Linear motion under a variable force

CIE A-Level Further Mathematics· 25 min read

1. Force as a function of time F(t)★★☆☆☆⏱ 8 min

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For forces that change explicitly with time, Newton's second law $F = ma$ can be rewritten as $a = F (t) / m$ . Since acceleration is the rate of change of velocity, you can integrate directly with respect to time to find velocity, then integrate velocity again to find displacement.

📘 Definition

Impulse of a time-varying force

$J = \int_{t_{1}}^{t_{2}} F (t) d t$

The total change in momentum of a particle acted on by $F (t)$ over the interval from $t_{1}$ to $t_{2}$

Example:

A force $F (t) = 2 t$ N acting for 3 seconds delivers total impulse of 9 Ns

📐 Worked Example

A particle of mass 0.5 kg moves along a straight line, acted on by force $F (t) = 6 t + 2$ N, where $t$ is time in seconds. Given initial velocity $u = 2$ m s⁻¹ at $t = 0$ , find the velocity at $t = 3$ .

1
Apply Newton's second law to isolate acceleration:
$0.5a = 6t + 2 \implies a = 12t + 4$
2
Use $a = d v / d t$ and separate variables for integration:
$\int dv = \int (12t + 4) dt$
3
Evaluate the indefinite integral and apply initial condition $v = 2$ at $t = 0$ :
$v = 6t^2 + 4t + 2$
4
Substitute $t = 3$ to find final velocity:
$v = 6(9) + 4(3) + 2 = 68 \text{ m s}^{-1}$

✓ Quick check

Test your understanding of F(t) motion

A 2kg particle has force $F (t) = 4 t$ N acting on it, starting from rest. What is its velocity at $t = 2$ ?
- A) 2 m/s
- B) 4 m/s
- C) 8 m/s
- D) 16 m/s
Reveal answer
B —
Acceleration $a = F / m = 2 t$ , integrating from $t = 0$ to $t = 2$ gives $v = t^{2} = 4$ m/s

Exam tip:

CIE often asks for maximum velocity from a time-dependent force that changes direction: set $a = d v / d t = 0$ to find the critical time value.

2. Force as a function of displacement F(x)★★★☆☆⏱ 9 min

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When force depends only on the position of the particle, using $a = d v / d t$ creates an equation with no explicit time term. Instead, use the chain rule identity $a = v \frac{d v}{d x}$ to write the differential equation in terms of $v$ and $x$ only.

🔬 Derivation

Goal:

Derive the work-energy principle for a displacement-dependent force

Starting from:

Newton's second law $F (x) = ma$

1
Substitute the chain rule acceleration identity:
$F(x) = m v \frac{dv}{dx}$
2
Rearrange to separate variables:
$F(x) dx = m v dv$
3
Integrate both sides between initial state $(x_{1}, v_{1})$ and final state $(x_{2}, v_{2})$ :
$\int_{x_1}^{x_2} F(x) dx = \int_{v_1}^{v_2} mv dv$

Result:

The integral of $F (x)$ over displacement equals the change in kinetic energy of the particle, no time calculations required.

📐 Worked Example

A particle of mass 1 kg moves along the x-axis, acted on by force $F (x) = 12 - 4 x$ N, where $x$ is displacement in metres. It starts from rest at $x = 0$ . Find its speed when $x = 3$ .

1
Substitute into Newton's second law with $a = v d v / d x$ :
$12 - 4x = v \frac{dv}{dx}$
2
Separate variables and set limits for integration:
$\int_0^3 (12 - 4x) dx = \int_0^v v dv$
3
Evaluate the left hand (force) integral:
$\left[12x - 2x^2\right]_0^3 = 36 - 18 = 18$
4
Evaluate the right hand (kinetic energy) integral:
$\left[\frac{1}{2}v^2\right]_0^v = \frac{1}{2}v^2$
5
Equate and solve for v:
$18 = \frac{1}{2}v^2 \implies v = 6 \text{ m s}^{-1}$

3. Force as a function of velocity F(v)★★★★☆⏱ 10 min

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This case is most common for motion with velocity-dependent resistance, such as air drag proportional to $v$ or $v^{2}$ . You select the acceleration identity based on what the question asks for: use $a = d v / d t$ to find time, or $a = v d v / d x$ to find displacement.

📐 Worked Example

A particle of mass 2 kg moves in a straight line, resisted by force $3 v$ N, where $v$ is velocity. No other forces act. Initial speed is 10 m s⁻¹. Find the time taken for speed to drop to 2 m s⁻¹.

1
Apply Newton's second law, with negative sign for resistive force opposing motion:
$-3v = 2 \frac{dv}{dt}$
2
Rearrange to form a separable differential equation:
$\int dt = -\frac{2}{3} \int \frac{dv}{v}$
3
Integrate and apply initial condition $t = 0, v = 10$ to find constant C:
$t = -\frac{2}{3} \ln v + \frac{2}{3} \ln 10$
4
Substitute $v = 2$ and simplify:
$t = \frac{2}{3} \left(\ln 10 - \ln 2\right) = \frac{2}{3} \ln 5 \approx 1.07 \text{ s}$

✓ Quick check

Quick check for F(v) motion

For the same particle above, what substitution would you use to find the distance travelled before speed drops to 2 m/s?
- A) a = dv/dt
- B) a = v dv/dx
- C) a = d²x/dt²
- D) F = ma directly
Reveal answer
B —
Since you are solving for displacement as a function of velocity, the v dv/dx identity is the most efficient approach.

4. Common Pitfalls

Wrong move:

Using a = dv/dt for F(x) problems and trying to integrate over time

Why:

F(x) has no explicit time dependence, so you cannot relate x to t without extra steps, leading to an unsolvable equation.

Correct move:

Always use the a = v dv/dx substitution for any force that is a function of displacement.

Wrong move:

Forgetting the negative sign for resistive forces in F(v) problems

Why:

This produces an exponentially increasing velocity instead of decreasing, a physically impossible result that loses most method marks.

Correct move:

Define the positive direction as the direction of motion, so all resistive forces take a negative sign.

Wrong move:

Applying SUVAT constant acceleration equations to variable force problems

Why:

SUVAT formulas only hold when acceleration is fixed, which is never true for any variable force scenario.

Correct move:

Solve the relevant differential equation via integration for the given force type.

Wrong move:

Using mass in grams instead of kilograms in F=ma

Why:

This gives acceleration values 1000 times larger than the correct answer, even if all integration steps are perfect.

Correct move:

Convert all mass values to kg before writing any equations.

Wrong move:

Skipping the constant of integration and not applying initial conditions

Why:

Your general solution will have an unknown constant, so you cannot get the numerical value the exam question asks for.

Correct move:

Immediately substitute the given initial value (e.g. v=u at t=0) right after integrating to find C.

5. Quick Reference Cheatsheet

Force type	Acceleration identity to use	Integral formula
F(t)	a = dv/dt	$v = \frac{1}{m} \int F (t) d t + C$
F(x)	a = v dv/dx	$\frac{1}{2} m v^{2} = \int F (x) d x + C$
F(v)	a = dv/dt (for time)	$t = m \int \frac{d v}{F ( v )} + C$
F(v)	a = v dv/dx (for displacement)	$x = m \int \frac{v d v}{F ( v )} + C$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · 4
F(x) motion with work energy
2023 · 3
Velocity-dependent resistance motion
2022 · 4
Impulse from time-varying force

What's Next

Mastering linear motion under variable force is critical to tackling subsequent Further Mechanics topics, including motion in a circle with variable speed, the derivation of simple harmonic motion, and vertical motion under gravity with air resistance. This skill also builds a foundation for extending the work-energy principle to 2D systems and rigid body dynamics, which appear in 15+ mark extended response questions on CIE A2 papers. Use the linked modules to reinforce your understanding and build problem-solving speed ahead of the exam.