Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define and apply the concept of acceleration.
• Solve problems using the equations of motion.
• Draw and interpret graphs of motion.
• Differentiate between instantaneous and average velocity.
• Explain the effect of gravity on motion.

Lesson summary

This week, we delve deeper into the world of motion! We're continuing our study of mechanics, specifically focusing on motion in one dimension. This is crucial because understanding how objects move is fundamental to understanding the world around us. From predicting the time it takes a taxi to reach its destination to understanding the trajectory of a cricket ball, motion is everywhere. Understanding the concepts we will cover this week will provide you with a strong foundation for more complex physics topics in the future, and give you the tools to analyze and predict real-world scenarios.

Lesson notes

2.1 Acceleration: Definition: Acceleration is the rate of change of velocity. It tells us how quickly the velocity of an object is changing. Unlike velocity, which tells us how fast an object is moving and in what direction, acceleration tells us how fast that "how fast and in what direction" is changing.

Formula: acceleration (a) = (change in velocity (Δv)) / (change in time (Δt)) or a = (vf - vi) / (tf - ti)

Where: vf = final velocity vi = initial velocity tf = final time ti = initial time Units: meters per second squared (m/s²) Important

Note: Acceleration is a vector quantity, meaning it has both magnitude (size) and direction. A positive acceleration means the velocity is increasing in the positive direction (or decreasing in the negative direction). A negative acceleration means the velocity is decreasing in the positive direction (or increasing in the negative direction). It's essentially deceleration if the acceleration opposes the direction of the motion. 2.2 Equations of Motion (Uniformly Accelerated Motion): These equations are only valid when the acceleration is constant (uniform). This is a crucial point! vf = vi + aΔt (Final velocity equals initial velocity plus acceleration times change in time) Δx = viΔt + ½a(Δt)² (Displacement equals initial velocity times change in time plus one-half acceleration times change in time squared) vf² = vi² + 2aΔx (Final velocity squared equals initial velocity squared plus two times acceleration times displacement) Δx = ½(vi + vf)Δt (Displacement equals one-half the sum of initial and final velocities times change in time)

Where: Δx = displacement (change in position) in meters (m) vi = initial velocity in meters per second (m/s) vf = final velocity in meters per second (m/s) a = acceleration in meters per second squared (m/s²) Δt = change in time in seconds (s) Important Considerations when using the Equations of Motion: Direction: Define a positive direction (e.g., upwards or to the right). Be consistent with the sign of velocity, acceleration, and displacement.

Initial Conditions: Carefully identify the initial velocity (vi) and initial position (often considered to be 0).

Choose the right equation: Select the equation that contains the known variables and the variable you need to find. 2.3 Graphs of Motion: Understanding graphs of motion is vital for visualizing and interpreting motion. Position vs. Time (x vs. t): The slope of the line represents the velocity. A straight line indicates constant velocity. A curved line indicates changing velocity (acceleration). The steeper the slope, the greater the velocity. Velocity vs. Time (v vs. t): The slope of the line represents the acceleration. A horizontal line indicates constant velocity (zero acceleration). A straight, sloped line indicates constant acceleration. The area under the line represents the displacement. Acceleration vs. Time (a vs. t): A horizontal line indicates constant acceleration. The area under the line represents the change in velocity. 2.4 Instantaneous vs.

Average Velocity: Average Velocity: The total displacement divided by the total time interval. v_avg = Δx/Δt Instantaneous Velocity: The velocity of an object at a specific instant in time. This is the velocity you would read on a speedometer at any given moment. On a position vs time graph, instantaneous velocity is the slope of the tangent line at a particular point. 2.5 The Effect of Gravity: Near the surface of the Earth, all objects experience a constant downward acceleration due to gravity. The acceleration due to gravity is denoted by g and has a value of approximately 9.8 m/s². We often round this to 10 m/s² for easier calculations (unless specified otherwise). When dealing with vertical motion (objects thrown upwards or falling downwards), remember to define a positive direction (usually upwards) and assign the correct sign to g (it will typically be negative if upwards is positive).

Worked example

Example 1: A taxi starts from rest and accelerates uniformly at 2 m/s² for 5 seconds. What is its final velocity?

Step 1: Identify Knowns and Unknowns:

vi = 0 m/s (starts from rest)

a = 2 m/s²

Δt = 5 s

vf = ? (unknown)

Step 2: Choose the Appropriate Equation:

vf = vi + aΔt

Step 3: Substitute and Solve:

vf = 0 + (2 m/s²)(5 s)

vf = 10 m/s

Example 2: A soccer ball is kicked vertically upwards with an initial velocity of 15 m/s. Neglecting air resistance, what is the maximum height reached by the ball? (Use g = 9.8 m/s²)

Step 1: Identify Knowns and Unknowns:

vi = 15 m/s (upwards – define upwards as positive)

a = -9.8 m/s² (downwards – acceleration due to gravity)

vf = 0 m/s (at maximum height, the ball momentarily stops)

Δx = ? (unknown – displacement, which is the maximum height)

Step 2: Choose the Appropriate Equation:

vf² = vi² + 2aΔx

Step 3: Substitute and Solve:

0² = (15 m/s)² + 2(-9.8 m/s²)Δx

0 = 225 m²/s² - 19.6 m/s² Δx

19.6 m/s² Δx = 225 m²/s²

Δx = 225 m²/s² / 19.6 m/s²

Δx = 11.48 m

Example 3: A train travels at a constant velocity of 20 m/s for 10 seconds, then decelerates uniformly at a rate of 1 m/s² for 5 seconds. What is the total distance traveled by the train?

Step 1: Break the problem into two parts:

Part 1: Constant velocity

vi = 20 m/s

a = 0 m/s²

Δt = 10 s

Δx = viΔt = (20 m/s)(10 s) = 200 m

Part 2: Deceleration

vi = 20 m/s

a = -1 m/s²

Δt = 5 s

Δx = viΔt + ½a(Δt)² = (20 m/s)(5 s) + ½(-1 m/s²)(5 s)² = 100 m - 12.5 m = 87.5 m

Step 2: Calculate the total distance:

Total distance = 200 m + 87.5 m = 287.5 m

Guided Practice (With Solutions)

Question 1: A bakkie accelerates from 10 m/s to 30 m/s in 4 seconds. What is its acceleration?

Solution:

Knowns: vi = 10 m/s, vf = 30 m/s, Δt = 4 s

Unknown: a = ?

Equation: a = (vf - vi) / Δt

a = (30 m/s - 10 m/s) / 4 s = 20 m/s / 4 s = 5 m/s²

Answer: The acceleration is 5 m/s².

Commentary: This is a direct application of the acceleration formula. Make sure the units are consistent.