# Linear motion under a variable force

> CIE A-Level Further Mathematics · CIE A-Level Further Maths 9231
> Source: https://www.owlsprep.com/study/cie-9231-u3-linear-motion-under-a-variable/

This module solves linear motion problems where force varies with time, velocity or displacement, including the v dv/dx substitution, work-energy applications, and exam-standard problem workflows.

**Prerequisites:** [Constant acceleration linear kinematics](https://www.owlsprep.com/study/cie-9702-mechanics-constant-acceleration/); [Separable first-order differential equations](https://www.owlsprep.com/study/cie-9231-fp1-separable-differential-equations/)

## Learning objectives

- Relate variable force to acceleration as a function of displacement, velocity or time
- Solve first-order differential equations for motion using the v dv/dx substitution
- Apply the work-energy principle to variable force linear motion problems
- Calculate impulse and change in momentum from time-dependent variable forces

## Force as a function of time F(t)

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.

**Impulse of a time-varying force** — The total change in momentum of a particle acted on by $F(t)$ over the interval from $t_1$ to $t_2$

*Notation:* $J = \int_{t_1}^{t_2} F(t) dt$

*Example:* A force $F(t) = 2t$ 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 $F(t) = 6t + 2$ 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 $a = dv/dt$ 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:

   $$v = 6(9) + 4(3) + 2 = 68 \text{ m s}^{-1}$$

**Check your understanding**

Test your understanding of F(t) motion

1. A 2kg particle has force $F(t) = 4t$ 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

   *Why:* Acceleration $a = F/m = 2t$, 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 $a = dv/dt = 0$ to find the critical time value.

*Calculator:* allowed

## Force as a function of displacement F(x)

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

**Derivation:** 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 $(x_2, v_2)$:

   $$\int_{x_1}^{x_2} F(x) dx = \int_{v_1}^{v_2} mv dv$$

*Conclusion:* 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 $F(x) = 12 - 4x$ 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 $a = v dv/dx$:

   $$12 - 4x = v \frac{dv}{dx}$$
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:

   $$18 = \frac{1}{2}v^2 \implies v = 6 \text{ m s}^{-1}$$

**Exam command terms**

CIE uses specific command terms for this topic that define your required working:

- **Show that the speed of the particle at position x is...** — You must derive the full differential equation and integrate step-by-step, you cannot just quote the work-energy formula for full marks

*Calculator:* allowed

## Force as a function of velocity F(v)

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 $a = dv/dt$ to find time, or $a = v dv/dx$ to find displacement.

> **Substitution Rule Mnemonic**
>
> Want Time? Use a = dv/dt. Want distance (x)? Use a = v dv/dx

**Worked example:** A particle of mass 2 kg moves in a straight line, resisted by force $3v$ 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:

   $$t = \frac{2}{3} \left(\ln 10 - \ln 2\right) = \frac{2}{3} \ln 5 \approx 1.07 \text{ s}$$

**Check your understanding**

Quick check for F(v) motion

1. 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

   *Why:* Since you are solving for displacement as a function of velocity, the v dv/dx identity is the most efficient approach.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Using a = dv/dt for F(x) problems and trying to integrate over time
  - Why it fails: F(x) has no explicit time dependence, so you cannot relate x to t without extra steps, leading to an unsolvable equation.
  - Correct: Always use the a = v dv/dx substitution for any force that is a function of displacement.
- **Wrong:** Forgetting the negative sign for resistive forces in F(v) problems
  - Why it fails: This produces an exponentially increasing velocity instead of decreasing, a physically impossible result that loses most method marks.
  - Correct: Define the positive direction as the direction of motion, so all resistive forces take a negative sign.
- **Wrong:** Applying SUVAT constant acceleration equations to variable force problems
  - Why it fails: SUVAT formulas only hold when acceleration is fixed, which is never true for any variable force scenario.
  - Correct: Solve the relevant differential equation via integration for the given force type.
- **Wrong:** Using mass in grams instead of kilograms in F=ma
  - Why it fails: This gives acceleration values 1000 times larger than the correct answer, even if all integration steps are perfect.
  - Correct: Convert all mass values to kg before writing any equations.
- **Wrong:** Skipping the constant of integration and not applying initial conditions
  - Why it fails: Your general solution will have an unknown constant, so you cannot get the numerical value the exam question asks for.
  - Correct: Immediately substitute the given initial value (e.g. v=u at t=0) right after integrating to find C.

## Cheatsheet

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

## 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.

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