Term: 1^{st} Term
Week: 6
Class: Senior Secondary School 1
Age: 15 years
Duration: 40 minutes of 5 periods each
Date:
Subject: Physics
Topic: Vector & scalar quantity
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 12
PRESENTATION 
TEACHER’S ACTIVITY 
STUDENT’S ACTIVITY 
STEP 1 INTRODUCTION 
The teacher reviews the previous lesson on Friction 
Students pay attention 
STEP 2 EXPLANATION 
He explains these terminologies; Distance & Displacement, Speed & Velocity, Acceleration & Retardation

Students pay attention and participates 
STEP 3 DEMONSTRATION 
He represents a Distance/Displacement  Time Graph and a Speed/Velocity  Time Graph

Students pay attention and participate 
STEP 4 NOTE TAKING 
The teacher writes a summarized note on the board 
The students copy the note in their books 
NOTE
VECTOR AND SCALAR QUANTITY
LINEAR MOTION
Terminologies used in linear motion:
𝑠𝑝𝑒𝑒𝑑=𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑣= 𝑠
𝑡
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦=𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑣= 𝑠
𝑡
NOTE: Velocity is often used interchangeably with speed during calculations
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
𝑎= Δ𝑣
Δ𝑡
𝑎= 𝑣−𝑢
𝑡2−𝑡1
𝑣=𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑢=𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
Equation of Uniformly Accelerated motion
S = (v+u) t 7
2
v = u + at 8
v^{2} = u^{2} + 2 aS 9
S = ut + ½ at^{2} 10
Equations (7) to (10) are called equations of uniformly accelerated motion and could be used to solve problems associated with uniformly accelerated motion
where u initial velocity, v – final velocity, a – acceleration, S – distance covered and t – time
Example
A car moves from rest with an acceleration of 0.2mls^{2}. Find its velocity when it has moved a distance of 50m.
Solution
Given:
a = 0.2mls^{2}, S = 50m, u = 0m/s, v =?
v^{2} = u^{2} + 2 as
v^{2}= 0^{2} + (2x0.2x50) = 20
v = √20 m/s
Distance/Displacement Time Graph
The slope of this time graph gives speed/velocity.
For a uniform speed/velocity, the time graph is given below:
If the velocity is non – uniform, the velocity at a point is the gradient or slope of the tangent at that point.
S(m)
t (s)
Speed/Velocity  Time Graph
The slope of the speed/velocitytime graph gives acceleration.
Example
A car starts from rest and accelerates uniformly until it reaches a velocity of 30mls after 5 seconds. It travels with uniform velocity for 15 seconds and is then brought to rest in 10s with a uniform retardation. Determine
Solution
The velocity – time diagram for the journey is shown above, from this diagram
a.) the acceleration = slope of OA
= AE/EO
= (300) /(50)=30/5
= 6mls^{2}
b.) the retardation = slope of BC = CB / CD
= (030) / (3020) = 30/10
= 3mls^{2} (the negative sign indicate that the body is retarding)
c.) Distance traveled after 5s = area of A E O
= ½ x b x h
= ½ x 5 x 30
= 75m
d.) total distance covered = area of the trapezium OABC
= ½ (AB + OC) AE
= ½ (15 + 30) 30
= 675m.
Using equations of motion:
a.) U = O, V = 3, t = 5 V = u + t
a = vu/t = 30 – 0 / 5 a = 30/5 = 6ms^{2}
b.) a = (v – u) / t
a = (030) / 10
a = 3 mls^{2}
c.) S = (u + v) 5 2
S= (30 x 5)/2
S= 75m
d.) To determine the total distance traveled we need to find the various distance for the three stages of the journey and then add them.
For the 1^{st} part S= 75m from (c) above
For the 2^{nd} stage: where it moves with uniform velocity.
S = vt
For the last stage S = ½ (u + v) t
Total distance = 75 + 450 + 100 = 675m
EVALUATION: 1. Define the following terms as used in linear motion:
(i) the velocity in ms^{1}
(ii) the retardation
(b) a body at rest is given an initial uniform acceleration of 8.0ms^{2} for 30 seconds after which the acceleration is reduced to 5.0ms^{2} for the next 20 seconds. The body maintains the speed for 60seconds after which it is brought to rest in 20 seconds. Draw the velocitytime graph of the motion using the information given above
(c) using the graph, calculate the:
(i) maximum speed attained during the motion;
(ii) average retardation as the body is being brought to rest
(iii) total distance travelled during the first 50s;
(iv) average speed during the same interval as in (iii)
(i) calculate the acceleration during the first 10s
(ii) calculate the deceleration during the last 5s
(iii) sketch a graph of the motion and calculate the total distance covered throughout the motion
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively