Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Rectilinear Acceleration.
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Rectilinear Acceleration.
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Subject: Physics
Class: Senior Secondary 1
Term: 3rd Term
Week: 3
Theme: Interaction Of Matter, Space And Time
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This topic introduces teachers to the fundamental concepts of rectilinear acceleration, building upon students' prior knowledge of speed, velocity, and rectilinear motion. Understanding acceleration is crucial for analysing the motion of objects in a straight line, which forms the basis for more complex kinematics and dynamics in physics. It is vital for Nigerian learners to grasp these concepts as they apply directly to everyday experiences such as vehicle movement, sports, and general observation of motion in their environment.
Learner-Friendly Objectives: Upon completion of this lesson, students will be able to: Describe and provide examples of uniform motion.
acceleration.
Solution: Given: Initial velocity, $u = 0 \, \text{m/s}$ Final velocity, $v = 20 \, \text{m/s}$ Time taken, $t = 10 \, \text{s}$ Formula: $a = \frac{v - u}{t}$ Substitute the values: $a = \frac{20 \, \text{m/s} - 0 \, \text{m/s}}{10 \, \text{s}}$ $a = \frac{20 \, \text{m/s}}{10 \, \text{s}}$ $a = 2 \, \text{m/s}^2$ The acceleration of the BRT bus is $2 \, \text{m/s}^2$.
Example 2: Calculating Negative Acceleration (Deceleration) A Keke Napep travelling at 15 m/s applies its brakes and comes to a stop (final velocity = 0 m/s) in 3 seconds. Calculate its acceleration.
Solution: Given: Initial velocity, $u = 15 \, \text{m/s}$ Final velocity, $v = 0 \, \text{m/s}$ Time taken, $t = 3 \, \text{s}$ Formula: $a = \frac{v - u}{t}$ Substitute the values: $a = \frac{0 \, \text{m/s} - 15 \, \text{m/s}}{3 \, \text{s}}$ $a = \frac{-15 \, \text{m/s}}{3 \, \text{s}}$ $a = -5 \, \text{m/s}^2$ The acceleration of the Keke Napep is $-5 \, \text{m/s}^2$. The negative sign indicates deceleration.
Example 3: Determining Acceleration from a Velocity-Time Graph A velocity-time graph for a motorist in Abuja is shown below. Determine the acceleration of the motorist during the first 4 seconds. ``` ^ Velocity (m/s) | 60| (4s, 60m/s) | / | / | / | / | / | / | / | / | / ---------------------> Time (s) 0 (0s, 0m/s) ``` Solution: To determine acceleration from a v-t graph, calculate the slope of the graph. From the graph, during the first 4 seconds: Initial point: $(t_1, v_1) = (0 \, \text{s}, 0 \, \text{m/s})$ Final point: $(t_2, v_2) = (4 \, \text{s}, 60 \, \text{m/s})$ Slope (acceleration) $a = \frac{v_2 - v_1}{t_2 - t_1}$ $a = \frac{60 \, \text{m/s} - 0 \, \text{m/s}}{4 \, \text{s} - 0 \, \text{s}}$ $a = \frac{60 \, \text{m/s}}{4 \, \text{s}}$ $a = 15 \, \text{m/s}^2$ The acceleration of the motorist during the first 4 seconds is $15 \, \text{m/s}^2$. 2.1 Rectilinear Motion Rectilinear motion refers to the motion of an object along a straight line. This lesson focuses exclusively on motion in one dimension. 2.2 Recap: Speed and Velocity Speed: The rate at which an object covers distance. It is a scalar quantity (only magnitude).
Formula: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$ Unit: metres per second (m/s).
Velocity: The rate at which an object changes its displacement. It is a vector quantity (magnitude and direction).
Formula: $\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}$ Unit: metres per second (m/s). Crucially, a change in velocity can mean a change in speed, a change in direction, or both. For rectilinear motion, a change in velocity primarily refers to a change in speed. 2.3 Uniform Motion (Constant Velocity)
Definition: An object is said to be in uniform motion if it covers equal displacements in equal intervals of time. In this case, its velocity remains constant, both in magnitude and direction.
Implication: For uniform motion, the acceleration of the object is zero.
Graphical Representation: Position-Time Graph: A straight line with a non-zero slope. The slope represents the constant velocity.
Velocity-Time Graph: A horizontal straight line parallel to the time axis. This indicates that velocity is not changing over time. 2.4 Non-Uniform Motion Definition: An object is in non-uniform motion if its velocity changes over time. This change can be an increase in speed, a decrease in speed, or a change in direction. Since this topic focuses on rectilinear motion, the change is primarily in speed.
