9231 Further Mathematics — Paper 3: Mechanics
t v v_T dv/dt a = v dv/dx when F=F(x) F = m dv/dt when F=F(t)
9231 Further Maths › Paper 3: Mechanics › Lesson 05

Linear Motion Under a
Variable Force

Newton's Second Law as a differential equation — separation of variables, terminal velocity, and resistance problems.

🎯 Cambridge 9231 P3 📐 Syllabus Section 3.5 ⭐ Calculus-Heavy Topic ⏱ ~100 min
05Lesson

1. Newton's Second Law — The Foundation

When force varies with time, position, or velocity, the equation F = ma becomes a differential equation. Solving it gives velocity and displacement as functions of the independent variable.

⭐ The Master Equation
F = m a = m dv/dt = m v dv/dx

Choose the form that matches what F depends on.

The Three Forms of Acceleration

All three are equivalent — choose the right one for the problem
a = dv/dtrate of change of velocity with timeuse when F = F(t)
a = v dv/dxvelocity × rate of change of v with positionuse when F = F(x)
a = d²x/dt²second derivative of displacementuse when F = F(t) and want x(t)
💡 Why v dv/dx?

By the chain rule: dv/dt = (dv/dx)(dx/dt) = v dv/dx. This is the key substitution that lets us connect velocity directly to position — eliminating time from the equation entirely. Use this form whenever the force is a function of displacement x.

2. Three Problem Types — Method Selector

Every variable force problem falls into one of three categories. Recognising the type immediately tells you which form of Newton's law and which integration strategy to use.

Type 1

Force depends on Time

F = F(t) — force is a function of time only.

Strategy: use F = m dv/dt, separate variables, integrate both sides.

m dv/dt = F(t)
⟹ ∫m dv = ∫F(t) dt
Type 2

Force depends on Position

F = F(x) — force depends on how far the particle has moved.

Strategy: use F = mv dv/dx, separate v and x, integrate.

mv dv/dx = F(x)
⟹ ∫mv dv = ∫F(x) dx
Type 3

Force depends on Velocity

F = F(v) — resistance / drag proportional to v or v².

Strategy: use m dv/dt = F(v) or mv dv/dx = F(v), separate variables, partial fractions if needed.

m dv/dt = F(v)
⟹ ∫m dv/F(v) = ∫dt

3. Resistance Proportional to Velocity

A very common 9231 exam model: a particle moves against a resistive force proportional to its speed — R = kv. This models viscous drag at low speeds.

Horizontal Motion — Decelerating Particle

A particle of mass m, initial speed u, subject to resistance force kv (opposing motion):

Newton's 2nd Law (taking positive direction as motion)
m dv/dt= −kv
m/v dv= −k dtseparate variables
m ln v= −kt + Cintegrate both sides
v= u e^(−kt/m)apply v = u at t = 0

To find displacement x in terms of t, integrate again. To find v in terms of x, use mv dv/dx = −kv:

m dv/dx= −kv cancels!
v= u − kx/m
x_max= mu/kwhen v = 0

4. Terminal Velocity

When a particle falls under gravity with air resistance, it accelerates until the net force becomes zero. The constant velocity reached at this point is called terminal velocity.

⭐ Terminal Velocity Condition

At terminal velocity vT, acceleration = 0, so:

Driving force = Resistance force
mg = kvT  →  vT = mg/k

For resistance proportional to v². If R = kv²:  vT = √(mg/k)

Falling Under Gravity with Resistance kv

Taking downward as positive, with resistance opposing motion (upward):

Equation of motion (downward positive)
m dv/dt= mg − kv
dv/(mg−kv)= dt/mseparate variables
−(1/k) ln(mg−kv)= t/m + C
v= (mg/k)(1 − e^(−kt/m))with v = 0 at t = 0
v → mg/k= vT as t → ∞approaches terminal velocity

5. Resistance Proportional to v²

At higher speeds, drag is proportional to v² — a more realistic model for air resistance.

