Take A Break!

1. Take A Break! What about this??? Which one is false?

2. Take A Break! Aim & Throw where????

3. Kinematics

4. Assessment Objectives: (a) define displacement, speed, velocity and acceleration. (b) use graphical methods to represent displacement, speed, velocity and acceleration. (c) find the distance traveled by calculating the area under a velocity-time graph. (d) use the slope of a displacement-time graph to find the velocity. (e) use the slope of a velocity-time graph to find the acceleration.

5. Assessment Objectives: (f) derive, from the definitions of velocity and acceleration, equations which represent uniformly-accelerated motion in a straight line. (g) use equations which represent uniformly-accelerated motion in a straight line, including falling in uniform gravitational field without air resistance. (h) Describe qualitatively the motion of bodies falling in a uniform gravitational field with air resistance. (i) describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction.

6. Introduction This is a branch of mechanics which deals with the description of the motion of objects, without references to the forces which act on the system. In kinematics, we will study the following two types of motion: (a) one dimensional motion, i.e. along a straight line (b) two dimensional motion – projectile motion.

7. Note: Displacement is the shortest distance between the initial and final position of the body. Displacement Definition: Displacement is the distance travelled along a specific direction. • vector quantity. • symbol is s. • SI unit is the metre, m.

8. 10 m 10 m (f) (f) (i) (i) Displacement 10 m Distance = 10 m Displacement ??? Displaced 10 m to the left Displaced 10 m to the right

9. Note: <speed> is not always equal to <velocity>. Velocity Definition: Velocity is the rate of change of distance along a specific direction, or simply the rate of change of displacement. • vector quantity. • SI units are m s-1 • symbol is u (initial speed) or v (final speed).

10. <speed> = total distance / time Velocity (Quick check!) A car travels from point A to point B with the following speeds: 20 m s-1 for 2.0 s 40 m s-1 for 2.0 s 60 m s-1 for 6.0 s A B What is the average speed of this car for this journey? C 40 m s-1 A 12 m s-1 D 48 m s-1 B 13.3 m s-1 

11. <speed> = 480 x 2 / 20 Velocity (Quick check!) Then same car now travels back from B to A with the following speeds: A B 60 m s-1 for 4.0 s 40 m s-1 for 6.0 s What is the average speed of this car for the whole journey? C 40 m s-1 A 24 m s-1 B 48 m s-1 D 96 m s-1 

12. <v> = displacement / time=0/20 Velocity (Quick check!) What is the average velocity of this car for the whole journey? A B C 24 m s-1 A 0 m s-1  B 48 m s-1 D 96 m s-1

13. Note: The acceleration of a body is constant or uniform if its velocity changes at a constant rate. Acceleration Definition: Acceleration is the rate of change of velocity. • vector quantity. • SI units are m s-2 • symbol is a. more

14. v accelerating a v decelerating a Retardation / Deceleration Retardation or deceleration is a term used to describe a decrease in the magnitude of velocity with time. Occurs when velocity and acceleration are in opposite directions.

15. Retardation / Deceleration Are these objects undergoing constant velocity? • a change in magnitude e.g. body speeding up or slowing down. (b) a change in both magnitude and direction. (c) a change in direction only e.g. circular motion.

16. v t Constant acceleration  no acceleration Uniformly accelerated motion Refers to motion where the velocity of a body is changing at a steady rate. (constant acceleration) Eg) • a toy car sliding down a slope • a ball being thrown in the air.

17. Acceleration due to gravity, g ( constant acceleration) Acceleration due to gravity is produced by the gravitational field of the earth and it is always directed downward relative to the earth’s surface. It can be taken to be constant at 9.81 m s-2(unless otherwise stated)

18. s s v v t v a t t Relationship between s, v, a s v = ds / dt (gradient) s =  v dt a = dv / dt (gradient) v =  a dt

19. v s v a a v t t t Relationship between s, v, a s v = ds / dt (gradient) s =  v dt a = dv / dt (gradient) v =  a dt

20. Displacement-Time Graph Useful information from this graph: • (i) instantaneous displacement (the displacement of a body at any instant of time) • instantaneous velocity (gradient of the graph at a particular point) • average velocity

21. Displacement-Time Graph eg) Consider a car traveling along the x-axis. What deduction can be made about the car’s motion if its displacement-time relation is represented by (i) Graph 1 and (ii) Graph 2?

22. Displacement-Time Graph Graph 1

23. (c) Average velocity = <u>= Displacement-Time Graph Graph 1: (a) The displacement x carries only positive values. This implies that the car moves along the positive x direction. (b) The displacement x increases uniformly with time, thus the velocity of the car remains constant.

24. Displacement-Time Graph Graph 2

25. Displacement-Time Graph Graph 2: (a) Since the displacement x remains positive throughout the motion, the car’s positions remain to the right of the origin O. (b) At points A and C, the car is stationary, since its displacement x is not changing at these two points. (c) Around B, the displacement x increases uniformly with time, so the car must be traveling in the positive x-direction with uniform velocity.

26. Displacement-Time Graph Graph 2: (d) After point C, the displacement x of the car is decreasing. This implies that the car is traveling in the opposite direction, i.e. the negative direction. (e) The gradient at D is greater than at E. Thus, the magnitude of the velocity at D is greater than at E. (f) Around E, the displacement x decreases with time at a decreasing rate up to F. This shows that the car eventually stops at F.

