Connect displacement, velocity and acceleration
Differentiate displacement to find velocity and acceleration, and integrate known rates using initial conditions to recover motion.
Section 14 · Calculus
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is velocity in terms of displacement s and time t?
What you will learn
- Use v = ds/dt and a = dv/dt
- Integrate acceleration or velocity and apply an initial condition
- Keep signed displacement and velocity distinct from distance and speed
Differentiate along the motion chain
For straight-line motion, velocity v = ds/dt and acceleration a = dv/dt = d²s/dt². A negative velocity means motion in the negative direction; a negative acceleration means velocity is decreasing in the chosen sign convention.
Use seconds and the stated length unit consistently. If s is in metres, v is in m/s and a is in m/s².
Integration recovers motion up to constants
If acceleration is known, integrate to find v(t) + C, then use a given velocity at a time to determine C. Integrate v to find s(t) + C and use a position condition.
Initial conditions are essential because a rate alone does not fix the starting velocity or position. Preserve signs when a particle reverses direction.
Pause and explain
If v = 6t − 4, what is acceleration?
Worked example
A particle has displacement s(t) = t³ − 3t² + 2t + 5 metres. Find v(t), a(t), and both at t = 2 seconds.
Show the worked solution
- Differentiate s: v(t) = 3t² − 6t + 2 m/s.
- Differentiate v: a(t) = 6t − 6 m/s².
- At t = 2, v = 2 m/s and a = 6 m/s².
Answer v(t) = 3t² − 6t + 2; a(t) = 6t − 6; at t = 2, v = 2 m/s and a = 6 m/s².
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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 has velocity v(t) = 6t − 4 m/s and s(0) = 3 m. Find s(t).
Give me a hint
Integrate velocity and use the starting displacement to find the constant.
Compare my reasoning
- s(t) = ∫(6t − 4)dt = 3t² − 4t + C.
- Substitute t = 0 and s = 3 to get C = 3.
- State the position function.
s(t) = 3t² − 4t + 3 metres.
Look for these in your work
- Velocity integrated
- Initial condition applied
- Position function stated
Acceleration is a(t) = 4t m/s², with v(0) = −2 m/s and s(0) = 1 m. Find v(t) and s(t).
Give me a hint
Integrate acceleration first, then velocity, using each initial condition.
Compare my reasoning
- v(t) = 2t² + C; v(0) = −2 gives C = −2.
- s(t) = (2/3)t³ − 2t + D; s(0) = 1 gives D = 1.
- Keep units tied to each quantity.
v(t) = 2t² − 2 m/s; s(t) = (2/3)t³ − 2t + 1 m.
Look for these in your work
- First integration constant found
- Second integration constant found
- Both initial conditions checked
A particle starts at s = 0 with velocity v = 5 m/s and constant acceleration −2 m/s². Find its velocity and displacement after 3 seconds.
Give me a hint
Integrate or use the constant-acceleration formulas consistent with the derivative model.
Compare my reasoning
- v(t) = 5 − 2t, so v(3) = −1 m/s.
- s(t) = 5t − t² using s(0) = 0.
- s(3) = 15 − 9 = 6 m.
After 3 s, velocity is −1 m/s and displacement is 6 m.
Look for these in your work
- Acceleration sign retained
- Velocity evaluated
- Displacement integrated from its initial condition
Common mistakes
- Confusing velocity with speed by removing its sign
- Differentiating with respect to the wrong variable
- Integrating without using the initial condition
- Using position as if it were displacement from every starting point
What can you explain now?
If s is measured in metres and t in seconds, what are the units of acceleration?
Compare with the explanation
m/s²
Differentiate position twice with respect to time.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do differentiation and integration connect s, v and a, and why are initial conditions needed?
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.