This lesson covers the full Cambridge 9709 Mechanics 2 kinematics syllabus. You will learn to handle motion where acceleration is not constant — using calculus, vector methods, relative motion and Simple Harmonic Motion.
Topics Covered:
1. Variable acceleration using calculus (v = dx/dt, a = dv/dt)
2. Velocity as a function of displacement (a = v dv/dx)
3. 2D motion with position vectors
4. Relative velocity and closest approach
5. Simple Harmonic Motion (SHM)
How to use this lesson: Work through Learn 1–5 in order, study the Examples, then test yourself with Ex 1–5. Use Practice and Challenge for extra drill, then attempt Exam Style and Past Paper Qs under timed conditions.
Learn 1: Variable Acceleration Using Calculus
Core Relationships
When acceleration is a function of time, we use calculus to link displacement, velocity and acceleration.
v = dx/dt | a = dv/dt = d²x/dt²
From acceleration to velocity
Integrate a(t) with respect to t to obtain v(t). Add a constant of integration C determined by an initial condition (usually v = v₀ when t = 0).
v(t) = ∫ a(t) dt + C₁
From velocity to displacement
Integrate v(t) with respect to t to obtain x(t). Add another constant of integration determined by the initial position.
x(t) = ∫ v(t) dt + C₂
Initial Conditions: You must always use a given condition to find each constant. Common conditions:
• At t = 0, x = 0 (starts at origin)
• At t = 0, v = u (initial velocity u)
• At t = T, x = d (known position at time T)
Distance vs Displacement
Warning: To find total distance travelled, check whether the particle changes direction (v = 0). Split the integral at turning points and sum magnitudes.
Example Walkthrough
A particle moves so that a = 6t − 4 ms⁻². At t = 0, v = 2 ms⁻¹ and x = 0.
Step 1: Integrate a to find v. v = ∫(6t − 4) dt = 3t² − 4t + C₁. At t=0, v=2 → C₁=2. ∴ v = 3t² − 4t + 2 M1 A1
Step 2: Integrate v to find x. x = ∫(3t² − 4t + 2) dt = t³ − 2t² + 2t + C₂. At t=0, x=0 → C₂=0. ∴ x = t³ − 2t² + 2t M1 A1
Always write "+ C" when integrating, then substitute the condition on the very next line to find C's value.
Learn 2: Velocity as a Function of Displacement
The Chain Rule Connection
When acceleration is given as a function of displacement x rather than time t, we cannot integrate directly with respect to t. Instead we use:
a = v · dv/dx
Derivation: a = dv/dt. By the chain rule: dv/dt = (dv/dx)·(dx/dt) = v·dv/dx. So a = v dv/dx whenever v and x are related.
Finding v² as a function of x
Rewrite as: v dv/dx = f(x). Integrate both sides with respect to x:
½v² = ∫ f(x) dx + C
Use the initial condition (v = v₀ when x = x₀) to find C.
Polynomial Example
If a = 3x² − 2x and v = 4 when x = 1:
Step 1: v dv/dx = 3x² − 2x ∫v dv = ∫(3x² − 2x)dx ½v² = x³ − x² + C M1
Step 3: Find v at x = 2: v² = 16 − 8 + 16 = 24 → v = 2√6 ms⁻¹ A1
Exponential Example
If a = 2e^(−x) and v = 0 when x = 0:
½v² = ∫2e^(−x) dx = −2e^(−x) + C. At x=0, v=0: 0 = −2 + C → C = 2. ∴ v² = 4 − 4e^(−x) = 4(1 − e^(−x)) M1 A1
Note: Always check that v² ≥ 0 for the domain of x you are considering. If v² < 0, the particle cannot reach that position.
Learn 3: 2D Motion with Vectors
Position, Velocity and Acceleration
r = xi + yj | v = dr/dt | a = dv/dt
Differentiate the position vector component by component.
If r = (3t² + 1)i + (2t³ − t)j, then:
v = 6ti + (6t² − 1)j
a = 6i + 12tj
Speed and Direction
speed = |v| = √(vₓ² + vᵧ²)
The angle of motion with the positive x-axis: θ = arctan(vᵧ / vₓ).
