Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Equation of uniformly accelerated motion
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Equation of uniformly accelerated motion
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Subject: Physics
Class: Senior Secondary 2
Term: 1st Term
Week: 3
Theme: Interaction Of Matter, Space And Time
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`v = u + at` (missing `s`) `s = ut + 1⁄2at2` (missing `v`) `v2 = u2 + 2as` (missing `t`)
Important Note on Signs: In these equations, the vector quantities (displacement, velocity, acceleration) must be assigned positive or negative signs depending on their direction relative to a chosen positive reference direction. For example, if motion to the right is positive, then motion to the left is negative. Upwards could be positive, downwards negative (e.g., for free fall).
A velocity-time graph plots velocity on the y-axis against time on the x-axis. For an object undergoing uniform acceleration, the graph is a straight line. Consider an object starting with an initial velocity `u` at time `t = 0`, and uniformly accelerating to a final velocity `v` at time `t`. [Teacher
Note: It is crucial to sketch this graph clearly on the board as the derivations rely on it.] Sketch of the v-t graph: Draw a Cartesian plane with 'Velocity (m/s)' on the y-axis and 'Time (s)' on the x-axis. Mark a point `A` on the y-axis corresponding to `u` (initial velocity). Draw a straight line from `A` upwards and to the right, ending at point `B`. From point `B`, drop a perpendicular to the x-axis, meeting it at `C`. The value on the x-axis at `C` is `t`. From point `B`, draw a horizontal line to the y-axis, meeting it at `D`. The value on the y-axis at `D` is `v` (final velocity). Draw a horizontal line from `A` to meet the perpendicular line `BC` at `E`. Now, we have a trapezium OABC (where O is the origin) and a rectangle OAEC and a triangle ABE. `OA = u` `OC = t` `BC = v` `OE = t` `CE = OA = u` `BE = BC - CE = v - u` Interpretation of the v-t graph: Slope of the v-t graph = Acceleration (a): Slope `a = (change in velocity) / (change in time)` From the graph, slope of line `AB = (BE) / (AE)` `BE = v - u` `AE = OC = t` Therefore, `a = (v - u) / t` Area under the v-t graph = Displacement (s): The area under the graph OABC represents the total displacement `s`. This area can be calculated as the sum of the area of the rectangle OAEC and the area of the triangle ABE. Area of rectangle `OAEC = length × breadth = OA × OC = u × t = ut` Area of triangle `ABE = 1⁄2 × base × height = 1⁄2 × AE × BE = 1⁄2 × t × (v - u)` Using the v-t graph and its interpretations, we can deduce the three equations: Equation 1: v = u + at (Velocity-Time Relation) From the definition of acceleration or the slope of the v-t graph: `a = (v - u) / t` Multiply both sides by `t`: `at = v - u` Rearrange to solve for `v`: `v = u + at` Significance: This equation relates the final velocity (`v`) to the initial velocity (`u`), acceleration (`a`), and time (`t`). It does not involve displacement.
Equation 2: s = ut + 1⁄2at2 (Displacement-Time Relation)
From the area under the v-t graph: `s = Area of rectangle OAEC + Area of triangle ABE` `s = (OA × OC) + (1⁄2 × AE × BE)` `s = (u × t) + (1⁄2 × t × (v - u))` From Equation 1, we know that `v - u = at`. Substitute this into the equation for `s`: `s = ut + 1⁄2 × t × (at)` `s = ut + 1⁄2at2` Significance: This equation relates displacement (`s`) to initial velocity (`u`), acceleration (`a`), and time (`t`). It does not involve final velocity.
Equation 3: v2 = u2 + 2as (Velocity-Displacement Relation) We can also calculate the area under the v-t graph (displacement `s`) using the formula for the area of a trapezium: `s = 1⁄2 × (sum of parallel sides) × height` `s = 1⁄2 × (OA + BC) × OC` `s = 1⁄2 × (u + v) × t` From Equation 1 (`v = u + at`), we can express `t` as: `t = (v - u) / a` Substitute this expression for `t` into the trapezium area equation: `s = 1⁄2 × (u + v) × ((v - u) / a)` Rearrange the terms: `s = ( (v + u) × (v - u) ) / (2a)` Using the algebraic identity `(x + y)(x - y) = x2 - y2`, we get: `s = (v2 - u2) / (2a)` Multiply both sides by `2a`: `2as = v2 - u2` Rearrange to solve for `v2`: `v2 = u2 + 2as` Significance: This equation relates final velocity (`v`) to initial velocity (`u`), acceleration (`a`), and displacement (`s`). It does not involve time. This section provides the foundational knowledge and detailed derivations required for the topic. Teacher Activities (T.A.): Introduction & Recap (5 mins): T.A. initiates a brief recap of previous concepts: distance, displacement, speed, velocity, and acceleration. T.A. defines uniform acceleration and distinguishes it from non-uniform acceleration. T.A. states the lesson topic and explains its relevance (e.g., in vehicle dynamics, sports).
Introduction to v-t Graph (10 mins): T.A. explains the components of a v-t graph (axes, units). T.A. sketches a v-t graph for an object starting with initial velocity `u` and accelerating uniformly to `v` over time `t`. T.A. labels all relevant points (O, A, B, C, E) and values (`u`, `v`, `t`).
