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### chapter-1

1. 1. Chapter 1 Motion in a straight line 1-2 Displacement vs Distance Average Velocity 1-3 INSTANTANEOUS VELOCITY 1 - 4 Acceleration 1.6 the acceleration of gravity and falling objects 10/19/2011 Norah Ali Almoneef 1
2. 2. 1-2 Displacement vs Distance Average Velocity • Displacement is a vector that points from an object’s initial position to its final position • and has a magnitude that equals the shortest distance between the two positions. _Only depends on the initial and final positions – Independent of actual paths between the initial and final positions • Distance is a scalar – Depends on the initial and final positions as well as the actual path between them 10/19/2011 Norah Ali Almoneef 2
3. 3. Displacement This type of x(t) plot shows the position of an object at any time, e.g., Position at t=3 s, x(3) = 1 m x (m) 3 4 t (s) Displacement between t=1 s and t=5 s ∆x = 1.0 m - 2.0 m = -1.0 m 10/19/2011 Norah Ali Almoneef -3 3
4. 4. Given the train’s initial position and its final position what is the displacement of the train? What is the distance traveled by the train ? Displacement = ∆x = x − x f 10/19/2011 i Norah Ali Almoneef 4
5. 5. Example: A boy travels from D to A,A to B .B to C.C to D Displacement from D to D ( which are initial and final points ) = 0 Distance traveled = 8 +4+8+4 = 24 m 10/19/2011 Norah Ali Almoneef 5
6. 6. Example : Distance = 4 m + 3 m =7 m Displacement = 5 m 10/19/2011 Norah Ali Almoneef 6
7. 7. Speed and Velocity The average speed being the distance traveled divided by the time required to cover the distance: How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m /s? Distance = 5400 s x 2.22 m / s = 11988 10/19/2011 Norah Ali Almoneef m 7
8. 8. Speed Speed can be defined in a couple of ways: How fast something is moving The distance covered in a certain amount of time The rate of change of the position of an object Units for speed are: This is the standard unit 10/19/2011 miles / hour (mi/hr) kilometers / hour (km/hr) feet / second (ft/s) meters / second (m/s) Norah Ali Almoneef 8
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11. 11. example A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the velocity of the particle when t = 4 s. dx v= = 3t 2 − 6t dt At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s 10/19/2011 Norah Ali Almoneef 11
12. 12. example What is 10/19/2011 Norah Ali Almoneef 12
13. 13. example From A to B What is A 10/19/2011 Norah Ali Almoneef B 13
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17. 17. 1-3 INSTANTANEOUS VELOCITY Instantaneous velocity – is how fast an object is moving at a particular instant. ∆ x dx example v = lim ∆→ t 0 ∆ t = dt The position of a particle moving on the x axis is given by x = 7.8 + 9.2 t – 2.1t2. What is its instantaneous velocity at t = 3.5 seconds v = 0 + 9.2 – (2)(2.1)t v = 0 + 9.2 – (2)(2.1)(3.5) = -5.5 m/s 10/19/2011 Norah Ali Almoneef 17
18. 18. 1 - 4 Acceleration Acceleration: is a rate at which a velocity is changing. Instantaneous acceleration = dv / dt = d2 x / d t2 10/19/2011 Norah Ali Almoneef 18
19. 19. Example A car’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocity of 26 m/s. What is the acceleration of the car? The car is increasing its velocity by 8 m/s for every second it is moving. 10/19/2011 Norah Ali Almoneef 19
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21. 21. • Instantaneous Acceleration • Instantaneous Acceleration ∆v dv a = lim = ∆t →0 ∆t dt • Suppose a particle is moving in a straight line so that its position is given by the relationship x = (2.10 m/s2)t3 + 2.8 m. Find its instantaneous acceleration at 5 seconds. v = dx / dt = (3)(2.1)t2 a = dv / dt = (2)(3)(2.1)t at t= 5s a = (2)(3)(2.1)(5) = 63 m/s2 10/19/2011 Norah Ali Almoneef 21
22. 22. example 10/19/2011 Norah Ali Almoneef 22
23. 23. example A bullet train starts from rest from a station and travels along a straight horizontal track towards another station. The graph in fig. shows how the speed of the train varies withtime over the whole journey. Determine: (a) the total distance covered by the train, (b) the average speed of the train. A) Total distance travelled 40 + 0 40 + 0 x= × 2 + 40 × 10 + ×4 2 2 x = 40 + 400 + 80 = 520m OR Total distance travelled = ‘area under the graph’ = (1/2)(10 + 16)(40) = 520 m Speed / ms-1 40 0 Time 2 12 16 Average speed = (total distance) / (total time ) = 520 / 16 = 32.5 ms-1 10/19/2011 Norah Ali Almoneef 23
