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1. 2.1b MechanicsOn the move Breithaupt pages 112 to 129 April 11th, 2010

2. AQA AS Specification

3. Distance (x) and Displacement (s) Distance (x) • the length of the path moved by an object • scalar quantity • SI unit: metre (m) Displacement (s) • the length and direction of the straight line drawn from object’s initial position to its final position • vector quantity • SI unit: metre (m)

4. Speed (v) average speed = distance change time taken vav = Δx / Δt scalar quantity SI unit: ms -1 Instantaneous speed (v) is the rate of change of distance with time: v = dx / dt

5. Velocity (v) average velocity = displacement change time taken vav = Δs / Δt vector quantity direction: same as the displacement change SI unit: ms -1 Instantaneous velocity (v) is the rate of change of displacement with time: v = ds / dt

6. Speed and Velocity Conversions 1 kilometre per hour (km h-1) = 1000 m h-1 = 1000 / 3600 ms-1 1 km h-1 = 0.28 ms-1 and 1 ms-1 = 3.6 km h-1 Also: 100 km h-1 = 28 ms-1 = approx 63 m.p.h

7. Complete 20 1400 0.20 40 20 8 3.03

8. car A: distance = 6km Chertsey displacement = 2km EAST Weybridge car B: distance = 4km Speed and Velocity Question Two cars (A and B) travel from Chertsey to Weybridge by the routes shown opposite. If both cars take 30 minutes to complete their journeys calculate their individual average speeds and velocities.

9. Car A speed = 6km / 0.5h = 12km h-1 velocity = 2km EAST / 0.5h = 4km h-1 EAST Car B speed = 4km / 0.5h = 8km h-1 velocity = 2km EAST / 0.5h = 4km h-1 EAST

10. Acceleration (a) average acceleration = velocity change time taken aav = Δv / Δt vector quantity direction: same as the velocity change SI unit: ms -2 Instantaneous acceleration (a) is the rate of change of velocity with time: a = dv / dt

11. Notes: 1. Change in velocity: = final velocity (v) – initial velocity (u) so: aav = (v – u) / Δt 2. Uniform acceleration: This is where the acceleration remains constant over a period of time. 3. Deceleration: This is where the magnitude of the velocity is decreasing with time.

12. Acceleration due to gravity (g) An example of uniform acceleration. In equations ‘a’ is substituted by ‘g’ On average at sea level: g = 9.81 ms-2 downwards g is often approximated to 10 ms -2 YOU ARE EXPECTED TO USE 9.81 IN EXAMINATIONS!!

13. Question Calculate the average acceleration of a car that moves from rest (0 ms-1) to 30 ms-1 over a time of 8 seconds. aav = (v – u) / Δt = (30 – 0) / 8 average acceleration = 3.75 ms-2

14. Complete 45 3 30 - 5 - 60

15. Distance-time graphs The gradientof a distance-time graph is equal to the speed

16. Displacement-time graphs The gradient of a displacement-time graph is equal to the velocity The graph opposite shows how the displacement of an object thrown upwards varies in time. Note how the gradient falls from a high positive value to zero (at maximum height) to a large negative value. Estimate the initial velocity of the object. Initial gradient = (5 – 0)m / (0.5 – 0)s = 10 ms-1 Initial velocity = 10 ms-1

17. s / m D C B E A t / s Question Describe the motion shown by the displacement-time graph below: O → A: acceleration from rest A → B: constant velocity B → C: deceleration to rest C → D: rest (no motion) D → E: acceleration from rest back towards the starting point

18. velocity Velocity-time graphs With velocity-time graphs: gradient = acceleration a = (v – u) / t The area under the ‘curve’ = displacement s = [u x t] + [½ (v – u) x t]

19. v / ms-1 A B v1 t / s C F v2 D E Question 1 Describe the motion shown by the velocity-time graph below: O → A: UNIFORM POSITIVE acceleration from rest to velocity v1. A → B: constant velocity v1. B → C → D : UNIFORM NEGATIVE acceleration from v1to negative velocity v2. At C: The body reverses direction D → E: constant negative velocity v2. E → F: NON-UNIFORM POSITIVE acceleration to rest

20. v / ms-1 12 t / s T 4 6 11 -10 Question 2 The graph shows the velocity-time graph of a car. Calculate or state: (a) the acceleration of the car during the first 4 seconds. (b) the displacement of the car after 6 seconds. (c) time T. (d) the displacement after 11 seconds. (e) the average velocity of the car over 11 seconds.

21. v / ms-1 12 t / s T 4 6 11 -10 Question 2 a) the acceleration of the car during the first 4 seconds. acceleration = gradient = (12 - 0)ms-1 / (4 – 0)s = 12 / 4 acceleration = 3 ms-2

22. Question 2 v / ms-1 (b) the displacement of the car after 6 seconds. displacement = area = area A + area B = ½ (12 x 4) + (12 x 2) = 24 + 24 displacement = 48 m 12 area A area B t / s T 4 6 11 -10

23. Question 2 v / ms-1 (c) time T. By similar triangles: (T - 6):(11 - T) = 12:10 i.e. (T - 6) / (11 - T) = 12 / 10 (T - 6) / (11 - T) = 1.20 (T - 6) = 1.20 (11 – T) T - 6 = 13.2 – 1.2T 2.2T = 19.2 T = 19.2 / 2.2 T = 8.73 seconds Note: T can also be found by scale drawing or by using the equations of uniform acceleration (see later). 12 area A area B t / s T 4 6 11 -10

24. Question 2 v / ms-1 (d) the displacement after 11 seconds. displacement = area = area A + area B + area C – area D = 24 + 24 + ½ (12 x 2.73) – ½ (10 x 3.27) = 24 + 24 + 16.38 – 11.35 = 53.03 displacement = 53.0 m 12 area A area B area C t / s T 4 6 11 area D -10

25. v / ms-1 12 area A area B area C t / s T 4 6 11 area D -10 Question 2 (e) the average velocity of the car over 11 seconds. average velocity = displacement / time = 53.03 / 11 average velocity = 4.82 ms-1

26. s / ms-1 64 53 48 24 4 6 11 T t / s Question 3 Sketch the displacement-time graph for the car of question 2. displacement-time co-ordinates:

27. displacement h time gradients = - 9.8 ms-2 Question 4 Sketch displacement and velocity time graphs for a bouncing ball. Take the initial displacement of the ball to be h at time t = 0. Use the same time axis for both curves and show at least three bounces. velocity

28. v = u + at v2 = u2 + 2as s = ½ (u + v) t s = ut + ½ at2 v = FINAL velocity u = INITIAL velocity a = acceleration t = time for the velocity change s = displacement during the velocity change The equations of uniform acceleration THESE EQUATIONS ONLY APPLY WHEN THE ACCELERATION REMAINS CONSTANT

29. Question 1 Calculate the final velocity of a car that accelerates at 2ms -2 from an initial velocity of 3ms -1 for 5 seconds. v = u + at v = 3 + (2 x 5) = 3 + 10 final velocity = 13 ms-1

30. Question 2 Calculate the stopping distance of a car that is decelerated at 2.5 ms -2 from an initial velocity of 20 ms -1. v2 = u2 + 2as 0 = 202 + (2 x - 2.5 x s) 0 = 400 + - 5s - 400 = - 5s - 400 / - 5 = s stopping distance = 80 m

31. Question 3 A stone is dropped from the edge of a cliff. If it accelerates downwards at 9.81 ms -2 and reaches the bottom after 1.5s calculate the height of the cliff. s = ut + ½ at2 s = (0 x 1.5) + ½ (9.81 x (1.5)2) s = ½ (9.81 x 2.25) cliff height = 11.0 m

32. Question 4 Calculate the time taken for a car to accelerate uniformly from 5 ms -1 to 12 ms -1 over a distance of 30m. s = ½ (u + v) t 30 = ½ (5 + 12) x t 30 = 8.5 x t 30 ÷ 8.5 = t time = 3.53 s

33. Question 5 A ball is thrown upwards against gravity with an initial speed of 8 ms -1. What is the maximum height reached by the ball? v2 = u2 + 2as where: s = height upwards u = 8 ms -1 upwards v = 0 ms -1 (at maximum height) a = - 9.81 ms -2 (acceleration is downwards)

34. Question 5 continued v2 = u2 + 2as 0 = (8)2 + 2 (-9.81 x s) 0 = 64 - 19.62 x s - 64 = - 19.62 x s - 64 / - 19.62 = s maximum ball height = + 3.26 m

35. Calculate the ? quantities 16 45 16 4

36. Calculate the other quantities 128 6 2 0.5

37. Projectile motion This is where a body is moving in two dimensions. For example a stone being thrown across a stretch of water has both horizontal and vertical motion. The motion of the body in two such mutually perpendicular directions can be treated independently.

38. Example 1 A stone is thrown horizontally at a speed of 8.0 ms-1 from the top of a vertical cliff. If the stone falls vertically by 30m calculate the time taken for the stone to reach the bottom of the cliff and the horizontal distance travelled by the stone (called the ‘range’). Neglect the effect of air resistance. height of fall path of stone range

39. Example 1 Stage 1 Consider vertical motion only s = ut + ½ at2 30 = (0 x t) + ½ (9.81 x (t)2) 30 = ½ (9.81 x (t)2) 30 = 4.905 x t2 t2= 6.116 time of fall = 2.47 s height of fall path of stone range

40. Example 1 Stage 2 Consider horizontal motion only During the time 2.47 seconds the stone moves horizontally at a constant speed of 8.0 ms-1 speed = distance / time becomes: distance = speed x time = 8.0 x 2.47 = 19.8 range = 19.8 m height of fall path of stone range

41. Further Questions (a) Repeat this example this time for a cliff of height 40m with a stone thrown horizontally at 20 ms-1. time of fall = 2.83 s range = 56.6 m (b) How would these values be changed if air resistance was significant? time of fall - longer range - smaller height of fall path of stone range

42. path of shell maximum height 30° range Example 2 A shell is fired at 200 ms-1 at an angle of 30 degrees to the horizontal. Neglecting air resistance calculate: (a) the maximum height reached by the shell (b) the time of flight (c) the range

43. Example 2 Stage 1 - Part (a) Consider vertical motion only At the maximum height, s The final VERTICAL velocity, v = 0. v2 = u2 + 2as 0 = (200 sin 30°)2+ (2 x - 9.81 x s) [upwards +ve] 0 = (200 x 0.500)2+ (-19.62 x s) 0 = (100)2+ (-19.62 x s) 0 = 10000- 19.62s - 10000 = - 19.62s s = 10000/ 19.62 s = 509.7 maximum height = 510 m

44. Example 2 Stage 2 – Part (b) Consider vertical motion only v = u + at 0 = (200 sin 30°) + (- 9.81 x t) 0 = 100 - 9.81t -100 = - 9.81t t = 100 / 9.81 t = 10.19 Time to reach maximum height = 10.19 s If air resistance can be neglected then this is also the time for the shell to fall to the ground again. Hence time of flight = 2 x 10.19 time of flight = 20.4 seconds

45. Example 2 Stage 3 – Part (c) Consider horizontal motion only During the time 20.38 seconds the shell moves horizontally at a constant speed of (200 cos 30°) ms-1 speed = distance / time becomes: distance = speed x time = (200 cos 30°) x 20.38 = (200 x 0.8660) x 20.38 = 173.2 x 20.38 = 3530 range = 3530 m (3.53 km)

46. path of shell maximum height 45° range Question Repeat example 2 this time for a firing angle of 45°. sin 45° = 0.7071; 200 x sin 45° = 141.4 maximum height = 1020 m time of flight = 28.8 s range = 4072 m (4.07 km) Note: 45° yields the maximum range in this situation.

47. The Moving Man- PhET - Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you. Maze Game- PhET - Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can. Motion in 2D- PhET - Learn about velocity and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle). See the velocity and acceleration vectors change as the ball moves. Motion with constant acceleration- Fendt Bouncing ball with motion graphs- netfirms Displacement-time graph with set velocities - NTNU Displacement & Aceleration-time graphs with set velocities- NTNU Displacement & Velocity-time graphs with set accelerations- NTNU Football distance-time graphs- eChalk Motion graphs with tiger- NTNU Two dogs running with graphs- NTNU Motion graphs test - NTNU BBC KS3 Bitesize Revision: Speed - includes formula triangle applet Projectile Motion- PhET- Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target. Projectile motion - with or without air drag - NTNU Projectile motion - NTNU Projectile motion x- with variable height of projection - netfirms Projectile Motion- Fendt Projectile motion- Virgina Golf stroke projectile challenge- Explore Science Shoot the monkey- Explore Science Canon & target projectile challenge- Sean Russell Slug projectile motion game- 7stones Bombs released from an aeroplane- NTNU Internet Links

48. Define what is meant by: (a) displacement; (b) speed & (c) velocity. Explain the difference between speed and velocity. Define what is meant by acceleration. What is the difference between UNIFORM and NON-UNIFORM acceleration? Illustrate your answer with sketch graphs. State the equations of constant acceleration. Explain what each of the five symbols used means. What is meant by ‘free fall’? What is the average value of the acceleration of free fall near the Earth’s surface? What information is given by the gradients of (a) distance-time; (b) displacement-time; & (c) velocity-time graphs? What information is given by the area under the curves of (a) speed-time & (b) velocity-time graphs? List and explain the three principles applying to the motion of all projectiles. Repeat the worked example on page 127 this time with the object projected horizontally at a speed of 20 ms-1 from the top of a tower of height 40 m. Core Notes from Breithaupt pages 112 to 129

49. Notes from Breithaupt pages 112 & 113 • Define what is meant by: (a) displacement; (b) speed & (c) velocity. • Explain the difference between speed and velocity. • A ball is thrown vertically upwards and then caught when it falls down again. Sketch distance-time and displacement-time graphs of the ball’s motion and explain why these graphs are different from each other. • Try the summary questions on page 113

50. Notes from Breithaupt pages 114 & 115 • Define what is meant by acceleration. • What is the difference between UNIFORM and NON-UNIFORM acceleration? Illustrate your answer with sketch graphs. • Repeat the worked example on page 115 this time with the vehicle moving initially at 12 ms-1 applying its brakes for 20s. • Try the summary questions on page 115