Finding When Velocity Has a Given Direction
Set up the condition on the ratio vᵧ/vₓ. For example, "moving parallel to i" means vᵧ = 0; "moving at 45° above horizontal" means vᵧ = vₓ.
Example: v = (2t − 3)i + (t² − 4)j. Find t when motion is parallel to i. Need vᵧ = 0: t² − 4 = 0 → t = 2 (taking t > 0) M1 A1
Integrating Vector Functions
Integrate each component separately. Each component has its own constant of integration, found from initial conditions.
r = ∫v dt = (∫vₓ dt)i + (∫vᵧ dt)j
Write the vector constants as Ci + Dj where C and D are found from the initial position r₀.
Learn 4: Relative Motion
Relative Velocity
vAB = vA − vB
This is the velocity of A as seen by an observer moving with B. If vAB = 0 then A and B have the same velocity.
Relative Position
rAB(t) = rA(t) − rB(t)
Closest Approach
The particles are closest when |rAB(t)|² is minimised. This is easier than minimising |rAB| directly.
Method: 1. Find rAB(t) = rA(t) − rB(t) 2. Form |rAB|² = (x-component)² + (y-component)² 3. Differentiate with respect to t and set = 0 4. Solve for t, then find minimum distance M1 M1 A1
Interception
A intercepts B when rA(t) = rB(t), i.e. rAB(t) = 0. Set each component equal to zero and solve simultaneously.
Example: A is at (1+2t)i + (3t)j; B is at (4t−2)i + (1+t)j.
x: 1+2t = 4t−2 → t = 1.5
y: 3(1.5) = 4.5; 1+1.5 = 2.5 ← not equal, so no interception at t = 1.5. They do not meet.
Key: Both components must give the same t for interception to occur. If they give different values of t, the paths cross but the particles do not meet simultaneously.
Learn 5: Simple Harmonic Motion (SHM)
Defining Equation
a = −ω²x
A particle executes SHM if and only if its acceleration is proportional to displacement from a fixed point and directed towards it.
General Solution
x = A cos(ωt + φ)
A = amplitude (maximum displacement from centre), ω = angular frequency (rad s⁻¹), φ = phase angle (determined by initial conditions).
Equivalently: x = a sin(ωt + φ), or x = P cos(ωt) + Q sin(ωt). Choose the form that fits initial conditions most cleanly.
✓ Find where v = 0, split integral, sum magnitudes of each section
Mistake 4: Relative Velocity Direction Error
✗ vAB = vB − vA (wrong sign)
✓ vAB = vA − vB (velocity of A relative to B = A's velocity minus B's velocity)
Mistake 5: Confusing ω and T in SHM
✗ T = ω, or f = ω (treating angular frequency as period or frequency)
✓ T = 2π/ω and f = ω/(2π). If a = −9x then ω = 3, T = 2π/3
Mistake 6: Sign Error When Differentiating Vectors
✗ Differentiating r = (3t²+1)i + (2−t)j to get v = 6ti + (2)j (wrong sign for j component)
✓ v = d/dt(2−t)j = −j, so v = 6ti − j
Mistake 7: Using v² = ω²(A²−x²) with Wrong ω
✗ Reading a = −4x and saying ω = 4, then using v² = 4(A²−x²)
✓ a = −ω²x, so a = −4x gives ω² = 4, ω = 2. Then v² = 4(A²−x²) but only if ω² = 4 ✓
Key Formulas
Formula
Description
Notes
v = dx/dt
Velocity from position
Differentiate x(t)
a = dv/dt
Acceleration from velocity
Differentiate v(t)
a = d²x/dt²
Acceleration from position
Second derivative
v = ∫a dt + C
Velocity by integration
Need initial v
x = ∫v dt + C
Position by integration
Need initial x
a = v dv/dx
Acceleration when a = f(x)
Chain rule
½v² = ∫f(x) dx + C
v² from a = f(x)
Use initial (x,v)
v = dr/dt
Velocity vector
Differentiate each component
a = dv/dt
Acceleration vector
Differentiate each component
|v| = √(vₓ²+vᵧ²)
Speed
Magnitude of velocity vector
θ = arctan(vᵧ/vₓ)
Direction of motion
Angle with x-axis
vAB = vA − vB
Relative velocity
Velocity of A relative to B
rAB = rA − rB
Relative position
Position of A relative to B
a = −ω²x
SHM defining equation
Towards centre
x = A cos(ωt + φ)
SHM displacement
A = amplitude
v² = ω²(A² − x²)
SHM velocity-displacement
Max |v| = Aω at x=0
T = 2π/ω
SHM period
Seconds per oscillation
Max |a| = Aω²
Maximum acceleration
At x = ±A
Proof Bank
Proof 1: a = v dv/dx from the Chain Rule
We know: a = dv/dt (by definition of acceleration).
By the chain rule for differentiation:
dv/dt = (dv/dx) · (dx/dt)
But dx/dt = v (by definition of velocity), so:
a = dv/dt = v · dv/dx □
This identity is useful whenever acceleration is expressed as a function of displacement rather than time, allowing us to find v as a function of x without knowing t explicitly.
Proof 2: General Solution of SHM ODE
The SHM equation is: d²x/dt² = −ω²x, i.e. ẍ + ω²x = 0.
This is a second-order linear ODE with constant coefficients.
Auxiliary equation: m² + ω² = 0 → m = ±ωi (pure imaginary roots).
For roots m = ±αi the general solution is:
x = P cos(αt) + Q sin(αt)
With α = ω:
x = P cos(ωt) + Q sin(ωt)
This can be written as x = A cos(ωt + φ) where:
A = √(P² + Q²) and tan φ = −Q/P □
But x = A cos(ωt+φ), so cos²(ωt+φ) = x²/A²:
v² = A²ω²(1 − x²/A²) = ω²(A² − x²) □
This formula is especially useful when finding the speed at a specific displacement, or finding the amplitude given a speed and displacement.
SHM Visualiser — Spring-Mass System
1.52.0
x = 0.00 m
v = 0.00 ms⁻¹
a = 0.00 ms⁻²
Phase: 0°
T = 3.14 s | Max speed = 3.00 ms⁻¹ | Max |a| = 6.00 ms⁻²
Exercise 1 — Variable Acceleration (Time)
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Exercise 2 — v as Function of x
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Exercise 3 — 2D Vector Motion
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Exercise 4 — Relative Motion
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Exercise 5 — Simple Harmonic Motion
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Practice — 30 Questions (Mixed)
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Challenge — 15 Hard Questions
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Exam Style Questions
Q1 [5 marks]
A particle moves along the x-axis. Its velocity is v = 3t² − 12t + 9 ms⁻¹. At t = 0 the particle is at the origin. Find the total distance travelled in the first 3 seconds.
M1: Find when v=0: 3(t²−4t+3)=3(t−1)(t−3)=0 → t=1, t=3.
M1: x = ∫v dt = t³−6t²+9t (+C; at t=0, x=0 → C=0).
A1: x(1) = 1−6+9 = 4; x(3) = 27−54+27 = 0; x(0) = 0.
M1: Distance = |4−0| + |0−4| = 4 + 4 = 8 m.
A1: Total distance = 8 m.
Q2 [6 marks]
A particle moves in a straight line with acceleration a = (2x + 3) ms⁻², where x is displacement from O. When x = 1, v = 2 ms⁻¹. Find v when x = 4, and the displacement when v = 10 ms⁻¹.
M1: v dv/dx = 2x+3. Integrate: ½v² = x²+3x+C.
M1 A1: At x=1, v=2: 2 = 1+3+C → C=−2. So v² = 2x²+6x−4.
A1: At x=4: v² = 32+24−4 = 52 → v = 2√13 ≈ 7.21 ms⁻¹.
M1 A1: v=10: 100 = 2x²+6x−4 → 2x²+6x−104=0 → x²+3x−52=0 → x = (−3+√(9+208))/2 = (−3+√217)/2 ≈ 5.86 m.
Q3 [7 marks]
At time t, the position of a particle is r = (t²−3t)i + (2t−t²)j m. Find: (a) the velocity when the particle is moving parallel to i, (b) the speed when t = 4, (c) the time when the particle is at rest.
M1: v = (2t−3)i + (2−2t)j.
A1(a): Parallel to i means j-component = 0: 2−2t=0 → t=1. v = (2−3)i = −i ms⁻¹.
M1 A1(b): At t=4: v = 5i + (−6)j. Speed = √(25+36) = √61 ≈ 7.81 ms⁻¹.
