Skip to main content
MathidiaMathematics Academy
Variable acceleration
Lesson 67 of 75
Easy practice

Integrating motion

Objective. Recover velocity and displacement using initial conditions.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Integrating motion reverses differentiation to recover velocity from acceleration and displacement from velocity, using v = ∫a dt and s = ∫v dt, with the constant of integration found from given initial conditions such as the velocity or position at t = 0. This is essential when acceleration varies with time rather than being constant, since suvat formulas no longer apply. For example, if a = 6t and v = 4 when t = 0, then v = 3t² + c, and substituting gives c = 4, so v = 3t² + 4.

Worked example

A particle has acceleration a = 4t − 2 m/s² and initial velocity v = 3 m/s at t = 0. Find an expression for the velocity of the particle at time t.

4 … 2
Illustrated card showing 4 … 2
  1. 1v = ∫a dt = ∫(4t − 2) dt
  2. 2v = 2t² − 2t + c
  3. 3At t = 0, v = 3, so c = 3
  4. 4v = 2t² − 2t + 3

Answer: v = 2t² − 2t + 3

Concept mastery

Based on your past attempts at this lesson — the question types to practise again come first.

Complete a practice set to see your concept breakdown here.

Practice

Practice happens on its own screen, one question at a time. Answers stay hidden until you submit yours, and you can stop and pick up at the same question later.

Start practising

Common mistake. Students omit the constant of integration or forget to use the initial condition to find it, leaving an incomplete or wrong expression. Always add +c when integrating and substitute given initial values to determine it.

Your progress

Tick each step as you finish it. Your exact reading position is saved automatically, so Continue drops you back on the same line.

0 of 5 steps complete

Your place saves automatically as you read.

My dashboard