Linear motion under a variable force
Translate a changing force into a differential equation and solve separable models using time or displacement.
Section Paper 3 · Further Mechanics · Topics 3.1–3.6
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Model linear particle motion under a variable force
- Use a=v dv/dx when position is the independent variable
- Solve the separable differential equations within the 9709 calculus scope
Choose time or displacement as the independent variable
Newton’s second law gives m dv/dt=F. If force depends on position and the problem asks for speed against displacement, use a=dv/dt=(dv/dx)(dx/dt)=v dv/dx.
Write the force direction and sign before separating variables. An opposing resistance has the opposite sign to velocity; an applied force can vary with t, x or v.
A solved differential equation still needs its initial condition
Separate variables and integrate both sides, then use the given initial position or velocity to determine the constant.
Check the model's range and interpretation: a formula may approach a limiting speed or zero without reaching it in finite time. Only separable differential equations are required in this syllabus topic.
Pause and explain
Which habit best checks your work on linear motion under a variable force?
Worked example
A particle of mass m moves with resistance force −cv, where c>0. Find v as a function of time if v(0)=u.
Show the worked solution
- Newton’s law gives m dv/dt=−cv.
- Separate: dv/v=−(c/m)dt, so ln v=−ct/m+C.
- Use v(0)=u to obtain C=ln u.
Answer v=u e^(−ct/m).
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle of mass 2 kg experiences a force F=6−2v N in its direction of motion. Form the differential equation for v(t).
Give me a hint
Apply F=m dv/dt and keep the dependence on v.
Compare my reasoning
- Newton’s law gives 2 dv/dt=6−2v.
- Divide by 2: dv/dt=3−v.
- The equilibrium speed is the value for which acceleration is zero.
dv/dt=3−v; the limiting speed is 3 m s⁻¹.
Look for these in your work
- Mass included
- Force expression preserved
- Equilibrium speed identified
A particle starts at rest and moves under a force F=4x N along a straight line. Its mass is 2 kg. Find v² in terms of x, taking x=0 initially.
Give me a hint
Use a=v dv/dx and integrate with the initial condition.
Compare my reasoning
- 2v dv/dx=4x, so v dv=2x dx.
- Integrating gives v²/2=x²+C.
- At x=0, v=0, so C=0 and v²=2x².
v²=2x², for motion in the positive direction v=√2 x.
Look for these in your work
- Chain-rule acceleration used
- Integration constant found
- Physical velocity branch stated
A resistance law is F=−cv for c>0. Explain why the minus sign is needed when the particle moves in the positive direction.
Give me a hint
Compare force direction with velocity direction.
Compare my reasoning
- Positive velocity means motion in the positive direction.
- Resistance opposes motion, so its force is negative.
- A negative acceleration reduces the speed, consistent with the model.
The force must oppose the velocity; for positive v it is negative, so F=−cv.
Look for these in your work
- Velocity direction specified
- Resistance direction opposed
- Acceleration interpretation linked
Common mistakes
- Using a=dv/dx instead of v dv/dx
- Giving a resistance force the same sign as the motion
- Forgetting the constant fixed by initial data
What can you explain now?
What is a in terms of v and x when velocity depends on position?
Compare with the explanation
v dv/dx
Use the chain rule with v=dx/dt.
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Make it stick
When should you write acceleration as dv/dt, and when is v dv/dx more useful?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.