Falling particle, resistance kv², downward positive
m dv/dt= mg − kv²
vT= √(mg/k)terminal velocity
∫m dv/(mg−kv²)= ∫dtuse partial fractions
📐 Partial Fractions for v² Resistance

The integral ∫dv/(a²−v²) uses the factorisation (a−v)(a+v):

1/(a²−v²)= (1/2a)[1/(a−v) + 1/(a+v)]
∫dv/(a²−v²)= (1/2a) ln|(a+v)/(a−v)| + C

6. Work-Energy Theorem for Variable Forces

The work done by a variable force F(x) on a particle moving from x₁ to x₂ equals the change in kinetic energy:

W= ∫F(x) dx from x₁ to x₂work done by variable force
W= ΔKE = ½mv₂² − ½mv₁²work-energy theorem
⚠ Key Exam Distinction
  • When asked for velocity as a function of time: use m dv/dt = F, integrate w.r.t. t
  • When asked for velocity as a function of position: use mv dv/dx = F, integrate w.r.t. x
  • When asked for displacement as a function of time: find v(t) first, then integrate v = dx/dt
  • Always state your initial conditions when finding the constant of integration
Example 1 Force Function of Time — Finding v(t) and x(t) ★☆☆ Standard
A particle of mass 2 kg moves along the x-axis from rest at the origin. A force F = 6t − 2 N acts on it (t in seconds). Find (a) the velocity at time t, (b) the displacement when t = 3 s.
1
Newton's 2nd Law → dv/dt
2 dv/dt= 6t − 2
dv/dt= 3t − 1
2
Part (a) — integrate for v(t)
v= ∫(3t − 1) dt = 3t²/2 − t + C
v=0 at t=0:C = 0
v(t)= 3t²/2 − t= 1.5t² − t m/s
3
Part (b) — integrate for x(t)
x= ∫v dt = ∫(1.5t² − t) dt = 0.5t³ − 0.5t² + C
x=0 at t=0:C = 0
x(3)= 0.5(27) − 0.5(9) = 13.5 − 4.5 = 9 m
Example 2 Force Function of Position — Using v dv/dx ★★☆ Challenging
A particle of mass 0.5 kg moves along the x-axis. It passes through the origin with velocity 8 m/s. A retarding force F = −2x N acts on it. Find (a) the velocity when x = 3 m, (b) the maximum displacement from the origin.
1
Use mv dv/dx = F(x)
0.5 v dv/dx= −2x
v dv= −4x dx
∫v dv= ∫−4x dx
v²/2= −2x² + C
2
Apply initial condition v = 8 at x = 0
64/2 = 0 + C→ C = 32
v²= 64 − 4x²
3
Part (a): v at x = 3
v²= 64 − 4(9) = 64 − 36 = 28
v= √28 = 2√7 ≈ 5.29 m/s
Part (b): maximum displacement when v = 0
0 = 64 − 4x²→ x² = 16 → x = 4 m
Example 3 Resistance Proportional to v — Terminal Velocity ★★☆ Challenging
A particle of mass 0.4 kg falls from rest under gravity. Air resistance equals 0.8v N, where v is the speed in m/s. Find (a) the terminal velocity, (b) v as a function of t, (c) the distance fallen when v = 3 m/s. (g = 10 m/s²)
1
Part (a) — Terminal velocity
vT= mg/k = (0.4 × 10)/0.8 = 5 m/s
2
Part (b) — v(t) using m dv/dt = mg − kv
0.4 dv/dt= 4 − 0.8v
dv/(5 − v)= 2 dtdividing by 0.4, then rearranging
−ln|5 − v|= 2t + C
v = 0 at t = 0:C = −ln 5
v(t)= 5(1 − e^(−2t))m/s
3
Part (c) — distance when v = 3, using mv dv/dx = mg − kv
0.4v dv/dx= 4 − 0.8v
v dv/(5−v)= 2 dx

Use algebraic division: v/(5−v) = −1 + 5/(5−v)

