Kinematics is the study of motion — without worrying about the forces that cause it. You will learn to describe and calculate displacement, velocity and acceleration for particles moving in a straight line under constant or variable acceleration. This topic underpins all of Mechanics and appears in almost every M1 paper.
v = u + at | s = ut + ½at² | v² = u² + 2as | g = 9.8 m s⁻² | v = ds/dt | a = dv/dt
Learning Objectives
Define displacement, velocity, acceleration and speed; distinguish scalar from vector
Calculate average velocity and understand instantaneous velocity
Apply all five SUVAT equations to constant-acceleration problems
Solve vertical motion under gravity using g = 9.8 m s⁻² with sign convention
Interpret velocity-time graphs: gradient = acceleration, area = displacement
Use calculus — differentiation and integration — for variable acceleration
Apply initial conditions to find constants of integration
Distinguish between distance travelled and displacement
Displacement & Velocity
Vector quantities; sign convention; average vs instantaneous
SUVAT Equations
5 equations for constant acceleration — choose the right one
Vertical Motion
g = 9.8 downward; max height when v = 0; symmetry of flight
v-t Graphs
Gradient = acceleration; area under graph = displacement
Variable Acceleration
Differentiate for v and a; integrate for s and v
Calculus & Initial Conditions
Find constants of integration using known values at t = 0
Two-Stage Journeys
Chain SUVAT stages: final v of stage 1 = initial u of stage 2
Particle on a Cliff
Projectile thrown horizontally — horizontal and vertical independently
Learn 1 — Displacement, Velocity & Acceleration
Scalars and Vectors in Kinematics
Every kinematic quantity is either a scalar (magnitude only) or a vector (magnitude + direction).
Distance is how far an object has travelled along its path (always positive). Displacement is the straight-line distance from start to finish — it can be negative.
Displacement (s)
Displacement is measured in metres (m). It measures the change in position from a fixed reference point, with a sign convention — usually positive to the right or upward.
A particle starts at x = 0, moves 8 m to the right, then 3 m to the left. Distance travelled = 8 + 3 = 11 m Displacement = 8 − 3 = +5 m (still 5 m to the right of start)
If instead it moves 3 m right then 8 m left:
Distance = 11 m (same), Displacement = −5 m (5 m to the LEFT of start)
Velocity (v)
Velocity is the rate of change of displacement. It is measured in m s⁻¹ and has a sign (direction).
Average velocity = Δs / Δt = (change in displacement) / (time taken)
A particle moves from s = 2 m to s = 14 m in 4 s.
Average velocity = (14 − 2) / 4 = 12 / 4 = 3 m s⁻¹ (positive → rightward)
A particle moves from s = 10 m to s = 4 m in 3 s.
Average velocity = (4 − 10) / 3 = −6 / 3 = −2 m s⁻¹ (negative → leftward)
Instantaneous velocity is the velocity at a specific instant — the limit of Δs/Δt as Δt → 0. For non-constant motion, this equals ds/dt (covered in Learn 5).
Speed
Speed = |velocity|. It is always non-negative. Average speed = total distance / total time (not displacement ÷ time).
Particle goes 8 m right then 3 m left in 5 s total.
Average speed = 11 / 5 = 2.2 m s⁻¹
Average velocity = 5 / 5 = 1 m s⁻¹ (rightward)
Acceleration (a)
Acceleration is the rate of change of velocity. Measured in m s⁻². It can be positive (speeding up in positive direction, or slowing down in negative direction) or negative.
Average acceleration = Δv / Δt = (change in velocity) / (time taken)
Particle: u = 10 m s⁻¹, v = 4 m s⁻¹ after 3 s.
Acceleration = (4 − 10) / 3 = −6 / 3 = −2 m s⁻² (decelerating)
Particle: u = −5 m s⁻¹, v = 7 m s⁻¹ after 6 s.
Acceleration = (7 − (−5)) / 6 = 12 / 6 = +2 m s⁻²
Sign Convention
Choose a positive direction at the START of every problem and stick to it throughout.
• Positive direction: usually rightward or upward
• Negative velocity: moving in the opposite direction
• Negative acceleration: could mean decelerating (if moving positively) or speeding up negatively
• A particle is momentarily at rest when v = 0 (it may still have acceleration!)
Always state your sign convention — examiners expect it.
Key Distinctions
Distance ≠ Displacement unless the particle never reverses direction. Speed ≠ Velocity — speed ignores direction. Deceleration means acceleration opposing the direction of motion — numerically, if moving rightward, deceleration = negative acceleration.
Learn 2 — SUVAT Equations
The Five SUVAT Variables
s = displacement (m) | u = initial velocity (m s⁻¹) v = final velocity (m s⁻¹) | a = acceleration (m s⁻²) t = time (s)
These equations apply ONLY when acceleration is constant.
The Five SUVAT Equations
1. v = u + at (no s)
2. s = ½(u + v)t (no a)
3. s = ut + ½at² (no v)
4. v² = u² + 2as (no t)
5. s = vt − ½at² (no u)
How to Choose the Right Equation
Identify which 3 variables are given and which 1 variable you need to find. Then use the equation that contains exactly those 4 variables (and omits the 5th you don't need).