M1(c): At rest: 2t−3=0 and 2−2t=0 → t=1.5 and t=1. Different values → particle never fully at rest simultaneously in both components.
Q4 [6 marks]
Ships A and B have velocities vA = (3i + 4j) kmh⁻¹ and vB = (i + 2j) kmh⁻¹. At t=0, rA = (2i + j) km and rB = (5i + 3j) km. Find the time and minimum distance of closest approach.
M1: rAB = rA−rB = (2+3t−5−t)i + (1+4t−3−2t)j = (2t−3)i + (2t−2)j.
M1: |rAB|² = (2t−3)²+(2t−2)² = 4t²−12t+9+4t²−8t+4 = 8t²−20t+13.
M1 A1: d/dt = 16t−20 = 0 → t = 1.25 h.
A1: |rAB|² = 8(1.5625)−25+13 = 12.5−25+13 = 0.5. Distance = √0.5 = 1/√2 ≈ 0.707 km.
A1: Closest at t = 1.25 h, distance ≈ 0.707 km.
Q5 [5 marks]
A particle performs SHM with a = −25x. It passes through x = 1.2 m with speed 3.5 ms⁻¹ moving away from the centre. Find the amplitude and the period.
B1: ω² = 25, ω = 5 rad s⁻¹.
M1: v² = ω²(A²−x²): 12.25 = 25(A²−1.44).
A1: A²−1.44 = 0.49 → A² = 1.93 → A = √1.93 ≈ 1.39 m.
M1 A1: T = 2π/5 ≈ 1.257 s.
Q6 [6 marks]
A particle's acceleration satisfies a = 6t − 2 ms⁻². Initially v = −5 ms⁻¹ and x = 3 m. Find: (a) the time when the particle is momentarily at rest, (b) the position at that time.
M1: v = 3t²−2t+C. At t=0, v=−5 → C=−5. ∴ v = 3t²−2t−5.
M1 A1: v=0: 3t²−2t−5=0 → (3t−5)(t+1)=0 → t=5/3 s (t>0).
M1: x = t³−t²−5t+D. At t=0, x=3 → D=3. ∴ x = t³−t²−5t+3.
A1: At t=5/3: x=(125/27)−(25/9)−(25/3)+3 = 125/27−75/27−225/27+81/27 = −94/27 ≈ −3.48 m.
Q7 [5 marks]
An object oscillates with SHM of period 4 s and amplitude 0.8 m. At t = 0, x = 0.8 m (at positive extreme). Find x and v at t = 1.5 s.
M1: T=4 → ω = 2π/4 = π/2.
B1: Starting at positive extreme → x = 0.8 cos(πt/2).
A1: At t=1.5: x = 0.8 cos(3π/4) = 0.8·(−√2/2) = −0.4√2 ≈ −0.566 m.
M1 A1: v = −0.8·(π/2)·sin(3π/4) = −0.4π·(√2/2) = −0.2π√2 ≈ −0.888 ms⁻¹.
Q8 [7 marks]
At time t ≥ 0, velocity v = (4−t²)i + (3t)j ms⁻¹. At t=0, position is 2i−j m. Find: (a) acceleration at t=2, (b) position when particle is moving parallel to j, (c) speed when t=3.
M1 A1(a): a = dv/dt = (−2t)i + 3j. At t=2: a = −4i+3j ms⁻².
M1(b): Parallel to j → i-component of v = 0: 4−t²=0 → t=2.
M1: r = ∫v dt = (4t−t³/3)i + (3t²/2)j + C. At t=0, r=2i−j → C=2i−j.
A1(b): At t=2: r = (8−8/3+2)i + (6−1)j = (22/3)i + 5j m.
M1 A1(c): At t=3: v = (4−9)i+9j = −5i+9j. Speed = √(25+81) = √106 ≈ 10.3 ms⁻¹.
Past Paper Questions (Adapted 9709 M2)
PP1 — 9709/43 adapted [6 marks]
A particle P moves along a straight line. At time t seconds, the velocity of P is v ms⁻¹, where v = 2t³ − 9t² + 12t − 4. Find the set of values of t for which P is moving in the positive direction, and find the total distance P travels in the interval 0 ≤ t ≤ 3.
M1: v = (2t−1)(t−2)² — factorise or use sign chart.
