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.
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:
- $60 \times \frac{1000}{3600} = 16.67 \, \text{m/s}
</li></ul><p><strong>2.DisplacementCalculation:</strong></p><ul><li><strong>AccelerationPhase:</strong>Areaoftriangle=½×base×height=½ × 10 s × 16.67 m/s = 83.35 m</li><li><strong>ConstantSpeedPhase:</strong>Areaofrectangle=length×width=15 s × 16.67 m/s = 250 m</li><li><strong>DecelerationPhase:</strong>Areaoftriangle=½×base×height=½ × 5 s × 16.67 m/s = 41.675 m</li></ul><p><strong>3.TotalDisplacement:</strong>83.35 m + 250 m + 41.675 m = 375.025 m</p><imgsrc="https://tutorchase−production.s3.eu−west−2.amazonaws.com/fce27bf7−d942−47b7−8ce0−977ab1a48c0e−file.png"alt="Velocity−timeGraph"style="width:600px;height:306px"width="600"height="306"><h3>Problem2:Runner′sDisplacement−TimeGraph</h3><p><strong>Problem:</strong>Arunner′sdisplacement−timegraphisastraightlinefromtheoriginwithaslopeof4m/s.Calculatethetotaldisplacementafter5seconds.</p><p><strong>Solution:</strong></p><p><strong>1.TotalDisplacementCalculation:</strong></p><ul><li>Displacement=Velocity×Time=4 m/s × 5 s = 20 m$