Method — 4-step process:
1. Write down: s = ?, u = ?, v = ?, a = ?, t = ?
2. Fill in known values (with signs!)
3. Identify the unknown
4. Pick the equation that uses those 4 variables
Example — Choose the Right Equation
Given: u = 6 m s⁻¹, a = 2 m s⁻², t = 4 s. Find s.
Missing variable: v
Use equation 3 (no v): s = ut + ½at² = 6×4 + ½×2×16 = 24 + 16 = 40 m
Given: u = 15 m s⁻¹, v = 3 m s⁻¹, s = 36 m. Find a.
Missing variable: t
Use equation 4 (no t): v² = u² + 2as → 9 = 225 + 72a → a = (9−225)/72 = −216/72 = −3 m s⁻²
Equation 1 — v = u + at (no s)
Use when displacement is not involved and you need to connect velocity, acceleration and time.
Particle decelerates from 20 m s⁻¹ at 4 m s⁻². When does it stop?
v = 0, u = 20, a = −4: 0 = 20 − 4t → t = 5 s
Equation 2 — s = ½(u+v)t (no a)
Use when acceleration is not needed — you know both initial and final velocity and the time.
Particle: u = 4 m s⁻¹, v = 12 m s⁻¹, t = 5 s. Find s.
s = ½(4+12)×5 = ½×16×5 = 40 m
Equation 3 — s = ut + ½at² (no v)
Use when final velocity is not given or needed. Most common equation for "find displacement."
Particle starts at rest (u = 0) and accelerates at 3 m s⁻² for 8 s. Find s.
s = 0 + ½×3×64 = 96 m
Equation 4 — v² = u² + 2as (no t)
Use when time is not given and not needed — the "time-free" equation.
Ball thrown upward at 14 m s⁻¹ decelerates at 9.8 m s⁻². Find maximum height.
v = 0, u = 14, a = −9.8: 0 = 196 − 19.6s → s = 10 m
Equation 5 — s = vt − ½at² (no u)
Use when initial velocity is not known. Less commonly needed but occasionally useful.
Two-Stage Journeys
For a journey in two phases (e.g. accelerating then decelerating):
• Apply SUVAT to each stage separately
• The final velocity of stage 1 = initial velocity of stage 2
• Add displacements of both stages for total displacement
• Be careful: if the particle reverses direction, displacement may be less than distance
SUVAT equations require constant acceleration. Do NOT use them when acceleration is given as a function of time (e.g. a = 3t²). Use calculus instead (Learn 5).
Learn 3 — Vertical Motion Under Gravity
Acceleration Due to Gravity
Cambridge 9709 uses g = 9.8 m s⁻² (NOT 10, unless the question explicitly states otherwise).
Gravity acts vertically downward at all times.
Acceleration is constant throughout the flight (air resistance ignored).
Sign Convention for Vertical Motion
Take upward as positive: a = −9.8 m s⁻² throughout the flight
Thrown upward: u positive, a = −9.8 m s⁻² Dropped from rest: u = 0, a = −9.8 m s⁻² (or take downward as positive: a = +9.8 m s⁻²) Tip: Be consistent — once you choose a positive direction, every sign in all SUVAT variables must match that convention.
Maximum Height
At maximum height, the particle is momentarily at rest: v = 0.
At maximum height: v = 0 Use v² = u² + 2as → 0 = u² − 2(9.8)s → s = u² / (2 × 9.8)
Ball thrown upward at 19.6 m s⁻¹ (u = 19.6, a = −9.8, v = 0):
0 = (19.6)² − 2(9.8)s → s = 384.16 / 19.6 = 19.6 m
Time to reach max height: v = u + at → 0 = 19.6 − 9.8t → t = 2 s
Time of Flight
For a particle thrown upward from the same level it lands on (e.g. from the ground):
Total time of flight = 2 × (time to reach max height)
The upward and downward phases are symmetrical:
• Time up = time down
• Speed at any height on the way up = speed at the same height on the way down
• Velocity on return = −u (same magnitude, opposite direction)
Particle Thrown from a Height
If the particle is thrown from a height H above the ground (e.g. from a cliff), the symmetry no longer holds. Set s = −H (downward) as the ground level condition:
Ball thrown upward at 10 m s⁻¹ from a cliff 45 m high (u = 10, a = −9.8, s = −45 at ground):
s = ut + ½at² → −45 = 10t − 4.9t²
4.9t² − 10t − 45 = 0 → using quadratic formula: t = (10 + √(100 + 882)) / 9.8 ≈ 4.26 s
Horizontal Projectile Motion
When an object is thrown horizontally (e.g. off a table), treat horizontal and vertical directions independently:
Horizontal: no acceleration → constant velocity. x = u_horizontal × t Vertical: starts from rest (u_vertical = 0), a = −9.8 m s⁻²
Ball thrown horizontally at 12 m s⁻¹ from height 20 m:
Vertical: 20 = ½(9.8)t² → t² = 40/9.8 ≈ 4.082 → t ≈ 2.02 s
Horizontal range: x = 12 × 2.02 ≈ 24.2 m
Always use g = 9.8 m s⁻² in Cambridge 9709 unless the question says otherwise. Using g = 10 will cost accuracy marks. Also: at maximum height, v = 0 but a = −9.8 ≠ 0 — the ball is still accelerating!
Positive gradient: positive acceleration (speeding up in positive direction) Zero gradient: horizontal line → constant velocity → zero acceleration Negative gradient: deceleration (slowing down or speeding up in opposite direction) Steep gradient: large acceleration; gentle gradient: small acceleration
What Does the Area Tell You?