Guiding Derivations (25 mins): Equation 1 (v = u + at): T.A. reminds students that the slope of a v-t graph represents acceleration. T.A. guides students through calculating the slope of line AB from the sketched graph: `a = (change in velocity) / (change in time) = (v - u) / t`. T.A. helps students rearrange to derive `v = u + at`. Equation 2 (s = ut + 1⁄2at2): T.A. reminds students that the area under a v-t graph represents displacement. T.A. guides students to divide the area under the graph OABC into a rectangle OAEC and a triangle ABE. T.A. guides them to calculate the area of the rectangle (`ut`) and the area of the triangle (`1⁄2t(v-u)`). T.A. prompts students to substitute `(v-u) = at` (from Equation 1) into the triangle area, leading to `1⁄2at2`. T.A. helps them sum the areas: `s = ut + 1⁄2at2`. Equation 3 (v2 = u2 + 2as): T.A. guides students to calculate the area under the graph as a trapezium OABC: `s = 1⁄2(u + v)t`. T.A. prompts students to express `t` in terms of `u, v, a` from Equation 1: `t = (v - u) / a`. T.A. assists in substituting `t` into the trapezium area formula and simplifying using `(v+u)(v-u) = v2-u2`, leading to `v2 = u2 + 2as`.
Application and Problem Solving (15 mins): T.A. presents the three worked examples from Section 2.5, explaining each step clearly and encouraging student participation. T.A. emphasizes selecting the appropriate equation based on the known and unknown variables.
Conclusion (5 mins): T.A. summarizes the three equations of motion and their applications. T.A. encourages questions and addresses any misconceptions. Student Activities (S.A.): Active Recall: Students participate in the recap session by answering questions on previous concepts.
Graph Sketching: Students sketch the v-t graph for uniformly accelerated motion in their notebooks as guided by the teacher.
Participatory Derivation: Students actively follow the teacher's guidance, performing calculations and algebraic manipulations in their notebooks to deduce the equations. They should attempt to anticipate the next step.
Problem Solving: Students work through the guided examples with the teacher, asking questions and verifying solutions.
Discussion: Students engage in discussions, clarifying doubts and contributing their understanding.
Example 1: Applying v = u + at A commercial bus (Danfo) at a bus stop in Lagos starts from rest and accelerates uniformly at 2 m/s2 for 5 seconds. What is its final velocity?
Given: Initial velocity, `u = 0 m/s` (starts from rest) Acceleration, `a = 2 m/s2` Time, `t = 5 s` Required: Final velocity, `v` Formula: `v = u + at` Solution: `v = 0 + (2 m/s2 × 5 s)` `v = 10 m/s` Answer: The final velocity of the Danfo bus is 10 m/s.
Example 2: Applying s = ut + 1⁄2at2 A delivery motorcycle (Okada) moving with an initial velocity of 8 m/s accelerates uniformly at 1.5 m/s
2. Calculate the distance it covers in 6 seconds.
Given: Initial velocity, `u = 8 m/s` Acceleration, `a = 1.5 m/s2` Time, `t = 6 s` Required: Displacement, `s` Formula: `s = ut + 1⁄2at2` Solution: `s = (8 m/s × 6 s) + (1⁄2 × 1.5 m/s2 × (6 s)2) ` `s = 48 m + (1⁄2 × 1.5 m/s2 × 36 s2) ` `s = 48 m + (0.75 m/s2 × 36 s2) ` `s = 48 m + 27 m` `s = 75 m` Answer: The motorcycle covers a distance of 75 m.
Example 3: Applying v2 = u2 + 2as A vehicle traveling on the Abuja-Kaduna highway needs to increase its speed. If it accelerates uniformly from 20 m/s to 30 m/s over a distance of 100 m, what is its acceleration?
Given: Initial velocity, `u = 20 m/s` Final velocity, `v = 30 m/s` Displacement, `s = 100 m` Required: Acceleration, `a` Formula: `v2 = u2 + 2as` Solution: `(30 m/s)2 = (20 m/s)2 + (2 × a × 100 m)` `900 m2/s2 = 400 m2/s2 + 200a m` `900 - 400 = 200a` `500 = 200a` `a = 500 / 200` `a = 2.5 m/s2` Answer: The acceleration of the vehicle is 2.5 m/s2.
Vehicle Safety and Road Design (Engineering): Understanding the equations of motion allows engineers to calculate braking distances for vehicles at different speeds and decelerations. This is critical for designing safe roads, determining speed limits, and placing road signs (e.g., "Sharp Bend Ahead", "Reduce Speed"). For instance, calculating the minimum stopping distance for a commercial bus or trailer on Nigerian highways helps in preventing accidents, especially in areas with high traffic or unexpected obstacles. Sports Performance Analysis (Physiology/Coaching): Coaches and sports scientists in Nigeria can use these equations to analyze the performance of athletes (e.g., sprinters, long jumpers). By measuring initial velocity, final velocity, and time, they can calculate acceleration and displacement to optimize training regimes. For example, analyzing a 100m sprinter's acceleration phase can help in refining their starting technique to improve overall race time. Construction and Building (Civil Engineering): Knowledge of uniform acceleration is applied in designing elevators, cranes, and other lifting mechanisms. Engineers calculate the forces required to accelerate loads safely up or down, ensuring that the components can withstand the stresses and that the motion is comfortable for occupants. This is relevant for high-rise buildings being constructed in cities like Lagos or Abuja.