24. 24. example : a car is traveling 30 m/s and approaches 10 m from an intersection when the driver sees a pedestrian and slams on his brakes and decelerates at a rate of 50 m/s2. (a) How long does it take the car to come to a stop? (b) how far does the car travel before coming to a stop? vf -vi= a t, where vo= 30 m/s, v = 0 m/s, and a = -50 m/s 2 t = (0 -30)/(-50) = 0.6 s Δx= vit + ½ a t2= (30)(0.6) + ½(-50)(0.6)2= 18 -9 = 9 m 10/19/2011 Norah Ali Almoneef 24
25. 25. 1.5 finding the motion of an object Equations of Kinematics for Constant Acceleration 10/19/2011 Norah Ali Almoneef 25
26. 26. example •   How long does it take a car going 30 m/sec to stop of it  decelerates at 7 m/sec2? 10/19/2011 Norah Ali Almoneef 26
27. 27. example - A car starting from rest attains a speed of 28 m/sec in 20 sec. Find the acceleration of the car and the distance it travels in this time. 10/19/2011 Norah Ali Almoneef 27
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29. 29. 1. Velocity & acceleration are both vectors. Are the velocity and the acceleration always in the same direction? NO!! If the object is slowing down, the acceleration vector is in the opposite direction of the velocity vector! 2. Velocity & acceleration are vectors. Is it possible for an object to have a zero acceleration and a non-zero velocity? YES!! If the object is moving at a constant velocity, the acceleration vector is zero! 10/19/2011 Norah Ali Almoneef 29
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31. 31. Examples : 1 ) What is the acceleration of a car that increased its speed from 10 m/s to 30 m/s in 4 seconds? a = (30 m/s – 10 m/s) ÷ 4s = 20 m/s ÷ 4s = 5 m/s2 2)the same car now slows down back to 10 m/s in 5 seconds. What is his acceleration? a = (10 m/s – 30 m/s) ÷ 5s = (- 20 m/s) ÷ 5s = - 4 m/s2 Means slowing down 10/19/2011 Norah Ali Almoneef 31
32. 32. Graphical Analysis *deduce from the shape of a speed-time graph when a body is: (i) at rest (ii) moving with uniform speed (iii) moving with uniform acceleration (iv) moving with non-uniform acceleration Velocity (ii) (iii) (i) at rest (ii) moving with uniform speed (iii) moving with uniform acceleration (iv) moving with non-uniform acceleration (iv) (i) 10/19/2011 Norah Ali Almoneef Time 32
33. 33. example 10/19/2011 Norah Ali Almoneef 33
34. 34. A bus stopped at a bus-stop for 10 seconds before accelerating to a velocity of 15 m/s in 4 seconds and then at a constant speed for the next 9 seconds. How does the graph look like? How far did the bus go in this 23 seconds? • Distance travelled in first 10 seconds is zero • Distance travelled in the next 4 seconds is           =  ½ x 4 x 15 = 30 m Velocity /m/s • Distance travelled in the final 9 seconds is         (ii)   = 9 x 15 = 135 m 15 • Total distance travelled = 165 m (iii) (i) 0 10 10/19/2011 14 23 Norah Ali Almoneef Time/s 34
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36. 36. 1.6 the acceleration of gravity and falling objects Objects thrown straight up The acceleration of a falling object is due to the force of gravity between the object and the earth. Galileo showed that falling objects accelerate equally, neglecting air resistance. Galileo found that all things fall at the same rate. On the surface of the earth, in a vacuum, all objects accelerate towards the surface of the earth at 9.8 m/s2. The acceleration of gravity (g) for objects in free fall at the earth's surface is 9.8 m/s2. ( down ward ) g actually changes as we move to higher altitudes 10/19/2011 Norah Ali Almoneef 36
37. 37. Equations of Kinematics for Constant Acceleration For free fall v = v0 + at v = v0 − gt 1 2 ∆x = v0t + at 2 2 2 v = v0 + 2a∆x 1 2 ∆y = v0t − gt 2 2 v 2 = v0 − 2g∆y 10/19/2011 Norah Ali Almoneef 37
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39. 39. A ball is dropped from a tall building and strikes the ground 4 seconds later. A ) what velocity does it strike the ground B ) what distance does it fall? 10/19/2011 Norah Ali Almoneef 39
40. 40. g=- 10/19/2011 Norah Ali Almoneef 40
41. 41. example How high can a human throw a ball if he can throw it with initial velocity 90 m / h?. 10/19/2011 Norah Ali Almoneef 41
42. 42. example 10/19/2011 Norah Ali Almoneef 42
43. 43. example 10/19/2011 Norah Ali Almoneef 43
44. 44. Notice in free fall 10/19/2011 Norah Ali Almoneef 44
45. 45. Word clues to numbers for problem solving • “free-fall”  acceleration due to gravity a=9.8m/s2, down • “at rest”  not moving v=0 • “dropped”  starts at rest and free-fall vi=0 and a=9.81m/s2, down • “constant velocity”  no acceleration a=0 • “stops”  final velocity is zero vf=0 10/19/2011 Norah Ali Almoneef 45
46. 46. Summary ∆x = x f − xi 1.Displacement: ∆x 2. Average velocity: v = ∆t v f − vi 4. Average a= acceleration ∆t : 10/19/2011 2. Time interval: 3.Instantaneou s velocity: v = Lim∆t →0 5.nstantaneous acceleration: Norah Ali Almoneef ∆x dx = ∆t dt a = Lim∆t →0 ∆v dv = ∆t dt 46