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CIE A-Level Maths Study Notes

3.2.1 Fundamental Kinematic Concepts

This comprehensive guide focuses on the fundamental principles of kinematics, with an emphasis on graphical representations of motion and the intricate relationship between kinematic quantities. The objective is to equip students with the skills necessary for precise analysis and application of these concepts in various contexts.

Kinematic Motion

Image courtesy of Byjus

Displacement, Velocity, and Acceleration:

  • Kinematics: Studies motion without focusing on forces.
  • Displacement: Change in position (direction matters, not the same as path length).
  • Velocity: How fast and in which direction something moves (change in displacement).
  • Acceleration: How quickly something speeds up, slows down, or changes direction (change in velocity).

Displacement-Time Graphs

  • Shows how position changes over time.
  • Slope = Velocity (steeper slope = higher velocity).
  • Horizontal line = No movement (zero velocity).

Velocity-Time Graphs

  • Shows how velocity changes over time.
  • Slope = Acceleration.
  • Area under graph = Total displacement.

Kinematic Calculations

  • Displacement from Velocity-Time Graphs: Calculate area under graph (shapes can vary).
  • Velocity from Displacement-Time Graphs: Slope gives velocity (straight line = constant velocity, curved line = changing velocity, may need calculus for exact value).

Example Questions

Problem 1: Car's Motion Analysis

Problem: A car accelerates uniformly from rest to 60 km/h in 10 seconds, maintains 60 km/h for 15 seconds, then decelerates to rest in 5 seconds. Sketch the velocity-time graph and calculate the total displacement.

Solution:

1. Velocity Conversion:

  • $60 \times \frac{1000}{3600} = 16.67 \, \text{m/s} </li></ul><p><strong>2.DisplacementCalculation:</strong></p><ul><li><strong>AccelerationPhase:</strong>Areaoftriangle=</li></ul><p><strong>2. Displacement Calculation:</strong></p><ul><li><strong>Acceleration Phase:</strong> Area of triangle = ½×base×height= × base × height = ½ × 10 s × 16.67 m/s = 83.35 m</li><li><strong>ConstantSpeedPhase:</strong>Areaofrectangle=length×width=</li><li><strong>Constant Speed Phase:</strong> Area of rectangle = length × width = 15 s × 16.67 m/s = 250 m</li><li><strong>DecelerationPhase:</strong>Areaoftriangle=½×base×height=</li><li><strong>Deceleration Phase:</strong> Area of triangle = ½ × base × height = ½ × 5 s × 16.67 m/s = 41.675 m</li></ul><p><strong>3.TotalDisplacement:</strong></li></ul><p><strong>3. Total Displacement:</strong> 83.35 m + 250 m + 41.675 m = 375.025 m</p><imgsrc="https://tutorchaseproduction.s3.euwest2.amazonaws.com/fce27bf7d94247b78ce0977ab1a48c0efile.png"alt="VelocitytimeGraph"style="width:600px;cursor:pointer"><h3>Problem2:RunnersDisplacementTimeGraph</h3><p><strong>Problem:</strong>Arunnersdisplacementtimegraphisastraightlinefromtheoriginwithaslopeof4m/s.Calculatethetotaldisplacementafter5seconds.</p><p><strong>Solution:</strong></p><p><strong>1.TotalDisplacementCalculation:</strong></p><ul><li>Displacement=Velocity×Time=</p><img src="https://tutorchase-production.s3.eu-west-2.amazonaws.com/fce27bf7-d942-47b7-8ce0-977ab1a48c0e-file.png" alt="Velocity-time Graph" style="width: 600px; cursor: pointer"><h3>Problem 2: Runner's Displacement-Time Graph</h3><p><strong>Problem:</strong> A runner's displacement-time graph is a straight line from the origin with a slope of 4 m/s. Calculate the total displacement after 5 seconds.</p><p><strong>Solution:</strong></p><p><strong>1. Total Displacement Calculation:</strong></p><ul><li>Displacement = Velocity × Time = 4 m/s × 5 s = 20 m$
 Displacement-time Graph
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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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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