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CIE A-Level Maths Study Notes

3.2.2 Graphical Analysis of Motion

In this section, we delve into the graphical analysis of motion, focusing on displacement-time and velocity-time graphs. These graphs provide a visual representation of motion, helping us understand how objects move over time.

Displacement-Time Graph

The displacement of an object is defined as its distance from the initial point in a specific direction. On a displacement-time graph, displacement is the dependent variable on the y-axis, while time is the independent variable on the x-axis. These graphs are also known as position-time graphs. They can depict three different scenarios:

1. Stationary Object: The graph shows a horizontal line, indicating that the object's position does not change over time. The slope is zero, signifying that the object's velocity is zero.

2. Constant Velocity: The graph is a straight line with a constant positive slope. This indicates that the object moves at a steady rate in a specific direction.

3. Constant Acceleration: The graph shows a curve, where the slope increases with time, indicating that the object's velocity is increasing at a constant rate.

The slope of a displacement-time graph is calculated as follows:

Slope=ΔdΔt\text{Slope} = \frac{\Delta d}{\Delta t}

Key takeaways from the displacement-time graph include:

  • The slope represents velocity.
  • A constant velocity is depicted by a straight line, whereas acceleration is shown by curved lines.
  • A positive slope indicates motion in a positive direction, a negative slope indicates motion in a negative direction, and a zero slope indicates that the object is at rest.
displacement - graph

Image courtesy of BYJUS

Example 1: Displacement-Time Graph Problem

A car starts from rest and moves in a straight line with a constant acceleration of 2m/s22 \, \text{m/s}^2 for 55 seconds. Plot the displacement-time graph and calculate the total displacement of the car during this period.

Solution:

1. Calculate Total Displacement:

Use the formula for displacement with constant acceleration, s=ut+12at2s = ut + \frac{1}{2}at^2, where u=0m/su = 0 \, \text{m/s} (initial velocity), a=2m/s2a = 2 \, \text{m/s}^2(acceleration), and t=5st = 5 \, \text{s} (time).

s=0×5+12×2×52=12×2×25=25ms = 0 \times 5 + \frac{1}{2} \times 2 \times 5^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{m}

2. Plot the Graph:

The displacement-time graph will start at the origin (since displacement is zero at t=0t=0) and curve upwards, reflecting constant acceleration. At t=5t=5, the displacement is 25m25 \, \text{m}.

displacement graph

Velocity-Time Graph

In a velocity-time graph, velocity is the dependent variable on the y-axis, and time is the independent variable on the x-axis. The slope of this graph is calculated as:

Slope=ΔvΔt\text{Slope} = \frac{\Delta v}{\Delta t}

This slope represents acceleration, leading to the following conclusions:

  • A steep slope indicates a rapid change in velocity.
  • A shallow slope indicates a slow change in velocity.
  • A negative slope indicates that the object is decelerating.
  • A positive slope indicates that the object is accelerating.
  • The area under the graph represents the total displacement of the object during the given time period.
velocity - graph

Image courtesy of BYJUS

Example 2: Velocity-Time Graph Problem

A cyclist moves with a constant velocity of 10m/s10 \, \text{m/s} for 44 seconds, then accelerates at 3m/s23 \, \text{m/s}^2 for the next 22 seconds. Plot the velocity-time graph and calculate the total displacement during these 66 seconds.

Solution:

1. Calculate Displacement for Each Phase:

  • Constant Velocity Phase (First (4) seconds):

Displacement is s1=v×t=10×4=40ms_1 = v \times t = 10 \times 4 = 40 \, \text{m}.

  • Acceleration Phase (Next (2) seconds):

Use s2=ut+12at2s_2 = ut + \frac{1}{2}at^2 where u=10m/su = 10 \, \text{m/s}, a=3m/s2a = 3 \, \text{m/s}^2, and t=2st = 2 \, \text{s}.

s2=10×2+12×3×22=20+6=26ms_2 = 10 \times 2 + \frac{1}{2} \times 3 \times 2^2 = 20 + 6 = 26 \, \text{m}

2. Calculate Total Displacement:

stotal=s1+s2=40m+26m=66ms_{total} = s_1 + s_2 = 40 \, \text{m} + 26 \, \text{m} = 66 \, \text{m}

3. Plot the Graph:

The velocity-time graph shows a horizontal line at 10m/s10 \, \text{m/s} for the first 44seconds, then a straight line sloping upwards for the next 22 seconds, reflecting the acceleration.

velocity graph
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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