4PH1

Movement & Position

Forces & Motion · 1 question type

Practise

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 12% of your exam marks.

stable
High
Stable12%

Distance-time graphs, speed calculations and velocity appear in nearly every series.

What the area means

  • The area enclosed between a velocity-time line and the time axis equals the distance that has been covered during the interval (or, equivalently, the magnitude of the displacement if the motion runs in a single direction)
  • This is true because area = velocity × time on every thin vertical strip under the line, and velocity × time = distance moved in that strip

Splitting a multi-stage motion into shapes

  • A constant-velocity section is a horizontal line, so the area under it is a rectangle:

rectangle area = base × height

  • A section of constant acceleration or deceleration is a straight slope, so the area under it (down to the time axis) is a triangle:

triangle area = ½ × base × height

  • A section that combines a non-zero starting velocity with an acceleration gives a trapezium; split it into a rectangle and a triangle and add the two
  • The total distance for a multi-stage motion is found by adding together the area of every enclosed region one stage at a time
Velocity-time graph with the region under the line divided into a triangle beneath the sloping part labelled ½ × base × height and a rectangle beneath the flat part labelled base × height, with a note that distance = area
Source: Area under velocity graph by Save My Exams
Worked example

Finding distance from the area under a velocity–time graph

A velocity–time graph shows an object accelerating uniformly from rest (0 m/s) to 6.0 m/s over 4.0 s, then the recording stops.

Solution:

  • The region under the line is a triangle (straight slope from zero)
  • Area of triangle = ½ × base × height = ½ × 4.0 × 6.0
  • Distance = 12 m