Chapter 1 Motion in a straight line 1

  • Slides: 46
Download presentation
Chapter 1 Motion in a straight line 1 -2 Displacement vs Distance Average Velocity

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

1 -2 Displacement vs Distance Average Velocity • Displacement is a vector that points

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

Displacement This type of x(t) plot shows the position of an object at any

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 Dx = 1. 0 m - 2. 0 m = -1. 0 m 10/19/2011 Norah Ali Almoneef -3 3

Given the train’s initial position and its final position what is the displacement of

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 = 10/19/2011 Norah Ali Almoneef 4

Example: A boy travels from D to A, A to B. B to C.

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

Example : Distance = 4 m + 3 m =7 m Displacement = 5

Example : Distance = 4 m + 3 m =7 m Displacement = 5 m 10/19/2011 Norah Ali Almoneef 6

Speed and Velocity The average speed being the distance traveled divided by the time

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 m 10/19/2011 Norah Ali Almoneef 7

Speed can be defined in a couple of ways: How fast something is moving

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

10/19/2011 Norah Ali Almoneef 9

10/19/2011 Norah Ali Almoneef 9

10/19/2011 Norah Ali Almoneef 10

10/19/2011 Norah Ali Almoneef 10

example A particle moves along a straight line such that its position is defined

example A particle moves along a straight line such that its position is defined by s = (t 3 – 3 t 2 + 2 ) m. Determine the velocity of the particle when t = 4 s. At t = 4 s, the velocity = 3 (4)2 – 6(4) = 24 m/s 10/19/2011 Norah Ali Almoneef 11

example What is 10/19/2011 Norah Ali Almoneef 12

example What is 10/19/2011 Norah Ali Almoneef 12

example From A to B What is A 10/19/2011 Norah Ali Almoneef B 13

example From A to B What is A 10/19/2011 Norah Ali Almoneef B 13

10/19/2011 Norah Ali Almoneef 14

10/19/2011 Norah Ali Almoneef 14

10/19/2011 Norah Ali Almoneef 15

10/19/2011 Norah Ali Almoneef 15

10/19/2011 Norah Ali Almoneef 16

10/19/2011 Norah Ali Almoneef 16

1 -3 INSTANTANEOUS VELOCITY Instantaneous velocity – is how fast an object is moving

1 -3 INSTANTANEOUS VELOCITY Instantaneous velocity – is how fast an object is moving at a particular instant. example The position of a particle moving on the x axis is given by x = 7. 8 + 9. 2 t – 2. 1 t 2. What is its instantaneous velocity at t = 3. 5 seconds v = 0 + 9. 2 – (3)(2. 1)t 2 v = 0 + 9. 2 – (3)(2. 1)(3. 5) = -68 m/s 10/19/2011 Norah Ali Almoneef 17

 1 - 4 Acceleration: is a rate at which a velocity is changing.

1 - 4 Acceleration: is a rate at which a velocity is changing. Instantaneous acceleration = dv / dt = d 2 x / d t 2 10/19/2011 Norah Ali Almoneef 18

Example A car’s velocity at the top of a hill is 10 m/s. Two

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

10/19/2011 Norah Ali Almoneef 20

10/19/2011 Norah Ali Almoneef 20

 • Instantaneous Acceleration • Suppose a particle is moving in a straight line

• Instantaneous Acceleration • Suppose a particle is moving in a straight line so that its position is given by the relationship x = (2. 10 m/s 2)t 3 + 2. 8 m. Find its instantaneous acceleration at 5 seconds. v = dx / dt = (3)(2. 1)t 2 a = dv / dt = (2)(3)(2. 1)t at t= 5 s a = (2)(3)(2. 1)(5) = 63 m/s 2 10/19/2011 Norah Ali Almoneef 21

example 10/19/2011 Norah Ali Almoneef 22

example 10/19/2011 Norah Ali Almoneef 22

example A bullet train starts from rest from a station and travels along a

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 Speed / ms-1 40 OR Total distance travelled = ‘area under the graph’ = (1/2)(10 + 16)(40) = 520 m 0 Time 2 12 16 Average speed = (total distance) / (total time ) = 520 / 16 = 32. 5 ms-1 10/19/2011 Norah Ali Almoneef 23

example : a car is traveling 30 m/s and approaches 10 m from an

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/s 2. (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 t 2= (30)(0. 6) + ½(-50)(0. 6)2= 18 -9 = 9 m 10/19/2011 Norah Ali Almoneef 24