Implication: Non-uniform motion implies the presence of acceleration. 2.5 Acceleration Definition: Acceleration is the rate of change of velocity. It quantifies how quickly an object's velocity is changing.
Formula: $a = \frac{\text{Change in velocity}}{\text{Time taken}}$ If $u$ is the initial velocity and $v$ is the final velocity over a time interval $t$, then: $a = \frac{v - u}{t}$ Unit: Metres per second squared (m/s2). This unit signifies (m/s) per second.
Vector Quantity: Acceleration is a vector quantity. Its direction is the same as the direction of the change in velocity.
Types of Acceleration: Positive Acceleration: Occurs when the velocity of an object increases in the direction of motion. For example, a car speeding up. Negative Acceleration (Deceleration or Retardation): Occurs when the velocity of an object decreases (it slows down). The acceleration vector is in the opposite direction to the velocity vector. For example, a car braking.
Uniform (Constant)
Acceleration: Occurs when the velocity changes by equal amounts in equal time intervals. The acceleration remains constant throughout the motion. 2.6 Velocity-Time Graphs Velocity-time graphs are powerful tools for analysing motion.
Interpretation of Slope: The slope (gradient) of a velocity-time graph represents the acceleration of the object. Slope = $\frac{\text{Change in velocity}}{\text{Change in time}} = \frac{\Delta v}{\Delta t} = a$ Horizontal Line: Represents zero slope, hence zero acceleration (uniform velocity).
Straight Line with Positive Slope: Represents uniform positive acceleration.
Straight Line with Negative Slope: Represents uniform negative acceleration (deceleration).
Curved Line: Represents non-uniform acceleration (acceleration is changing). This is beyond the scope of SS1 but useful for teacher awareness.
Interpretation of Area: The area under a velocity-time graph represents the displacement of the object. (This is typically introduced after acceleration but is a key feature of v-t graphs). Worked
Examples: Example 1: Calculating Positive Acceleration A BRT bus starts from rest (initial velocity = 0 m/s) at a bus stop in Lagos and reaches a velocity of 20 m/s in 10 seconds. Calculate its acceleration.
Solution:* Given: Initial velocity, $u = 0 \, \text{m/s}$ Final velocity, $v = 20 \, \text{m/s}$ Time taken, $t = 10 \, \text{s}$ Formula: $a = \frac{v - u}{t}$ Substitute the values: $a = \frac{20 \, \text{m/s} - 0 \, \text{m/s}}{10 \, \text{s}}$ $a = \frac{20 \, \text{m/s}}{10 \, \text{s}}$ $a = 2 \, \text{m/s}^2$ The acceleration of the BRT bus is $2 \, \text{m/s}^2$.
Example 2: Calculating Negative Acceleration (Deceleration) A Keke Napep travelling at 15 m/s applies its brakes and comes to a stop (final velocity = 0 m/s) 3.1 Teacher Activities Introduction: Begin by eliciting students' prior knowledge of speed, velocity, and different types of motion. Ask questions about how vehicles start, speed up, and slow down in their daily experiences.
Concept Explanation (Uniform Motion): Clearly define uniform motion using simple examples (e.g., a car moving at a constant speed on a highway without traffic). Illustrate with a simple position-time graph (straight line) and a velocity-time graph (horizontal line). Emphasise zero acceleration for uniform motion.
Concept Explanation (Acceleration): Introduce acceleration as the rate of change of velocity. Derive the formula $a = (v-u)/t$ step-by-step, explaining each variable and its unit. Discuss positive and negative acceleration (deceleration) with practical examples (e.g., car accelerating vs. car braking, a stone falling vs. a ball thrown upwards). Emphasise acceleration as a vector quantity.
Velocity-Time Graphs: Explain how to plot a velocity-time graph. Demonstrate how the slope of a v-t graph represents acceleration using simple straight-line examples (constant velocity, constant positive acceleration, constant negative acceleration). Work through Example 3 on the board, showing how to extract data points and calculate the slope.
Problem Solving: Solve the worked examples (Examples 1 and 2) collaboratively with students, ensuring they understand each step, unit conversion (if needed), and the significance of the sign of acceleration.
Guided Practice: Lead students through the guided practice questions, offering support and clarifying misconceptions. 3.2 Student Activities Discussion and Brainstorming: Students participate in discussions about various types of motion and provide real-life examples of objects speeding up or slowing down.
Note-Taking: Students take notes on definitions, formulae, and key concepts.
Problem Solving: Students attempt to solve numerical problems (e.g., Examples 1 and 2) independently or in pairs after the teacher's demonstration.