A1: v>0 when t>1/2 and t≠2, i.e. t∈(1/2, 2)∪(2, ∞) within [0,3]: {t: 1/2<t≤3, t≠2}.
M1: x = ∫v dt = t⁴/2−3t³+6t²−4t(+C). x(0)=0, x(1/2)=..., x(2)=..., x(3)=...
A1: x(1/2) = 1/32−3/8+6/4−2 = −27/16; x(2)=8−24+24−8=0; x(3)=81/2−81+54−12=13.5−81+54−12=−25.5 → recheck: 40.5−81+54−12=1.5.
M1: Distance = |x(1/2)−x(0)| + |x(2)−x(1/2)| + |x(3)−x(2)|
A1: = 27/16 + 27/16 + 3/2 = 54/16 + 24/16 = 78/16 = 39/8 = 4.875 m.
PP2 — 9709/42 adapted [7 marks]
At time t = 0, particles A and B have position vectors (6i+2j) m and (2i+4j) m respectively. A has constant velocity (−i+3j) ms⁻¹ and B has constant velocity (2i−j) ms⁻¹. Find the time when the particles are closest together and the minimum distance between them.
M1: rA = (6−t)i+(2+3t)j; rB = (2+2t)i+(4−t)j.
M1: rAB = (4−3t)i+(−2+4t)j.
M1: |rAB|² = (4−3t)²+(4t−2)² = 16−24t+9t²+16t²−16t+4 = 25t²−40t+20.
M1 A1: d/dt = 50t−40=0 → t = 4/5 = 0.8 s.
A1: |rAB|² = 25(0.64)−40(0.8)+20 = 16−32+20 = 4. Distance = 2 m.
A1: Minimum distance = 2 m at t = 0.8 s.
PP3 — 9709/41 adapted [5 marks]
A particle moves with SHM about a centre O. The particle has speed 6 ms⁻¹ when its displacement from O is 0.5 m and speed 4 ms⁻¹ when its displacement is 1 m. Find the amplitude and the period of the motion.
A particle moves along a straight line so that at time t its acceleration is a = e^(2t) − 3 ms⁻². When t = 0, v = 1 ms⁻¹ and x = 0. Find v and x when t = ln 2.
M1: v = ∫(e^(2t)−3) dt = ½e^(2t)−3t+C.
A1: At t=0, v=1: 1=½+C → C=½. ∴ v = ½e^(2t)−3t+½.
M1 A1: At t=ln2: v = ½e^(2ln2)−3ln2+½ = ½(4)−3ln2+½ = 2.5−3ln2 ≈ 0.421 ms⁻¹.
M1: x = ∫v dt = ¼e^(2t)−(3/2)t²+½t+D. At t=0, x=0: D=−¼.
A1: At t=ln2: x = ¼(4)−(3/2)(ln2)²+½ln2−¼ = 1−(3/2)(ln2)²+½ln2−¼ = 0.75+(ln2)(½−(3/2)ln2) ≈ 0.75+0.693(0.5−1.040) ≈ 0.75−0.374 ≈ 0.376 m.
PP5 — 9709/42 adapted [7 marks]
A particle moves along the x-axis. At time t, a = 12t − 18 ms⁻². At t = 0, v = 9 ms⁻¹, x = 0. (a) Find the times when the particle is instantaneously at rest. (b) Find the displacement when t = 4. (c) Show that the particle passes through the origin again and find this time.
M1: v = 6t²−18t+C. C=9. ∴ v = 6t²−18t+9.
A1(a): 6t²−18t+9=0 → 2t²−6t+3=0 → t=(6±√(36−24))/4=(6±√12)/4=(3±√3)/2.
t₁=(3−√3)/2≈0.634 s, t₂=(3+√3)/2≈2.366 s.
M1: x = 2t³−9t²+9t(+C, C=0).
A1(b): x(4)=128−144+36=20 m.
M1(c): x=0: 2t³−9t²+9t=0 → t(2t²−9t+9)=0 → t(2t−3)(t−3)=0.
t=0 (start), t=3/2, t=3.
A1: Passes through origin again at t=3 s (x=54−81+27=0 ✓) and also t=1.5 s.
A1: First return at t = 1.5 s, second at t = 3 s.