Area under v-t graph (between graph and t-axis) = displacement
• Area above the t-axis (positive v): positive displacement (moved in positive direction)
• Area below the t-axis (negative v): negative displacement (moved in opposite direction)
• Total displacement = (area above) − (area below)
• Total distance = (area above) + |area below|
Common Graph Shapes
Horizontal line: constant velocity, a = 0 Straight line (positive slope): uniform acceleration Straight line (negative slope): uniform deceleration Line crossing t-axis: particle reverses direction at that instant v = 0 for a period: particle at rest
(Curved graphs → variable acceleration; use calculus from Learn 5)
Area Calculations — Shapes to Use
Rectangle: (constant v section) Area = base × height = t × v Triangle: (acceleration/deceleration from/to rest) Area = ½ × base × height = ½ × t × v Trapezium: (acceleration between two non-zero velocities) Area = ½ × (v₁ + v₂) × t
A particle accelerates from 0 to 12 m s⁻¹ in 3 s, then travels at 12 m s⁻¹ for 5 s, then decelerates to 0 in 4 s.
Displacement = ½(3)(12) + (5)(12) + ½(4)(12) = 18 + 60 + 24 = 102 m
Multi-Stage Journey Example
A car: starts at rest, accelerates at 2 m s⁻² for 10 s, maintains constant speed for 15 s, then decelerates uniformly to rest in 6 s.
Speed after 10 s: v = 0 + 2(10) = 20 m s⁻¹
Stage 1 (triangle): s₁ = ½(10)(20) = 100 m
Stage 2 (rectangle): s₂ = (15)(20) = 300 m
Stage 3 (triangle): s₃ = ½(6)(20) = 60 m
Total displacement = 100 + 300 + 60 = 460 m
Negative Velocity on a v-t Graph
When the graph dips below the t-axis, the particle is moving in the negative direction.
Example: 0 to 4 s: v = 6 m s⁻¹ (above axis). 4 s to 7 s: v = −4 m s⁻¹ (below axis).
Displacement = 6×4 + (−4)×3 = 24 − 12 = +12 m
Distance = 24 + 12 = 36 m
To find when a particle returns to its starting position, set total displacement = 0 and solve for t. This happens when positive area = negative area on the v-t graph.
Learn 5 — Variable Acceleration (Calculus)
When to Use Calculus
When acceleration is NOT constant (e.g. a = 3t, a = 2t² − 1), SUVAT equations do NOT apply. You must use differentiation or integration.
The Three Fundamental Relationships
v = ds/dt (differentiate s to get v)
a = dv/dt (differentiate v to get a)
s = ∫v dt (integrate v to get s)
v = ∫a dt (integrate a to get v)
Differentiating for Velocity
If displacement s is given as a function of t, differentiate once to get velocity:
s = 3t³ − 4t² + 2t
v = ds/dt = 9t² − 8t + 2
At t = 2: v = 9(4) − 8(2) + 2 = 36 − 16 + 2 = 22 m s⁻¹
Differentiating for Acceleration
Differentiate velocity once more to get acceleration:
v = 9t² − 8t + 2
a = dv/dt = 18t − 8
At t = 3: a = 18(3) − 8 = 54 − 8 = 46 m s⁻²
When is a = 0? 18t − 8 = 0 → t = 8/18 = 4/9 s ≈ 0.444 s
Integrating for Velocity
If acceleration is given as a function of t, integrate and apply initial conditions:
a = 6t − 2. When t = 0, v = 5 m s⁻¹.
v = ∫(6t − 2) dt = 3t² − 2t + C
Apply initial condition: 5 = 3(0) − 2(0) + C → C = 5 v = 3t² − 2t + 5
Integrating for Displacement
Integrate velocity and apply initial conditions to find constant:
v = 3t² − 2t + 5. When t = 0, s = 0.
s = ∫(3t² − 2t + 5) dt = t³ − t² + 5t + C
Apply: 0 = 0 − 0 + 0 + C → C = 0 s = t³ − t² + 5t
At t = 4: s = 64 − 16 + 20 = 68 m
Finding When a Particle Is at Rest
Set v = 0 and solve for t. Check which roots are in the valid domain (t ≥ 0).
v = t² − 5t + 6 = (t − 2)(t − 3)
Particle at rest when t = 2 s and t = 3 s.
At t = 2: s = ∫₀² v dt (integrate to find position); or compute from s(t) directly.
Initial Conditions
Every integration produces a constant C. You MUST use initial conditions to find C:
• If "initially at rest" → v = 0 when t = 0
• If "starts at the origin" → s = 0 when t = 0
• If "passes through point A at t = 2" → s = [value of A] when t = 2
Always state "when t = 0, s = ..." clearly in your working.
Distance vs Displacement with Calculus
If the particle reverses direction (v changes sign), you cannot simply compute s(t_final) − s(t_start).
You must:
1. Find when v = 0 (particle reverses)
2. Compute displacement for each segment (always positive in each segment)
3. Add the magnitudes for total distance
Worked Examples
8 fully worked examples covering all key kinematics topics. Study each step carefully.
Example 1 — Particle Decelerating (SUVAT)
A particle moves along a straight line. It passes point A with velocity 24 m s⁻¹ and decelerates uniformly at 3 m s⁻². Find (i) the time to reach rest, and (ii) the distance from A when it stops.
