Lesson Notes By Weeks and Term v4 - SHS 2
KINEMATICS
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KINEMATICS
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Subject: Physics
Class: SHS 2
Term: 1st Term
Week: 7
Grade code: 2.1.2.LI.4
Strand code: 1
Sub-strand code: 3
Content standard code: 2.1.2.CS.1
Indicator code: 2.1.2.LI.4
Theme: MECHANICS AND MATTER
Subtheme: KINEMATICS
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This lesson introduces the fundamental equations used to describe the motion of objects moving in a straight line with constant acceleration. Kinematics is the study of motion without considering the forces that cause it. We see it all around us in Ghana: a tro-tro speeding up from a bus stop, a footballer kicking a ball, or even a mango falling from a tree. Understanding these principles is the first step to becoming an engineer, a pilot, an architect, or a sports scientist. By mastering these equations, we can predict an object's future position and speed.
A. Fundamental Concepts (Recap) Before we dive into the equations, let's refresh our memory on some key terms. Displacement (s): The change in position of an object. It is a vector quantity, meaning it has both magnitude (how far) and direction. Measured in metres (m). Initial Velocity (u): The velocity of an object at the start of the time interval (t=0). It is a vector. Measured in metres per second (m/s). If an object "starts from rest", its initial velocity `u = 0`. Final Velocity (v): The velocity of an object at the end of the time interval. It is a vector. Measured in metres per second (m/s). If an object "comes to a stop", its final velocity `v = 0`. Acceleration (a): The rate of change of velocity. It is a vector. Measured in metres per second squared (m/s²). Uniform/Constant Acceleration: This means the velocity changes by the same amount every second. Positive acceleration: Speeding up. Negative acceleration (or deceleration): Slowing down. Time (t): The duration of the motion. It is a scalar. Measured in seconds (s). B. The Four Equations of Uniformly Accelerated Motion These equations are the tools we use to solve problems where acceleration is constant. They are often called the "suvat" equations, named after the variables they contain. `v = u + at` What it means: The final velocity is the initial velocity plus the change in velocity due to acceleration over time. When to use it: Use this equation when you don't know and don't need the displacement (`s`). `s = ut + ½at²` What it means: The displacement is the sum of the distance covered if there were no acceleration (`ut`) and the extra distance covered due to acceleration (`½at²`). When to use it: Use this when you don't know and don't need the final velocity (`v`). `v² = u² + 2as` What it means: This relates the final and initial velocities to the acceleration and displacement without involving time. When to use it: Use this when you don't know and don't need the time (`t`). `s = ½(u + v)t` What it means: The displacement is the average velocity (`(u+v)/2`) multiplied by the time. When to use it: Use this when you don't know and don't need the acceleration (`a`). C. Worked Examples
Example 1: A Sprinter's Race An athlete at the Inter-Colleges competition starts from rest and accelerates uniformly at 2.5 m/s² for the first 6 seconds of a 100m race. What is her final velocity after these 6 seconds? Step 1: Identify the known variables. Starts from rest, so initial velocity `u = 0` m/s. Acceleration `a = 2.5` m/s². Time `t = 6` s. Step 2: Identify the unknown variable. We need to find the final velocity, `v`. Step 3: Choose the correct equation. We have `u`, `a`, `t` and need `v`. The equation that links these four is `v = u + at`. Step 4: Substitute the values and solve. `v = 0 + (2.5 m/s² * 6 s)` `v = 15` m/s Step 5: State the final answer with units. The athlete's final velocity after 6 seconds is 15 m/s.
Example 2: A Car Braking A taxi driver travelling at 72 km/h on the N1 highway sees a traffic jam ahead and applies the brakes. The car decelerates uniformly and comes to a stop in 50 metres. Calculate the deceleration of the car. Step 1: Convert units and identify knowns. Initial velocity `u = 72` km/h. We must convert this to m/s. `u = (72 * 1000 m) / (3600 s) = 20` m/s. Comes to a stop, so final velocity `v = 0` m/s. Displacement `s = 50` m. Step 2: Identify the unknown variable. We need to find the acceleration, `a`. (Since it's slowing down, we expect a negative answer). Step 3: Choose the correct equation. We have `u`, `v`, `s` and need `a`. The time `t` is not involved. The best equation is `v² = u² + 2as`. Step 4: Substitute and solve. `0² = 20² + 2 * a * 50` `0 = 400 + 100a` Rearrange the equation to solve for `a`: `-400 = 100a` `a = -400 / 100` `a = -4` m/s² Step 5: State the final answer with units. The deceleration of the car is 4 m/s² (the negative sign indicates it is deceleration).
Guided Practice (With Solutions)