Functions of time

Objective. Analyse turning points, rest and direction changes.

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

Learn: the key idea

Functions of time analysis uses calculus to study a particle's motion by finding when it is instantaneously at rest (v = 0), when its velocity is a maximum or minimum (a = 0), and when it changes direction (v changes sign). Solving v(t) = 0 locates rest points, substituting into s(t) gives position at that instant, and checking the sign of v either side (or the sign of a at that point) confirms whether direction actually reverses there. For example, a particle with v = (t−2)(t−5) is at rest at t = 2 and t = 5, and its direction is reversed between these since v is negative in between but positive outside.

Worked example

A particle moves with velocity v = t² − 6t + 8 m/s. Find the times at which the particle is instantaneously at rest.

1 $Set v = 0: t² - 6t + 8 = 0$
2 $Factorise: (t - 2)(t - 4) = 0$
3 $Solve: t = 2 or t = 4$
4 $The particle is at rest at t = 2 s and t = 4 s$

Answer: t = 2 s and t = 4 s

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 assume v = 0 always means a change of direction, but if v touches zero and returns to the same sign (e.g. a repeated root), the particle merely pauses without reversing. Always check the sign of v on both sides of the root to confirm direction change.

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