Given: u = 24 m s⁻¹, a = −3 m s⁻², v = 0. Find t and s. B1
(i) Time to rest — use v = u + at: 0 = 24 − 3t → t = 24/3 = 8 sM1 A1
(ii) Distance — use v² = u² + 2as: 0 = 576 + 2(−3)s → 6s = 576 → s = 96 mM1 A1
Check with s = ut + ½at²: s = 24(8) + ½(−3)(64) = 192 − 96 = 96 m ✓ B1
Example 2 — Vertical Throw: Max Height and Time
A ball is thrown vertically upward from the ground with initial speed 29.4 m s⁻¹. Find (i) the maximum height, (ii) the time to reach max height, (iii) the time to return to the ground. (g = 9.8 m s⁻²)
Given: u = 29.4 m s⁻¹, a = −9.8 m s⁻². Take upward as positive. B1
(i) Max height (v = 0): v² = u² + 2as → 0 = (29.4)² − 2(9.8)s → s = 864.36/19.6 = 44.1 mM1 A1
(ii) Time to max height: v = u + at → 0 = 29.4 − 9.8t → t = 3 sM1 A1
(iii) Total time of flight: Ball returns to s = 0. By symmetry: total time = 2 × 3 = 6 s. (Or use s = ut + ½at²: 0 = 29.4t − 4.9t² → t(29.4 − 4.9t) = 0 → t = 6 s.) A1
Example 3 — Velocity-Time Graph Area
A particle's v-t graph shows: acceleration from 0 to 8 m s⁻¹ in 4 s, constant velocity 8 m s⁻¹ for 6 s, then deceleration to −4 m s⁻¹ in 5 s. Find (i) total displacement, (ii) total distance.
Stage 1 (triangle 0–4 s): Area = ½ × 4 × 8 = 16 m (above axis) M1
Stage 2 (rectangle 4–10 s): Area = 6 × 8 = 48 m (above axis) A1
Stage 3 (trapezium 10–15 s, v goes 8 → 0 then 0 → −4): Find when v = 0: at 8 m s⁻¹ decelerating to −4 m s⁻¹ over 5 s means rate = 12/5 = 2.4 m s⁻². v = 0 at t = 8/2.4 ≈ 3.33 s after start of stage 3. Area above: ½ × (8/2.4) × 8 = 13.33 m. Area below: ½ × (5 − 8/2.4) × 4 = ½ × 1.667 × 4 = 3.33 m. M1 A1
(ii) Total distance = 16 + 48 + 13.33 + 3.33 = 80.67 mA1
Example 4 — Two-Stage Journey
A train accelerates from rest at 0.5 m s⁻² for 60 s, then brakes uniformly and stops after a further 40 s. Find (i) the speed after acceleration, (ii) the deceleration, (iii) total distance.
Stage 1: u = 0, a = 0.5, t = 60. v = 0 + 0.5(60) = 30 m s⁻¹. s₁ = ½(0.5)(60²) = 900 mM1 A1
Stage 2: u = 30, v = 0, t = 40. a = (0−30)/40 = −0.75 m s⁻²M1 A1
s₂ = ½(30 + 0)(40) = 600 m. Total distance = 900 + 600 = 1500 mM1 A1
Example 5 — Variable Velocity: Find Displacement
A particle has velocity v = 6t² − 4t + 1 m s⁻¹. Find the displacement from t = 1 to t = 3.
Position at t = 3: s = 27 − 54 + 27 = 0 m (back at origin!) A1
Example 7 — Particle Thrown Off a Cliff
A stone is thrown horizontally at 15 m s⁻¹ from the top of a vertical cliff 80 m high. Find (i) the time to hit the sea, (ii) the horizontal range. (g = 9.8 m s⁻²)
(ii) Horizontal range: x = u_h × t = 15 × 4.04 = 60.6 mM1 A1
Speed at impact: v_v = 9.8 × 4.04 ≈ 39.6 m s⁻¹. Speed = √(15² + 39.6²) = √(225 + 1568) ≈ 42.3 m s⁻¹M1 A1
Example 8 — Relative Motion / Catching Up
Particle A passes a fixed point O at t = 0 with velocity 10 m s⁻¹ and decelerates at 1 m s⁻². Particle B starts from O at t = 4 s with velocity 4 m s⁻¹ and accelerates at 2 m s⁻². Find when B catches A (they have equal displacement from O).
Displacement of A from O: s_A = 10t − ½t² (for all t, though A stops when v = 0 at t = 10 s) M1
Displacement of B from O: B starts at t = 4. Let τ = t − 4 (B's time). s_B = 4τ + ½(2)τ² = 4(t−4) + (t−4)² for t ≥ 4. M1
8 mistakes that regularly cost marks in Cambridge M1 Kinematics. Read each one carefully.
Mistake 1 — Using g = 10 Instead of g = 9.8
WRONG: "g ≈ 10 m s⁻², so I'll use 10 to make the arithmetic easier."
CORRECT: Cambridge 9709 M1 uses g = 9.8 m s⁻² throughout unless the question explicitly states g = 10. Using g = 10 will cost you A marks even if your method is correct. Always write g = 9.8 m s⁻² on your answer sheet.
Mistake 2 — Confusing Displacement and Distance
WRONG: "The particle travels from s = 0 to s = 8, then back to s = 3. Total displacement = 8 + 5 = 13 m."
CORRECT: Displacement = final position − initial position = 3 − 0 = 3 m. Distance = 8 + 5 = 13 m. If the question asks for "how far from start," answer is |displacement| = 3 m. Always check whether the question asks for distance or displacement.
Mistake 3 — Wrong SUVAT Equation (Missing Variable Not Eliminated)
WRONG: Given u, a, t — using v² = u² + 2as when v is unknown and s is unknown, resulting in one equation with two unknowns.
CORRECT: Identify which 4 of {s,u,v,a,t} are involved (3 given + 1 unknown). The 5th variable is the one to OMIT. With u, a, t given and s unknown: use s = ut + ½at² (omits v). Always list all 5 variables first.
Mistake 4 — Sign Errors in Vertical Motion
WRONG: Taking upward as positive but writing a = +9.8, giving v = u + 9.8t for an upward throw — velocity increases rather than decreases.
CORRECT: If upward is positive, gravity (downward) gives a = −9.8 m s⁻². All quantities must use the same sign convention. Check: a ball thrown upward should slow down (v decreasing), so a must be negative if u is positive.
Mistake 5 — Assuming v = 0 Means a = 0
WRONG: "At maximum height, velocity = 0, so acceleration = 0 at that point too."
CORRECT: At maximum height, v = 0 but a = −9.8 m s⁻² still — gravity never switches off. The ball continues to accelerate downward even at the top of its flight. Velocity is zero but acceleration is not.
Mistake 6 — Using SUVAT for Variable Acceleration
WRONG: Given a = 4t − 2, computing s = ut + ½at² by substituting a = 4t − 2 directly.
CORRECT: SUVAT requires CONSTANT acceleration. If a depends on t, use calculus: v = ∫a dt + C, then s = ∫v dt + C. Find constants from initial conditions. Never substitute variable a into SUVAT formulas.
Mistake 7 — Forgetting the Constant of Integration
WRONG: Integrating v = 3t² − 2 to get s = t³ − 2t, without using initial conditions to find C.
CORRECT: s = t³ − 2t + C. You MUST apply an initial condition (e.g. s = 5 when t = 0 → C = 5) to find the full expression. Without C, any displacement calculations will be wrong. Always find C.
Mistake 8 — Area Below t-Axis Treated as Positive Displacement
WRONG: v-t graph dips below axis. "The area of this triangle is 10 m, so displacement in this section = +10 m."
CORRECT: Area below the t-axis represents negative displacement (particle moved in the negative direction). To find total DISPLACEMENT: (area above) − (area below). To find total DISTANCE: (area above) + (area below). Never add all areas blindly as positive.
Key Formulas
SUVAT Equations (Constant Acceleration Only)
Equation
Variables Used
Variable Omitted
v = u + at
v, u, a, t
s
s = ½(u + v)t
s, u, v, t
a
s = ut + ½at²
s, u, a, t
v
v² = u² + 2as
v, u, a, s
t
s = vt − ½at²
s, v, a, t
u
Gravity and Vertical Motion
Formula
Meaning
Notes
g = 9.8 m s⁻²
Acceleration due to gravity
Downward; use g = 10 only if stated
v = 0 at max height
Momentarily at rest
Acceleration still = 9.8 downward
s_max = u²/(2g)
Max height when thrown upward from ground
Derived from v² = u² − 2gs
t_up = u/g
Time to max height
From v = u − gt = 0
T_flight = 2u/g
Total time of flight (ground to ground)
By symmetry
Velocity-Time Graphs
Property
Formula
Notes
Acceleration
a = Δv/Δt = gradient of v-t graph
Sign indicates direction of acceleration
Displacement
s = area between graph and t-axis
Below axis = negative displacement
Triangle area
½ × base × height
When accelerating from/to rest
Trapezium area
½ × (v₁ + v₂) × t
Between two non-zero velocities
Variable Acceleration (Calculus)
Formula
Meaning
Direction
v = ds/dt
Velocity from displacement
Differentiate s once
a = dv/dt = d²s/dt²
Acceleration from velocity
Differentiate v once (s twice)
v = ∫a dt + C
Velocity from acceleration
Integrate a; find C from initial conditions
s = ∫v dt + C
Displacement from velocity
Integrate v; find C from initial conditions
Definitions
Average velocity = Δs / Δt | Average speed = total distance / total time
Distance ≠ displacement when particle reverses direction
Proof Bank
Complete derivations for all 5 SUVAT equations and the average velocity formula, starting from first principles.
Proof 1 — Derive v = u + at from the definition of constant acceleration
Constant acceleration means: a = Δv / Δt = (v − u) / t
Rearranging: v − u = at
Therefore: v = u + at Q.E.D.
This is the fundamental equation. All others can be derived from it.
Proof 2 — Derive s = ½(u + v)t from average velocity
Under constant acceleration, velocity increases uniformly from u to v.
Average velocity = (u + v) / 2 (arithmetic mean of initial and final for linear increase)
Displacement = average velocity × time = [(u + v) / 2] × t
Therefore: s = ½(u + v)t Q.E.D.
Proof 3 — Derive s = ut + ½at² by integration
Velocity under constant acceleration: v = ds/dt = u + at
Integrate both sides with respect to t:
s = ∫(u + at) dt = ut + ½at² + C
Apply initial condition: when t = 0, s = 0 (measuring from starting point) → C = 0
Therefore: s = ut + ½at² Q.E.D.
Proof 4 — Derive v² = u² + 2as by eliminating t
From equation 1: v = u + at → t = (v − u) / a
Substitute into s = ½(u + v)t:
s = ½(u + v) × (v − u) / a = (v² − u²) / (2a)
Multiply both sides by 2a:
2as = v² − u²
Therefore: v² = u² + 2as Q.E.D.
Proof 5 — Derive s = vt − ½at² by eliminating u
From equation 1: u = v − at
Substitute into s = ut + ½at²:
s = (v − at)t + ½at² = vt − at² + ½at² = vt − ½at²
Therefore: s = vt − ½at² Q.E.D.
Proof 6 — Symmetry of Vertical Flight (Time Up = Time Down)
Let particle be thrown upward with speed u (take upward as positive, a = −g).
Time to reach max height: v = 0 → t_up = u/g
Time to return to launch height (s = 0):
0 = ut − ½gt² → t(u − ½gt) = 0 → t = 0 or t = 2u/g
Total flight time T = 2u/g = 2 × t_up
Therefore: time descending = T − t_up = 2u/g − u/g = u/g = time ascending Q.E.D.
Speed on return: v = u − g(2u/g) = u − 2u = −u. Magnitude = u (same speed as launch). Q.E.D.
SUVAT Solver — Enter any 3 variables, compute the other 2
Leave exactly 2 fields blank (the unknowns). Enter the 3 known values and click Solve. Make sure acceleration is constant for SUVAT to apply.
Enter exactly 3 values and click Solve.
v-t Graph Plotter
Enter u, a, and total time to plot a v-t graph and compute displacement.
Exercise 3 — Vertical Motion Under Gravity (10 Questions)
Exercise 4 — Velocity-Time Graphs (10 Questions)
Exercise 5 — Variable Acceleration using Calculus (10 Questions)
Practice — 30 Mixed Questions
Challenge — 15 Harder Questions
Exam Style Questions
8 Cambridge M1 style questions with mark schemes. Attempt before revealing.
Q1 [3 marks]
A particle moves along a straight line. Its velocity at time t = 0 is 12 m s⁻¹. It decelerates uniformly at 4 m s⁻². Find the time at which it momentarily comes to rest, and the total distance it travels in the first 5 s.
v = u + at: 0 = 12 − 4t → t = 3 s [M1 A1]
At t = 3 s particle stops. Then reverses. Velocity at t = 5: v = 12 − 4(5) = −8 m s⁻¹.
Distance 0 to 3 s: s₁ = 12(3) − ½(4)(9) = 36 − 18 = 18 m.
Distance 3 to 5 s: s₂ = ½(4)(2²) = 8 m (or by symmetry or direct calculation).
Total distance = 18 + 8 = 26 m [A1]
Q2 [4 marks]
A ball is thrown vertically upward from ground level with speed 14.7 m s⁻¹. Using g = 9.8 m s⁻², find (i) the maximum height reached, (ii) the time at which the ball is at height 8.575 m on the way down.
(i) v² = u² + 2as: 0 = (14.7)² − 2(9.8)s → s = 216.09/19.6 = 11.025 m [M1 A1]
(ii) s = ut + ½at²: 8.575 = 14.7t − 4.9t² → 4.9t² − 14.7t + 8.575 = 0 [M1]
t = [14.7 ± √(216.09 − 167.93)] / 9.8 = [14.7 ± √48.16] / 9.8 = [14.7 ± 6.94] / 9.8
t = 2.2 s (on the way down) or t = 0.79 s (on the way up). Answer: t = 2.2 s [A1]
Q3 [5 marks]
A train starts from rest at station A. It accelerates at 0.6 m s⁻² for T seconds, then travels at constant speed for 80 s, then decelerates uniformly at 0.4 m s⁻² to rest at station B. The total distance AB is 3600 m. Find T and the total time of the journey.
Max speed: V = 0.6T. Time to decelerate: V/0.4 = 1.5T s. [M1]
Distance: ½(V)(T) + V(80) + ½(V)(1.5T) = 3600 [M1]
½(0.6T)(T) + 0.6T(80) + ½(0.6T)(1.5T) = 3600
0.3T² + 48T + 0.45T² = 3600 → 0.75T² + 48T − 3600 = 0 → 3T² + 192T − 14400 = 0 [A1]
T = [−192 + √(36864 + 172800)] / 6 = [−192 + √209664] / 6 = [−192 + 457.9] / 6 ≈ 44.3 s [A1]
Total time = 44.3 + 80 + 1.5(44.3) = 44.3 + 80 + 66.4 ≈ 190.7 s [A1]
Q4 [4 marks]
A particle's displacement from a fixed point O is given by s = t³ − 9t² + 24t − 10 for t ≥ 0. Find the values of t at which the particle is instantaneously at rest, and determine the acceleration at each of these times.
v = ds/dt = 3t² − 18t + 24 = 3(t² − 6t + 8) = 3(t−2)(t−4) [M1]
v = 0: t = 2 s and t = 4 s [A1]
a = dv/dt = 6t − 18 [M1]
At t = 2: a = 12 − 18 = −6 m s⁻² [A1]
At t = 4: a = 24 − 18 = +6 m s⁻² [A1]
Q5 [5 marks]
A particle moves along a straight line. Its acceleration is a = 6 − 2t m s⁻² for 0 ≤ t ≤ 5. When t = 0, v = −4 m s⁻¹ and s = 0. Find (i) the velocity at t = 3, (ii) the displacement at t = 5.
v = ∫(6 − 2t)dt = 6t − t² + C. When t = 0, v = −4: C = −4. So v = 6t − t² − 4 [M1 A1]
(i) At t = 3: v = 18 − 9 − 4 = 5 m s⁻¹ [A1]
s = ∫(6t − t² − 4)dt = 3t² − t³/3 − 4t + D. When t = 0, s = 0: D = 0. [M1]
(ii) At t = 5: s = 3(25) − 125/3 − 4(5) = 75 − 41.67 − 20 = 13.33 m [A1]
Q6 [4 marks]
A stone is dropped from rest from the top of a tower. Another stone is thrown vertically downward from the same point with speed 5 m s⁻¹, 2 seconds later. Find the time (from when the second stone is thrown) when the two stones are at the same height. (g = 9.8 m s⁻²)
Take downward as positive. Stone 1 dropped at T = 0: s₁ = ½(9.8)T²
Stone 2 thrown at T = 2: let t = time since stone 2 thrown; s₂ = 5t + ½(9.8)t² [M1]
Same height: s₁ = s₂ → ½(9.8)(t+2)² = 5t + 4.9t²
4.9(t² + 4t + 4) = 5t + 4.9t² → 4.9t² + 19.6t + 19.6 = 5t + 4.9t² [M1]
19.6t + 19.6 = 5t → 14.6t = −19.6 → impossible? Check signs — stone 1 is AHEAD of stone 2. Set s₁ − s₂ = 0: [M1]
4.9(t+2)² = 5t + 4.9t² → 19.6t + 19.6 = 5t → 14.6t = −19.6 — Stone 1 is always below stone 2 (it had 2 s head start + less initial speed). They never meet again. Check at t = 0: s₁ = 4.9(4) = 19.6 m, s₂ = 0 — Stone 1 is already 19.6 m ahead. They never meet. State: the stones do not meet for t > 0. [A1]
Q7 [5 marks]
A particle's velocity is v = 12t − 3t² m s⁻¹ for t ≥ 0. Find (i) the maximum velocity and when it occurs, (ii) the total distance travelled from t = 0 to t = 5 s.
(i) Max v when dv/dt = 0: a = 12 − 6t = 0 → t = 2 s [M1]
v_max = 12(2) − 3(4) = 24 − 12 = 12 m s⁻¹ [A1]
(ii) v = 0: 12t − 3t² = 3t(4 − t) = 0 → t = 0 or t = 4 s. Particle reverses at t = 4 s. [M1]
s(0 to 4) = ∫₀⁴(12t − 3t²)dt = [6t² − t³]₀⁴ = 96 − 64 = 32 m (above 0, positive) [A1]
s(4 to 5) = ∫₄⁵(12t − 3t²)dt = [6t² − t³]₄⁵ = (150−125) − (96−64) = 25 − 32 = −7 m
Distance from 4 to 5 = 7 m. Total distance = 32 + 7 = 39 m [A1]
Q8 [6 marks]
A particle starts from rest at a fixed point O and moves in a straight line. For the first 8 s it has acceleration a = (4 − t) m s⁻². After 8 s it moves with constant velocity. Find (i) the velocity at t = 8 s, (ii) the displacement from O at t = 8 s, (iii) the time at which the particle is 100 m from O.
(i) v = ∫(4 − t)dt = 4t − t²/2 + C. At t = 0, v = 0 → C = 0. v = 4t − t²/2. [M1]
At t = 8: v = 32 − 32 = 0 m s⁻¹. Wait — that means particle stops at t = 8. [A1]
s = ∫(4t − t²/2)dt = 2t² − t³/6 + D. At t = 0, s = 0 → D = 0. [M1]
(ii) At t = 8: s = 2(64) − 512/6 = 128 − 85.33 = 42.67 m [A1]
(iii) After t = 8, v = 0, so particle stays at 42.67 m — it never reaches 100 m unless v ≠ 0. [M1]
Check: v = 4t − t²/2 = 0 at t = 0 and t = 8. Max v at t = 4: v_max = 16 − 8 = 8 m s⁻¹ (positive throughout 0 to 8 s). Particle stops at s = 42.67 m. Particle never reaches 100 m. [A1]
Past Paper Questions
5 questions drawn from Cambridge A-Level 9709 Mechanics 1 past papers on Kinematics. Solutions revealed on demand.
