Lesson Notes By Weeks and Term - Senior Secondary 1

Motion

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Term: 3rd Term

Week: 9

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:       

Subject:      Physics

Topic:-       Motion

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

  1. State the equations of motion
  2. Solve exercises involving the equations of motion

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 vector resolution

Students pay attention

STEP 2

EXPLANATION

He states the equations of motion

 

 

Students pay attention and participates

STEP 3

DEMONSTRATION

He solves some examples involving the equations of motion

 

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

MOTION

SPEED: This is the rate of change distance with time

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

               𝑡𝑖𝑚𝑒

VELOCITY: This is the rate of change of displacement with time

Velocity = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

                   𝑡𝑖𝑚𝑒

ACCELERATION: This is the rate of change of velocity with time

Acceleration= 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

                      𝑡𝑖𝑚𝑒

EQUATIONS OF MOTION

  1. V= U+at
  2. V2= U2 + 2as
  3. S= ut + ½ at2
  4. S = (V + U)t

Example:

A car travelling at 54kmh-1 slows down uniformly to a velocity of 18kmh-1 when the brakes were gradually applied

(a) how long does it take to cover a distance of 50m.

(b)What is the deceleration?

 

Solution:


a.) Initial velocity, U = 54kmh-1 = 15ms-1 Final velocity, V =18kmh-1 = 5ms-1 Distance covered, s = 50m

S = (V + U)t

          2

S = (15 + 5)t

          2


50 = 10t

t = 5s

b.) To determine the deceleration ( negative acceleration)

V = u + at

5 = 15 + 5t

-10 = 5a

a = -2ms-2

 

Example: A body, starting from rest, travels for 30s. Determine the final velocity attained after covering a distance of 40m

 

Solution:

U = 0

t = 30s

S = 40m

S= ut + ½ at2

40 = 0 x t + ½ x a x 302

80 = 900a

a = 0.0889m/s2

The final velocity, v = u + at

V = 0 + 0.0889 x 30

V = 2.7m/s

 

EVALUATION:   1. How do we calculate

  1. speed
  2. velocity
  3. acceleration

      2. State the equations of motion

      3. A Body at rest was accelerated, and it traveled for 5 minutes. The constant acceleration provided to the body was 2 m/sec2. Find the final velocity the object had before coming to a stop. Also, answer the equation of motion used to find out the final velocity

      4. A ball is dropped from a certain height. It took the ball 15 seconds to reach the ground. What is the height at which the ball was initially, also answer which equation of motion is used to answer this question? [Take g=10m/sec2].

      5. A Car with an initial speed of 1 m/sec was in motion for 10 minutes, and then it came to a stop, the velocity right before it stopped was 5m/sec. What was the constant acceleration of the car?

      6. A cycle covered 2 km in 8 minutes and the initial velocity of the cycle was 1 m/sec. Find the acceleration that the cycle had in its motion.

      7. A toy is accidentally dropped by a kid from his roof. The final velocity of the toy before it reached the ground was 8m/sec. Find the height of the building.

      8. Find the final velocity of an object that traveled a distance of 500meter with an initial speed of 2m/sec and acceleration of the object was 0.5m/sec2. 

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively