1. 2.1 ANALYSING LINEAR MOTION

2. INTRODUCTION How fast ? (Speed / velocity) Does it change its speed ? (Acceleration / deceleration) How would you describe the motion in word ? How far does it travel ? (distance/displacement) Information required: How fast ? (Speed / velocity) How far does it travel ? (distance/displacement)

3. A straight line motion LINEAR MOTION Not a straight line motion NON LINEAR MOTION Total path travelled in a given time is the same as the shortest path Total path travelled in a given time is different from the shortest path

4. DISTANCE AND DISPLACEMENT DISTANCE AND DISPLACEMENT SPEED AND VELOCITY SPEED AND VELOCITY ACCELERATION AND DECELERATION ACCELERATION AND DECELERATION Learning area

5. DISTANCE AND DISPLACEMENT Pontian Kecil Desaru Johor Bahru Pontian Kecil Ayer Hitam Senai Kota Tinggi Mawai Benut How far is it from Johor Bahru to Desaru ? Distance = total path length =JB to Desaru via Kota Tinggi Displacement = shortest path length = JB direct to Desaru SCALA R VECTOR

6. SPEED AND VELOCITY start end path Distance = Displacement = Average Speed = Average Velocity = Time taken =

7. ACCELERATION AND DECELERATION Velocity increases Constant velocity Velocity decreases Acceleration = Rate of change of velocity = Change of velocity Time = final velocity – Initial velocity Time a = v – u t vector m s -2 Velocity increases = acceleration Velocity decreases = deceleration

8. Carry out Hands-on Activity 2.2 ( page 11 of the practical book) Aim : To differentiate between acceleration and deceleration Discussion : 1.(a) The speed of the trolley increases. (b) The speed of the trolley decreases. 2.Acceleration is the rate of increasing speed in a specified direction. Deceleration is the rate of decreasing speed in a specified direction.

9. Lesson 2

10. RELATING DISPLACEMENT, VELOCITY, ACCELERATION AND TIME a. Using ticker tape b. Using Equations of Motion Learning area

11. ticker timer ticker tape A.C. 50 Hz 50 dots made in 1 second Carbon disc

12. Time interval between two adjacent dots = 1/50 s = 0.02 s 1 tick = 0.02 s dots 1 tick Slow movement faster movement fastest movement

13. PREPARING A TAPE CHART (5 -TICKS STRIP) 0 510 First 5-tick strip 2 nd 5-tick strip Velocity, v (cm /s) Time / s

14. INFERENCE FROM TICKER TAPE AND CHART Zero acceleration constant velocity Constant acceleration Constant deceleration

15. Carry out Hands-on Activity 2.3 ( page 13 of the practical book) Aim : To use a ticker timer to identify the types of motion Discussion 2.3(A): 2. Spacing of the dots is further means a higher speed. Spacing of the dots is closer means a slower speed.

16. Discussion Hands-on Activity 2.3(B) ( page 13 of the practical book) Aim : To determine displacement, average velocity and acceleration Discussion 2.3(B): 1. Prepare a tape chart. 2. Determine average velocity using v = Total displacement time 3. Determine acceleration using a = final velocity – initial velocity time

17. Lesson 3

18. TO DETETMINE THE AVERAGE VELOCITY EXAMPLE The time for each 5-tick strip = 5 x 0.02 s = 0.1 s Length / cm Time / s 0 7 10 14 15 22 0.1 0.2 0.3 0.4 0.5 0.6 0.7 = (7 +10 +14 +15 +22 +14 +10) cm = 92 cm = 7 strips = 0.7 s Total displacement Total time taken Average velocity = displacement Time taken = 92 / 0.7 = 131.4 cm s -1

19. TO DETERMINE THE ACCELERATION EXAMPLE The time for each 10-tick strip = 10 x 0.02 s = 0.2 s 5.8 / 0.2 =29 cm s -1 27.3 / 0.2 = 136.5 Initial velocity, u Final velocity, v acceleration = v-u t = (136.5 – 29) cm s -1 1.2 s Length / cm Time / s 0 0.2 0.4 0.61 1.2 Time taken =(7-1 )strips = 6 x 0.2 s = 1.2 s 5.8 27.3 1.4 0.8 = 89.6 cm s -2

20. Lesson 4

21. s = Displacement u = Initial velocity v = Final velocity a = Constant acceleration t = Time interval THE EQUATIONS OF MOTION

22. EXAMPLE A car travelling at a velocity 10 m s -1 due north speeds up uniformly to a velocity of 25 m s -1 in 5 s. Calculate the acceleration of the car during these five seconds u = 10 m s -1, v = 25 m s -1, t = 5 s, a = ? v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t Using v = u + at 25 = 10 + a(5) 5 a = 15 a = 3 m s -2 Don’t forget the unit

23. EXAMPLE A rocket is uniformly accelerated from rest to a speed of 960 m s -1 in 1.5 minutes. Calculate the distance travelled. u = 0 m s -1, v = 960 m s -1, t = 1.5 x 60 = 90 s, s = ? v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t Using s = ½ (u + v)t s = ½ (0 + 960) 90 = 43 200 m What is the unit ?

24. EXAMPLE A particle travelling due east at 2 m s -1 is uniformly accelerated at 5 m s -2 for 4 s. Calculate the displacement of the particle. u = 2 m s -1, a = 5 m s -2, t = 4 s, s = ? v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t Using s = ut + ½ at 2 s = 2(4) + ½ (5)(4) 2 = 8 + 40 = 48 m What is the unit ?

25. EXAMPLE A trolley travelling with a velocity 2 m s -1 slides 10 m down a slope with a uniform acceleration. The final velocity is 8 m s -1. Calculate the acceleration. u =2 m s -1, v = 8 m s -1, s = 10 m, a = ? v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t Using v 2 = u 2 + 2as 8 2 = 2 2 + 2 a (10) 20 a = 64 – 4 = 60 a = 3 What is the unit ? m s -2

26. EXAMPLE teks book pg 27 u =0 m s -1, a = 2.5 m s -2, t = 10 s v = ?, s = ? Using v = u + at = 0 + (2.5)(10) = 25 m s -1 v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t Using s = ut + ½ at 2 = 0(10) + ½ (2.5)(10) 2 = 125 m

27. EXAMPLE teks book pg 27 u = 25m s -1, v = 0 m s -1, s = 50 m, a = ? Using v 2 = u 2 + 2as 0 = 25 2 + 2a (50) 0 = 625 + 100a a = - 625 100 = - 6.25 m s -2 v = u + at s = ut + ½ at 2 v 2 = u 2 + 2as s = ½ (u + v) t The negative sign shows deceleration.