∫[−1 + 5/(5−v)] dv= 2x
−v − 5 ln|5−v|= 2x + C
v=0, x=0:0 − 5 ln 5 = C → C = −5 ln 5
2x= −v + 5 ln(5/(5−v))
At v=3:2x = −3 + 5 ln(5/2) = −3 + 5(0.916) = 1.58
x= 0.790 m
Example 4 Resistance Proportional to v² — Partial Fractions ★★★ A* Level
A particle of mass m falls from rest. The resistance is mkv² where k is a positive constant. Show that the terminal velocity is V = 1/√k, and find an expression for v in terms of the distance fallen x.
1
Terminal velocity
At vT:mg = mkvT² → vT² = 1/k → vT = 1/√k ∎
2
Use mv dv/dx = mg − mkv²
v dv/dx= g(1 − kv²)dividing by m
v dv/(1−kv²)= g dx
−(1/2k) ln|1−kv²|= gx + C
v=0, x=0:C = 0
ln|1−kv²|= −2gkx
v²= (1/k)(1 − e^(−2gkx))v = √[(1−e^(−2gkx))/k]

As x → ∞, v → 1/√k = vT, confirming the result. ✓

Interactive Motion Plotter

Select a force model and visualise the resulting velocity-time graph, position-time graph, and terminal velocity behaviour in real time.

📈Variable Force Motion Plotter
1.0 kg
0.5
0 m/s

Velocity — Time

Position — Time

—Terminal v (m/s)
—Max Speed (m/s)
—x at t=5s (m)
—ODE Form

Practice Questions

Question 1 — Force Function of Time
A particle of mass 3 kg starts from rest at the origin. A force F = 12 − 6t acts on it for 0 ≤ t ≤ 4 s. Find (a) the velocity at time t, (b) the time when the particle momentarily comes to rest again, (c) the maximum displacement from the origin.
Use m dv/dt = F. Integrate to get v(t). For rest: set v = 0. For max displacement: integrate v = dx/dt from t = 0 to when v = 0 (that gives maximum x since particle then returns).
✓ Solution
3 dv/dt= 12 − 6t → dv/dt = 4 − 2t
(a) v= 4t − t² + C. v=0 at t=0 → C=0. v = 4t − t²
(b) v=0:t(4−t) = 0 → t = 4 s
(c) x= ∫₀⁴(4t−t²)dt = [2t²−t³/3]₀⁴ = 32 − 64/3 = 32/3 ≈ 10.7 m
Question 2 — Force Function of Position
A particle of mass 2 kg travels along the x-axis. At x = 0 its speed is 5 m/s. A force F = 3x + 4 N acts in the direction of motion. Find (a) the velocity when x = 2 m, (b) the kinetic energy gained over this distance.
Use mv dv/dx = F(x). Integrate: ∫2v dv = ∫(3x+4)dx. Apply initial condition. For KE gained, just find ½mv₂² − ½mv₁².
✓ Solution
2v dv/dx= 3x + 4
v²= ¾x² + 2x + C. v=5 at x=0: C=25
(a) v² at x=2= 3 + 4 + 25 = 32 → v = 4√2 ≈ 5.66 m/s
(b) ΔKE= ½(2)(32) − ½(2)(25) = 32 − 25 = 7 J

Check: work done = ∫₀²(3x+4)dx = [3x²/2+4x]₀² = 6+8 = 14. But W = ΔKE requires ΔKE = 7, not 14. Recheck: ΔKE = ½(2)(v₂²−v₁²) = v₂²−v₁² = 32−25 = 7 J. Work = ∫Fdx = 14 J ≠ 7 J. Discrepancy: check mass factor — ∫2v dv = v² so integrating gives v² − 25 = ∫(3x+4)/2 · 2 dx... Let me redo: 2v dv = (3x+4)dx → v² = (3x²/2 + 4x)|₀ˣ + 25. At x=2: v² = 6+8+25 = 39. v = √39 ≈ 6.24 m/s. ΔKE = ½(2)(39−25) = 14 J.

