Chapter 2 Solved Numericals: Kinematics

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Chapter Overview & SLOs

What is covered in these solved numericals? This section provides detailed step-by-step solutions for the numerical problems of Chapter 2, "Kinematics." The exercises focus on the practical application of kinematic equations and graphical analysis. How do we perform unit conversions for velocity and acceleration? The exercises begin with fundamental unit conversions, teaching you how to move between km/h and m/s for velocity, and km/h² and m/s² for acceleration: - Conversion factor for velocity: $1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{1}{3.6} \text{ m/s} \approx 0.278 \text{ m/s}$ - To convert km/h to m/s: Multiply by $\frac{5}{18}$ or divide by 3.6 - To convert m/s to km/h: Multiply by $\frac{18}{5}$ or multiply by 3.6 How do we calculate acceleration and deceleration? You will find step-by-step calculations for acceleration and deceleration using the first equation of motion ($a = \frac{v_f - v_i}{t}$): - Acceleration: Positive change in velocity over time (speeding up) - Deceleration: Negative change in velocity over time (slowing down), also called negative acceleration or retardation - Comparative problems between different vehicles (car vs bicycle, bus vs motorcycle) How do we solve vertical motion under gravity problems? A key focus is on vertical motion under gravity (freely falling bodies): - Objects thrown upward: Initial velocity upward ($v_i$ positive), gravitational acceleration downward ($g = -9.8 \text{ m/s}^2$). At maximum height, final velocity ($v_f$) = 0. Time to reach maximum height: $t = \frac{v_i}{g}$ - Objects falling downward: Initial velocity often 0, gravitational acceleration downward ($g = +9.8 \text{ m/s}^2$) - Using equations of motion adjusted for gravitational acceleration ($g \approx 9.8 \text{ m/s}^2$ or $10 \text{ m/s}^2$ for simplified calculations) How do we calculate distance from speed-time graphs? The numericals emphasize graphical analysis, where you learn to calculate total distance covered by finding the area under a speed-time graph: - Uniform velocity: Area = rectangle (length × width = velocity × time) - Uniform acceleration/deceleration: Area = triangle ($\frac{1}{2} \times \text{base} \times \text{height}$) - Combined motion: Area = sum of areas (rectangle + triangle) for different phases of motion - Distance = area under speed-time graph (regardless of shape) These solutions are strictly designed to help students master the mathematical requirements of the FBISE 2026 annual examination.

• How do we perform unit conversions for velocity and acceleration? Perform unit conversions for velocity between km/h and m/s using the conversion factor 1 km/h = 1/3.6 m/s (multiply by 5/18 to convert km/h to m/s, multiply by 18/5 to convert m/s to km/h), and for acceleration between km/h² and m/s² using appropriate conversion factors.
• How do we calculate and compare acceleration? Calculate and compare the acceleration of various objects (cars, bicycles, buses, motorcycles) using the formula a = (vf - vi)/t, where positive acceleration indicates speeding up and negative acceleration (deceleration) indicates slowing down over a specific time interval.
• How do we solve problems involving freely falling bodies? Solve problems involving freely falling bodies (vertical motion under gravity) by applying the three equations of motion adjusted for gravitational acceleration g = 9.8 m/s² (or 10 m/s²), including calculations for objects thrown upward (g negative, vf = 0 at maximum height) and objects falling downward (g positive, vi often 0).
• How do we calculate distance from speed-time graphs? Interpret speed-time graphs to identify phases of motion (uniform velocity, uniform acceleration, uniform deceleration) and calculate the total distance traveled using the area under the curve method, where distance = area of rectangle for uniform velocity (velocity × time) and area of triangle for acceleration (1/2 × base × height).

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