# Lesson Notes By Weeks and Term - Senior Secondary 1

Vector & scalar quantity

Term: 1st 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

1. Explain these terminologies; Distance & Displacement, Speed & Velocity, Acceleration & Retardation
2. Represent a Distance/Displacement - Time Graph
3. Represent a Speed/Velocity - Time Graph

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

 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:

1. Distance: This defined as the total length of path traversed. It is also the separation between two points. It is denoted as “s” or “x”. It is a scalar quantity. The SI unit of distance is meters (m)
2. Displacement: this is distance moved in a specified direction. It is denoted as “s” or “x”. It is a vector quantity. The SI unit of displacement is meters (m)
3. Speed: This is the rate of change of distance with time. It is a scalar quantity. Its SI unit is meter per seconds (m/s or ms-1)

𝑠𝑝𝑒𝑒𝑑=𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑

𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

𝑣=      𝑠

𝑡

1. Uniform speed: This is when the rate of change of distance with time is constant.
2. Velocity: This is the rate of change of displacement with time. It is a vector quantity. Its SI unit is meter per seconds (m/s or ms-1).

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦=𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

𝑣=      𝑠

𝑡

1. Uniform velocity: This is when the rate of change of displacement with time is constant.

NOTE: Velocity is often used interchangeably with speed during calculations

1. Acceleration: This is the increasing rate of change of velocity with time. It is a vector quantity. Its SI unit is meter per seconds-square (m/s2 or ms-2)

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒

𝑎=       Δ𝑣

Δ𝑡

𝑎=       𝑣𝑢

𝑡2−𝑡1

𝑣=𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑢=𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

1. Uniform acceleration: This is when the increasing rate of change of velocity with time is constant
2. Deceleration: This is the decreasing rate of change of velocity with time. It is a vector quantity. It is commonly referred to as negative acceleration or retardation.
3. Uniform deceleration: This is when the decreasing rate of change of velocity with time is constant

Equation of Uniformly Accelerated motion

S = (v+u) t                             7
2

v = u + at                              8

v2 = u2 + 2 aS                        9

S = ut + ½ at2                       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 0.2mls2. Find its velocity when it has moved a distance of 50m.

Solution

Given:

a = 0.2mls2, S = 50m, u = 0m/s, v =?

v2 = u2 + 2 as

v2= 02 + (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/velocity-time 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

1. the acceleration of the car
2. The retardation
3. The distance covered after 5s
4. The total distance covered (use both graphical and analytical method)

Solution

The velocity – time diagram for the journey is shown above, from this diagram

a.) the acceleration = slope of OA

= AE/EO

= (30-0) /(5-0)=30/5

= 6mls2

b.) the retardation = slope of BC = CB / CD

= (0-30) / (30-20) = -30/10

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 = v-u/t = 30 – 0 / 5 a = 30/5 = 6ms-2

b.) a = (v – u) / t

a = (0-30) / 10

a = -3 mls2

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 1st part    S= 75m from (c) above

For the 2nd stage: where it moves with uniform velocity.

S = vt

1. 30 x 15
2. 450m

For the last stage S = ½ (u + v) t

1. ½(30+0)10
2. 150m.

Total distance = 75 + 450 + 100 = 675m

EVALUATION:    1. Define the following terms as used in linear motion:

1. Acceleration
2. Speed
3. Displacement
1. A car moves with a velocity of 72kmhr-1. It is brought to rest in 10s. Find

(i) the velocity in ms-1

(ii) the retardation

1. State the differences & similarity between speed & velocity
2. (a) Explain the terms uniform acceleration and average speed

(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 velocity-time 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)

1. A car starts from rest and accelerates uniformly for 10s, until it attains a velocity of 25ms-1; it then travels with uniform velocity for 20s before decelerating uniformly to rest in 5s.

(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