Question 3 — Resistance kv, Terminal Velocity
A parachutist of total mass 80 kg falls from rest. Air resistance is 160v N where v is speed (m/s). Find (a) the terminal velocity, (b) v as a function of t, (c) the time to reach half the terminal velocity. (g = 10 m/s²)
v_T = mg/k. For v(t): separate variables in 80 dv/dt = 800 − 160v. The solution is v = v_T(1 − e^(−kt/m)). For time to reach v_T/2, set the expression equal to v_T/2 and solve for t.
✓ Solution
(a) vT= mg/k = (80×10)/160 = 5 m/s
80 dv/dt= 800 − 160v
dv/(5−v)= 2 dt
(b) v= 5(1 − e^(−2t)) m/s
(c) v = 2.5:2.5 = 5(1−e^(−2t)) → e^(−2t) = 0.5 → t = ln2/2 ≈ 0.347 s
Question 4 — Resistance kv, Find Distance
A bullet of mass 0.01 kg is fired horizontally into a fixed block with initial speed 400 m/s. The resistance force is 200v N (opposing motion). Find (a) v as a function of x, (b) the distance penetrated when the bullet comes to rest, (c) the time for the bullet to come to rest (show it takes infinite time).
For (a): use mv dv/dx = −kv → m dv/dx = −k. This simplifies to a simple integration. For (b): set v=0. For (c): use m dv/dt = −kv, find t → ∞ as v → 0.
✓ Solution
0.01 v dv/dx= −200v → 0.01 dv/dx = −200
(a) v= 400 − 20000x
(b) v=0:x = 400/20000 = 0.02 m = 2 cm
For (c):0.01 dv/dt = −200v → v = 400 e^(−20000t)
v→0only as t → ∞. Particle never truly stops!

This is the classic distinction: with resistance ∝ v, the particle travels a finite distance but takes infinite time — a beautiful and examinable result.

Formula Reference Sheet

Complete reference for Linear Motion Under a Variable Force — Cambridge 9231 P3, Section 3.5.

Forms of Newton's 2nd Law
F = m dv/dtforce as function of t
F = mv dv/dxforce as function of x
a = v dv/dxchain rule form
F = m d²x/dt²second derivative
Resistance ∝ v (linear drag)
v(t) = v₀ e^(−kt/m)decelerating from v₀
v(t) = v_T(1−e^(−kt/m))falling from rest
v_T = mg/kterminal velocity
x_max = mv₀/kfinite stop distance
Resistance ∝ v² (quadratic drag)
v_T = √(mg/k)terminal velocity
Use partial fractions:
∫dv/(a²−v²)= (1/2a)ln|(a+v)/(a−v)|
v² = v_T²(1−e^(−2gkx))/...
Key Integration Results
∫v dv = v²/2 + C
∫dv/v = ln|v| + C
∫dv/(a−v) = −ln|a−v| + C
∫dv/v² = −1/v + C
Work-Energy with Variable Force
W = ∫F(x) dxwork done
W = ΔKEwork-energy theorem
= ½mv₂² − ½mv₁²
Problem Type Selector
F = F(t)→ use dv/dt
F = F(x)→ use v dv/dx
F = F(v)→ separate, partial fractions
Want x(t)→ integrate v(t)
📋 Cambridge Exam Strategy — Variable Force
  • Identify the type first — write down what F depends on before setting up the ODE
  • When resistance is proportional to v, try mv dv/dx = −kv first — the v cancels and integration is trivial
  • For resistance ∝ v², you will need partial fractions — practice the standard split before the exam
  • Always check: finite distance, infinite time for resistance ∝ v; finite time to stop for resistance ∝ v⁰ (constant)
  • State constants of integration clearly and apply initial conditions explicitly — examiners deduct marks if IC is not shown
  • Terminal velocity is a sanity check — confirm v → v_T as t → ∞ in your solution
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