27. s t Examples of Displacement-Time Graphs Motion under constant acceleration s =  v dt v = ds / dt a = dv / dt v =  a dt

28. s t Examples of Displacement-Time Graphs Motion under constant deceleration

29. s t Examples of Displacement-Time Graphs Motion under constant velocity

30. Velocity-Time Graph Useful information from this graph: • (i) instantaneous velocity (the velocity of a body at any instant of time) • instantaneous acceleration (gradient of the graph at a particular point) • displacement (area under the graph)

31. v C B D Area 1 I F A E t Area 2 G H Velocity-Time Graph (Quick check!) eg) The motion of a body moving along a straight line is given as shown in the graph below. What deductions can be made about the body’s motion between points B and C?

32. v C B D Area 1 I F A E t Area 2 G H Velocity-Time Graph Deductions: (A) Body moving with velocity decreasing at a decreasing rate. (B) Body moving with velocity increasing at a decreasing rate. (C) Body moving with velocity decreasing at a increasing rate. (D) Body moving with velocity increasing at a increasing rate. (B)

33. v = u + at v2=u2 + 2as Equations of Motion Uniformly accelerated motion refers to motion of a body in which the acceleration is constant. Kinematics equations: Note: They are only valid for cases of uniform acceleration in a straight line. more

34. Sign Conventions (Quick check!) eg) A motorist travelling at 13 m s-1 approaches traffic lights which turn red when he is 25 m away from the stop line. His reaction time (time between seeing the red lights and applying the brakes) is 0.70 s and the condition of the road and his tyre is such that he cannot slow down at a rate of more than 4.5 m s-2. If he brakes fully, what is the total distance covered when he finally comes to a stop? (A) 22 m (B) 28 m (C) 36 m (D) 41 m

35. +ve Sign Conventions a 13 m s-1 (f) (i) s2 s1 25 m s1 = ut = 13 x 0.70 = 9.1 m v2 = u2 + 2as 02 = (+132) + 2(-4.5)s2 s2 = 18.8 m Total distance s = s1 + s2 = 9.1 + 18.8 = 27.9 m (B)

36. Fv (= kv) Fv (= kv) U U v v W W Terminal vel Low vel At terminal velocity: U + Fv = W Effect of Air Resistance U W Initially, v = 0 m s-1 Recall ?

37. Projectile Motion A projectile motion can be considered as a combination of twoindependentcomponents of motion: (i) horizontal motion with constant speed throughout, zero acceleration (since there is no acceleration horizontally), assuming there is no air resistance. more more (ii) vertical motion with uniform acceleration (for example due to gravity g). More (monkey) More (fire max range)

38. u cos u cos u cos u u sin u cos  u cos Projectile Motion Horizontally: The horizontal motion is a constant velocity motion. Hence the equation of motion is simply sx = (u cos) t

39. u u sin g  u cos Projectile Motion Vertically: The vertical motion is a uniformly accelerated motion (acc = g). The equations of motion are given by the kinematics equations e.g. v = usin -gt sy = (usin)t - ½(g)t2 v2 = (usin)2 - 2(g)sy

40. Projectile Motion Steps to success in projectile motion: • Draw a good diagram with all the data shown. • Draw the path of the motion of the body for better visualisation. • Decide on your sign convention. • Indicate your initial and final positions of the body on the path of the motion. • Decide on the direction of analysis (vertical or horizontal). • Use the appropriate kinematics equation.

41. 30 m s-1 55 m s Projectile Motion eg)A stone is projected horizontally with an initial velocity of 30 m s-1 above the horizontal from a cliff top which is 55 m above the sea level. What is the time taken for the stone to reach the sea?

42. which direction to analyse? • which equation to use? • sign convention? 30 m s-1 55 m s Projectile Motion (i) (f) Vertically: s = ut + ½ gt2 55 = 0 + ½ (9.81) t2 t = 3.34 s ( -3.34 s is inadmissible)

43. s 25 m s-1 Projectile Motion eg)A stone is projected vertically upwards with an initial velocity of 25 m s-1 above the ground. Determine the highest height reached by the stone.

44. which direction to analyse? • which equation to use? • sign convention? s 25 m s-1 Projectile Motion (f) Vertically: v2 = u2 + 2gs (i) 0 = 252 + 2(-9.81)s s = 31.9 m

45. 50 m s-1 28 s Projectile Motion eg)A stone is projected with an initial velocity of 50 m s-1 at angle of 28 above the horizontal as shown below. Calculate the maximum horizontal distance reached by the stone.

46. 50 sin 28o m s-1 50 cos 28o m s-1 Projectile Motion 50 m s-1 28 (f) (i) s vertically: s = ut + 1/2gt2 0 = (50 sin 28o)t – ½ (9.81)t2 t = 4.79 s (t=0 is inadmissible) horizontally: s = ut + 1/2gt2 = (50 cos 28o)4.79 – 0 = 211 m

47. 12 m s-1 30 75 m s Projectile Motion eg)A stone is projected with an initial velocity of 12 m s-1 at angle of 30 above the horizontal from a cliff top which is 75 m above the sea level. What is the time taken for the stone to reach the sea? (A) 3.0 s (B) 4.6 s (C) 6.0 s (D) 8.5 s (B)

48. +ve 12 12 sin30o 30o 12 m s-1 12 cos30o 30 Vertically: Using t = 4.6 s or – 3.3 s (inadmissible) 75 m s Projectile Motion (i) (f)

49. +ve Projectile Motion (b) Determine the position of the stone from the cliff when it reaches the sea. 12 m s-1 30 (i) 75 m (f) s horizontally: Using s = ut = 12 cos30o x 4.6 = 47.8 m (A) 47.8 m (B) 56.8 m (C) 70.2 m (D) 90.0 m (A)

50. The End