Average speed = total distance / total time = (200 + 100) / (25 + 15) = 300/40 = 7.5 m s⁻¹
• Average velocity = total displacement / total time = (200 − 100) / 40 = 100/40 = 2.5 m s⁻¹ north

Common Misconception: Average speed and average velocity are NOT the same when direction changes. Speed uses total distance (scalar); velocity uses net displacement (vector).

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7 Acceleration Due to Gravity

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):

• Dropped from rest: u = 0, a = g = 9.81 m s⁻² (downward)
• Thrown upward: take upward as positive, so a = −9.81 m s⁻²

Worked Example 7 — Free Fall

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 ✓

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Key Formulae Summary

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).