Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define and apply the equations of motion for uniformly accelerated motion.
• Solve problems for displacement, velocity, acceleration, and time.
• Interpret position, velocity, and acceleration vs. time graphs.
• Apply these concepts to real-world scenarios in South Africa.

Lesson summary

This week, we delve deeper into the fascinating world of motion, specifically focusing on motion in one dimension. This is a fundamental concept in physics and forms the basis for understanding more complex movements. Understanding motion allows us to analyze everything from the speed of a taxi traveling between Johannesburg and Pretoria to the trajectory of a soccer ball kicked during a PSL game. In South Africa, where transportation and sports play such significant roles, understanding motion is crucial for safe road usage, efficient logistics, and improved athletic performance.

Lesson notes

2.1 Uniformly Accelerated Motion Uniformly accelerated motion (also known as constant acceleration) occurs when an object's velocity changes at a constant rate. This means the acceleration remains the same throughout the motion. This is a simplified model, but it's a useful approximation for many real-world situations. 2.2 Equations of Motion (SUVAT Equations) These equations relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t) when the acceleration is constant. It's a good idea to remember the mnemonic SUVAT for these variables.

Equation 1: v = u + at (Final velocity = Initial velocity + (Acceleration x Time)) This equation relates final velocity, initial velocity, acceleration, and time. It's useful when you don't know the displacement.

Equation 2: s = ut + ½at² (Displacement = (Initial velocity x Time) + ½ (Acceleration x Time squared)) This equation relates displacement, initial velocity, acceleration, and time. It's useful when you don't know the final velocity.

Equation 3: v² = u² + 2as (Final velocity squared = Initial velocity squared + 2 x Acceleration x Displacement) This equation relates final velocity, initial velocity, acceleration, and displacement. It's useful when you don't know the time.

Equation 4: s = ½(u + v)t (Displacement = ½ (Initial velocity + Final velocity) x Time) This equation relates displacement, initial velocity, final velocity and time. It's useful when you don't know the acceleration.

Important Notes: These equations only apply when the acceleration is constant. Remember to use consistent units. In physics, we generally prefer meters (m) for displacement, meters per second (m/s) for velocity, and meters per second squared (m/s²) for acceleration, and seconds (s) for time. Direction is important! Choose a positive direction and stick to it. If an object is moving in the opposite direction to your chosen positive direction, its velocity or displacement should be negative. 2.3 Worked Examples Example 1: A taxi traveling at 20 m/s accelerates at a constant rate of 1.5 m/s² for 8 seconds. Calculate the final velocity of the taxi.

Step 1: Identify the known variables. u = 20 m/s (initial velocity) a = 1.5 m/s² (acceleration) t = 8 s (time) v = ? (final velocity – what we want to find)

Step 2: Choose the appropriate equation. Since we know u, a, and t, and we want to find v, we use equation 1: v = u + at Step 3: Substitute the values into the equation and solve. v = 20 + (1.5 x 8) v = 20 + 12 v = 32 m/s Answer: The final velocity of the taxi is 32 m/s.

Example 2: A soccer ball is kicked vertically upwards with an initial velocity of 15 m/s. Assuming the acceleration due to gravity is -9.8 m/s² (negative because it acts downwards), calculate the maximum height reached by the ball.

Step 1: Identify the known variables. u = 15 m/s (initial velocity) a = -9.8 m/s² (acceleration due to gravity) v = 0 m/s (final velocity at the maximum height – the ball momentarily stops before falling back down) s = ? (displacement – what we want to find)

Step 2: Choose the appropriate equation. Since we know u, a, and v, and we want to find s, we use equation 3: v² = u² + 2as Step 3: Substitute the values into the equation and solve. 0² = 15² + 2 x (-9.8) x s 0 = 225 - 19.6s 19.6s = 225 s = 225 / 19.6 s ≈ 11.48 m Answer: The maximum height reached by the soccer ball is approximately 11.48 meters.

Example 3: A train starts from rest (u = 0 m/s) and accelerates uniformly at 0.5 m/s² for 30 seconds. Calculate the distance the train travels during this time.

Step 1: Identify the known variables. u = 0 m/s (initial velocity) a = 0.5 m/s² (acceleration) t = 30 s (time) s = ? (displacement – what we want to find)

Step 2: Choose the appropriate equation. Since we know u, a, and t, and we want to find s, we use equation 2: s = ut + ½at² Step 3: Substitute the values into the equation and solve. s = (0 x 30) + ½ x 0.5 x (30)² s = 0 + 0.25 x 900 s = 225 m Answer: The train travels a distance of 225 meters. 2.4 Interpreting Graphs of Motion Position vs.

Time Graph: The gradient (slope) of a position vs. time graph represents the velocity of the object. A straight line indicates constant velocity, while a curved line indicates changing velocity (acceleration). Velocity vs.

Time Graph: The gradient (slope) of a velocity vs. time graph represents the acceleration of the object. A horizontal line indicates constant velocity (zero acceleration), while a sloping line indicates constant acceleration. The area under the velocity vs. time graph represents the displacement of the object. Acceleration vs.

Time Graph: For uniformly accelerated motion, the acceleration vs. time graph will be a horizontal line because the acceleration is constant. Guided Practice (With Solutions)

Question 1: A car accelerates from 10 m/s to 25 m/s in 5 seconds. Assuming constant acceleration, calculate the acceleration of the car.

Solution: u = 10 m/s v = 25 m/s t = 5 s a = ?