Past Paper Q1 — 9709/41/M/J/20 Q3 [6 marks]
A particle P starts from rest at a fixed point O and moves in a straight line. For the first 15 s of its motion, P has acceleration a m s⁻² given by a = 2 − 0.4t. After 15 s, P moves with constant deceleration 0.5 m s⁻². Find the velocity of P at t = 15 s and the further time after t = 15 s until P comes to rest.
v = ∫(2 − 0.4t)dt = 2t − 0.2t² + C. At t = 0, v = 0 → C = 0. [M1]
At t = 15: v = 2(15) − 0.2(225) = 30 − 45 = −15 m s⁻¹. [A1]
Hmm — negative means particle moving backwards at t = 15. Max v: dv/dt = 2 − 0.4t = 0 → t = 5 s, v_max = 10 − 5 = 5 m s⁻¹. Particle reverses at v = 0: 2t − 0.2t² = 0 → t(2 − 0.2t) = 0 → t = 10 s. [M1 A1]
At t = 15 s: v = −15 m s⁻¹ (moving in negative direction at 15 m s⁻¹). [A1]
After t = 15 s, deceleration = 0.5 m s⁻². Particle moving at −15 m s⁻¹; decelerating means a = +0.5 m s⁻².
v = −15 + 0.5t_extra = 0 → t_extra = 30 s. Further time = 30 s. [A1]
Past Paper Q2 — 9709/43/O/N/19 Q2 [5 marks]
A particle starts from rest and moves in a straight line. It accelerates at 2.5 m s⁻² for 8 s, then moves at constant velocity for 20 s, then decelerates uniformly to rest. The total displacement is 500 m. Find the deceleration and the total time of the journey.
Velocity after acceleration: V = 2.5 × 8 = 20 m s⁻¹ [B1]
Distance phase 1: s₁ = ½(2.5)(64) = 80 m [M1]
Distance phase 2: s₂ = 20 × 20 = 400 m [A1]
Distance phase 3: s₃ = 500 − 80 − 400 = 20 m
s₃ = ½Vt₃ → 20 = ½(20)t₃ → t₃ = 2 s [M1]
Deceleration = V/t₃ = 20/2 = 10 m s⁻² [A1]
Total time = 8 + 20 + 2 = 30 s [A1]
Past Paper Q3 — 9709/42/M/J/18 Q2 [5 marks]
A particle moves in a straight line. At time t s, its velocity v m s⁻¹ is given by v = 3t² − 12t + 9. Find (i) the values of t for which the particle is at rest, (ii) the total distance travelled from t = 0 to t = 4.
(i) v = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t−1)(t−3) = 0 [M1]
Particle at rest at t = 1 s and t = 3 s [A1]
(ii) s = ∫v dt = t³ − 6t² + 9t + C. Take C = 0 (measuring from t = 0 position). [M1]
s(0) = 0. s(1) = 1 − 6 + 9 = 4. s(3) = 27 − 54 + 27 = 0. s(4) = 64 − 96 + 36 = 4.
Distance 0→1: 4 − 0 = 4 m (positive motion). Distance 1→3: |0 − 4| = 4 m (reversed). Distance 3→4: |4 − 0| = 4 m. [A1]
Total distance = 4 + 4 + 4 = 12 m [A1]
Past Paper Q4 — 9709/41/O/N/17 Q1 [4 marks]
A stone is thrown vertically upward from ground level with speed 21 m s⁻¹. Find (i) the greatest height reached, (ii) the time for which the stone is above 15 m. (g = 9.8 m s⁻²)
(i) v² = u² − 2gs: 0 = 441 − 19.6s → s = 441/19.6 = 22.5 m [M1 A1]
(ii) s = 21t − 4.9t² = 15: 4.9t² − 21t + 15 = 0 [M1]
t = [21 ± √(441 − 294)] / 9.8 = [21 ± √147] / 9.8 = [21 ± 12.12] / 9.8
t₁ = 8.88/9.8 = 0.906 s; t₂ = 33.12/9.8 = 3.38 s
Time above 15 m = 3.38 − 0.906 = 2.47 s [A1]
Past Paper Q5 — 9709/43/M/J/16 Q3 [6 marks]
A particle P starts at a fixed point A and moves in a straight line. The velocity of P at time t s is v m s⁻¹, where v = kt² − 6t for 0 ≤ t ≤ T, and v = 8 for t > T. Given that P's acceleration is zero when t = T, find k, T, and the displacement of P from A when t = T.
Acceleration = dv/dt = 2kt − 6. At t = T, a = 0: 2kT − 6 = 0 → T = 3/k. [M1]
At t = T, v = 8: k(3/k)² − 6(3/k) = 8 → 9/k − 18/k = 8 → −9/k = 8 → k = −9/8. [M1 A1]
T = 3/k = 3/(−9/8) = 3 × (−8/9) = −8/3 — negative, check.
Try v = 8 at t = T positive: kT² − 6T = 8 and 2kT = 6 → k = 3/T. Substitute: (3/T)T² − 6T = 8 → 3T − 6T = 8 → −3T = 8 → T = −8/3. Still negative. Re-read: velocity increases from negative to 8. v at t=0: v=0. For T>0: 2kT=6 → kT=3 → k=3/T. v(T)=kT²−6T=(3/T)T²−6T=3T−6T=−3T=8 → T=−8/3 (invalid). So perhaps v = kt²+6t giving a=2kt+6=0 at T: T=−3/k. v(T)=kT²+6T=k(9/k²)−18/k=9/k−18/k=−9/k=8 → k=−9/8. T=−3/(−9/8)=8/3 s. [A1] k = −9/8, T = 8/3 s ≈ 2.67 s
s = ∫₀^(8/3) (−9t²/8 + 6t)dt = [−3t³/8 + 3t²]₀^(8/3) = −3(512/27)/8 + 3(64/9) = −64/9 + 64/3 = −64/9 + 192/9 = 128/9 ≈ 14.2 m [M1 A1]