A Detailed Lesson Plan In Science for Grade 9 Students Prepared by: Rolalen Joyce C. Paiton I. Objective General Objective: Describe the Uniformly Accelerated Motion (UAM) qualitatively and quantitatively. Specific Objective: Solve problems related to motion with uniform acceleration in horizontal dimension. II. Subject Matter A. Uniformly Accelerated Motion B. References: Science Learner’s Module Science Teacher’s Module University Physics by Young and Freedman C. Instructional Materials: Pictures, Visual Aids, PowerPoint, Videos D. Value Integration: Honesty E. Science Concepts: The simplest kind of accelerated motion is straight-line motion with constant acceleration. In this case, the velocity changes at the same rate throughout the motion. Uniformly Accelerated Motion occurs when a velocity increases by exactly the same amount during each time interval. Skills processed: Observing, Determining III. Procedure Teacher’s Activity Students’ Activity A. Preliminary Activities “Good morning class! How was “Good morning ma’am!” your weekend?” “As a part of my classroom management I want to assign an officer of the day. It would be alphabetical, so today the officer of the day is Brendon and for tomorrow the OD is Jhon Hardy. The OD is in charge for checking the “Okay ma’am.” attendance, he/she will be the one to lead the prayer and setup theTV. Also I’ll be giving points to those students who will participate in the class discussion.” (students pray) “Please stand everyone for the prayer and Brendon as the officers of the day please lead “None ma’am.” the prayer.” (students perform an energizer “Is anyone absent today?” activity) “It’s time for an energizer.” “We discussed about Uniformly Accelerated Motion.” B. Review Lesson “What have we discussed last meeting”? “Uniformly Accelerated Motion.is simply motion constant a with “Very good.” acceleration. “What is Uniformly Accelerated velocity changes at the same rate Motion?” throughout the motion.” In “Very good. You really listened to our discussion last meeting.” C. Lesson Proper (students answer) 1. Motivation “If you want to determine how far a passing vehicle would travel in a given amount of time, how will you do it?” “The best way to solve a problem is usually this case the determined by the information that is available to you.” 2. Presentation “Our topic Equations for for today is Uniformly Accelerated Motion” (Presentation of objectives) Objectives: The learners should be able to: • Solve problems related to motion with acceleration in uniform horizontal (students read the objectives) dimension. “Please read the objective.” “Thank you.” 3. Discussion “To be able to solve problems related to motion with uniform acceleration, in which the velocity may change but the acceleration is constant, we need to derive algebraic equations that describe this type of motion.” π£π − π£π π₯π£ π£π΄π£π = 2 = 2 “What is the formula for average velocity?” “Thank you.” “How about the formula for average acceleration?” ππ΄π£π = π₯π£ π₯π‘ = π£π − π£π π₯π‘ “Very good.” “Consider the defining equation for acceleration: π£π − π£π ππ΄π£π = π₯π‘ “If we rearrange this equation to solve for final velocity π£π, we get Equation 1: π£π = π£π + ππ₯π‘ (Equation 1)” “You may use Equation 1 in problems that do not directly involve displacement.” Substitute Equation 1 to π£π + π£π π£π΄π£π = 2 1 π£π΄π£π = (π£π + π£π) 2 1 π£π΄π£π = (π£π + π£π + ππ₯π‘) 2 1 π£π΄π£π = (2π£π + ππ₯π‘) 2 1 π£π΄π£π = (π£π + ππ₯π‘) 2 π₯π₯ Then substitute to π£π΄π£π = π‘ 1 We get π₯π₯ and = π£ π + ππ₯π‘ π‘ 2 multiply it by t. π₯π₯=π£ π‘+1 ππ₯π‘2 (Equation 2) π 2 “This is Equation 2, which allows you to determine the displacement of an object moving with uniform acceleration given a value for acceleration rather than a final velocity.” π£π−π£π π₯π‘= π ,π£ π΄π£π = π£π+π£π 2 π₯π₯=vt π₯π₯=( π£π+π£π π£π−π£π 2 π )( π₯π₯ = ( π£π2 3) ) π£π2 − π£π2 ) 2π 2ππ₯π₯ = π£π2 − π£π2 = π£π2 + 2ππ₯π₯(Equation π£π΄π£π = π£π+π£π 2 , π£π΄π£π = π£π + π£π 2 π₯π‘( π₯π₯ = ( π₯π₯ π‘ π£π+π£π 2 = = π₯π₯ π‘ , π₯π₯ π₯π‘ π£π + π£π 2 ) )π₯π‘ (Equation 4) “We can use Equation 4 to determine the displacement of an object undergoing that is uniform acceleration.” “Let us solve some sample problems. “ 1. A sports car approaches Given: a highway on-ramp at a vi= 20.0 m/s East ; velocity of 20.0 m/s East. If aAve =3.2 m/s2 East; the car accelerates at a rate Δt= 5s of 3.2 m/s2 East for 5.0 s, Required: what is the displacement of Δx the car? Analysis: Our first task is to determine which of the four equations of accelerated motion to use. Usually, we can solve a problem using only one of the four equations. We simply identify which equation contains all the variables for which we have given values and the unknown variable that we are asked to calculate. In the table, we see that Equation 2 has all the given variables and will allow us to solve for the unknown variable. Solution: π₯π₯=π£ π₯π‘+1 ππ₯π‘2 π 2 π π₯π₯=(20 )(5π )+ π 1 (3.2 2 π )(5π )2 π 2 π₯π₯=100m+40m π₯π₯=140m East Statement: During the 5.0 s time interval, the car is displaced 140 m East. 2. A sailboat accelerates uniformly from 6.0 m/s North to 8.0 m/s North at a rate of 0.50 m/s2 distance North. does the What boat travel? “What are the π π π π given “Given: vi= 6 , vf= 6 and quantities in the problem?” “What is required in this problem?” π aAve= 0.50 π 2 “Distance” “Base on the table which equation will allow us to solve for the unknown variable. “ First, we rearrange the equation to solve for Δx.” Solution: π£π2 = π£π2 + 2ππ₯π₯ π£π2 − π£π2 = 2ππ₯π₯ π£π2 − π£π2 π₯π₯ = 2π π π (8 )2 − (6 )2 π π π₯π₯ = π 2(0.50 π 2) π 2 π (64 ) − (36 )2 π π π₯π₯ = π 1 π 2 π₯π₯ = 28π Statement: The boat travels a distance of 28 m. 3. A dart is thrown at a target that is supported by a For a: Given: vi= 350 π πΈππ π‘ , π π πΈππ π‘, Δt=0.0050s wooden backstop. It strikes vi= 0 the backstop with an initial Required: aAve velocity of 350 m/s [E]. The “We may use the defining dart comes to rest in 0.0050 equation for acceleration: s. (a) What is the acceleration π ππ΄π£π = π£π−π£π π₯π‘ ” of the dart? (b) How far does the dart penetrate into the backstop? Solution: π£π − π£π ππ΄π£π = π₯π‘ π π 0 − 350 π ππ΄π£π = π 0.0050π π ππ΄π£π = −70000 2 π π ππ΄π£π = 70000 2 , πππ π‘ π “To solve (b), we have sufficient information to solve the problem using any equation displacement with in it. Generally speaking, in a two-part problem like this, it is a good idea to try to find an equation that uses only given information. Then, if we have made an error in calculating the first part (acceleration), our next calculation would be unaffected by the error. Therefore, we will use Equation 4 to solve (b), since it can be solved using only given information. Given: vi= 350 π πΈππ π‘ , π vi= 0 π πΈππ π‘, Δt=0.0050s π Required: Δd Solution: π£π + π£π π₯π₯ = ( )π₯π‘2 π π 0 + 350 π )(0.0050π ) π₯π₯ = ( π 2 π₯π₯ = 0.88 π, πΈππ π‘ 4. Generalization “What equation you may "π£π = π£π + ππ₯π‘ (Equation 1)” use in problems that do not directly involve displacement.” “What equation will allow you to determine the displacement of an object moving with uniform acceleration given a value for acceleration rather than a final velocity?” π₯π₯=π£ π‘+1 ππ₯π‘2 π 2 (Equation 2) “What equation you may use in problems that do not directly involve time?” “π£π2 = π£π2 + 2ππ₯π₯(Equation 3) “What equation you will use to determine the displacement of an object π₯π₯ = ( π£π+π£π 2 )π₯π‘ (Equation 4) that is undergoing uniform acceleration.” “Very good!” IV. Evaluation A car has uniformly accelerated from rest to a speed of 25m/s after travelling 75m. What is its acceleration? V. Assignment 1. A plane has a takeoff speed of 88.3 m/s from rest and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