Graph Sketching and Interpretation: Students practice sketching simple velocity-time graphs for given scenarios (e.g., constant velocity, constant acceleration) and determine acceleration from provided graphs.
Group Work: Students work in small groups to solve guided practice problems, fostering peer learning.
Questioning: Students ask questions for clarification during concept explanation and problem-solving sessions.
Question 1: A train departing from the railway station in Ibadan accelerates uniformly from 10 m/s to 30 m/s in 8 seconds. Calculate the acceleration of the train.
Solution: Given: Initial velocity, $u = 10 \, \text{m/s}$ Final velocity, $v = 30 \, \text{m/s}$ Time taken, $t = 8 \, \text{s}$ Formula: $a = \frac{v - u}{t}$ Substitute the values: $a = \frac{30 \, \text{m/s} - 10 \, \text{m/s}}{8 \, \text{s}}$ $a = \frac{20 \, \text{m/s}}{8 \, \text{s}}$ $a = 2.5 \, \text{m/s}^2$
Commentary: This question tests the basic calculation of positive acceleration, building confidence in formula application.
Question 2: An Okada rider travelling at 25 m/s suddenly spots a pothole and applies brakes, reducing his speed to 5 m/s in 4 seconds. What is the acceleration of the Okada?
Solution: Given: Initial velocity, $u = 25 \, \text{m/s}$ Final velocity, $v = 5 \, \text{m/s}$ Time taken, $t = 4 \, \text{s}$ Formula: $a = \frac{v - u}{t}$ Substitute the values: $a = \frac{5 \, \text{m/s} - 25 \, \text{m/s}}{4 \, \text{s}}$ $a = \frac{-20 \, \text{m/s}}{4 \, \text{s}}$ $a = -5 \, \text{m/s}^2$
Commentary: This question introduces negative acceleration (deceleration), demonstrating that a decrease in velocity results in a negative value for acceleration.
Question 3: Consider a vehicle's motion represented by the following velocity-time graph segment. ``` ^ Velocity (m/s) | 20| . | / | / | / | / | / | / | / 10| . | / | / |/ +---------------------> Time (s) 0 2 6 ``` Determine the acceleration of the vehicle between $t=2$ s and $t=6$ s.
Solution: From the graph: At $t_1 = 2 \, \text{s}$, $v_1 = 10 \, \text{m/s}$ At $t_2 = 6 \, \text{s}$, $v_2 = 20 \, \text{m/s}$ Formula for slope (acceleration): $a = \frac{v_2 - v_1}{t_2 - t_1}$ Substitute the values: $a = \frac{20 \, \text{m/s} - 10 \, \text{m/s}}{6 \, \text{s} - 2 \, \text{s}}$ $a = \frac{10 \, \text{m/s}}{4 \, \text{s}}$ $a = 2.5 \, \text{m/s}^2$
Commentary: This directly assesses the ability to determine acceleration from a velocity-time graph by calculating the slope of a given segment.
Example 1: Calculating Positive Acceleration
A BRT bus starts from rest (initial velocity = 0 m/s) at a bus stop in Lagos and reaches a velocity of 20 m/s in 10 seconds. Calculate its acceleration.
Solution:
Given:
Initial velocity, $u = 0 \, \text{m/s}$
Final velocity, $v = 20 \, \text{m/s}$
Time taken, $t = 10 \, \text{s}$
Formula: $a = \frac{v - u}{t}$
Substitute the values:
$a = \frac{20 \, \text{m/s} - 0 \, \text{m/s}}{10 \, \text{s}}$
$a = \frac{20 \, \text{m/s}}{10 \, \text{s}}$
$a = 2 \, \text{m/s}^2$
Vehicle Design and Safety (Road Safety Commission): Engineers use the principles of acceleration to design vehicle braking systems and airbags. Understanding deceleration rates is crucial for determining safe stopping distances. In Nigeria, this translates to road safety campaigns (e.g., FRSC) advising drivers to maintain safe distances, especially on busy expressways like the Lagos-Ibadan expressway, where sudden braking requires significant deceleration. Sports Performance (Athletics, Football): Athletes in Nigeria, from sprinters to long jumpers, apply and manage acceleration to improve their performance. A sprinter must achieve maximum acceleration from the starting block. A footballer needs to accelerate quickly to get past a defender or to intercept a pass. Coaches use knowledge of acceleration to train athletes for optimal performance. Traffic Management and Public Transportation: Planning traffic flow and public transport schedules (e.g., BRT system in Lagos) involves considering vehicle acceleration and deceleration capabilities. Efficient movement of large numbers of people requires understanding how quickly buses can accelerate from stops and decelerate for new stops, impacting journey times and congestion.