1. 5 finding the motion of an object Equations of Kinematics for Constant Acceleration

1. 5 finding the motion of an object Equations of Kinematics for Constant Acceleration 10/19/2011 Norah Ali Almoneef 25

example • How long does it take a car going 30 m/sec to stop

example • How long does it take a car going 30 m/sec to stop of it decelerates at 7 m/sec 2? 10/19/2011 Norah Ali Almoneef 26

example - A car starting from rest attains a speed of 28 m/sec in

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

10/19/2011 Norah Ali Almoneef 28

10/19/2011 Norah Ali Almoneef 28

1. Velocity & acceleration are both vectors. Are the velocity and the acceleration always

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

10/19/2011 Norah Ali Almoneef 30

10/19/2011 Norah Ali Almoneef 30

Examples : 1 ) What is the acceleration of a car that increased its

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) ÷ 4 s = 20 m/s ÷ 4 s = 5 m/s 2 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) ÷ 5 s = (- 20 m/s) ÷ 5 s = - 4 m/s 2 Means slowing down 10/19/2011 Norah Ali Almoneef 31

Graphical Analysis *deduce from the shape of a speed-time graph when a body is:

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) (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

example 10/19/2011 Norah Ali Almoneef 33

example 10/19/2011 Norah Ali Almoneef 33

A bus stopped at a bus-stop for 10 seconds before accelerating to a velocity

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 • Distance travelled in the final 9 seconds is Velocity /m/s = 9 x 15 = 135 m • Total distance travelled = 165 m (ii) 15 (iii) (i) 0 10 10/19/2011 14 23 Norah Ali Almoneef Time/s 34

10/19/2011 Norah Ali Almoneef 35

10/19/2011 Norah Ali Almoneef 35

1. 6 the acceleration of gravity and falling objects Objects thrown straight up The

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/s 2. The acceleration of gravity (g) for objects in free fall at the earth's surface is 9. 8 m/s 2. ( down ward ) g actually changes as we move to higher altitudes 10/19/2011 Norah Ali Almoneef 36

Equations of Kinematics for Constant Acceleration For free fall 10/19/2011 Norah Ali Almoneef 37

Equations of Kinematics for Constant Acceleration For free fall 10/19/2011 Norah Ali Almoneef 37

10/19/2011 Norah Ali Almoneef 38

10/19/2011 Norah Ali Almoneef 38

 A ball is dropped from a tall building and strikes the ground 4

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

g = - 10/19/2011 Norah Ali Almoneef 40

g = - 10/19/2011 Norah Ali Almoneef 40

example How high can a human throw a ball if he can throw it

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

example 10/19/2011 Norah Ali Almoneef 42

example 10/19/2011 Norah Ali Almoneef 42

example 10/19/2011 Norah Ali Almoneef 43

example 10/19/2011 Norah Ali Almoneef 43

Notice in free fall 10/19/2011 Norah Ali Almoneef 44

Notice in free fall 10/19/2011 Norah Ali Almoneef 44

Word clues to numbers for problem solving • “free-fall” acceleration due to gravity a=9.

Word clues to numbers for problem solving • “free-fall” acceleration due to gravity a=9. 8 m/s 2, down • “at rest” not moving v=0 • “dropped” starts at rest and free-fall vi=0 and a=9. 81 m/s 2, down • “constant velocity” no acceleration a=0 • “stops” final velocity is zero vf=0 10/19/2011 Norah Ali Almoneef 45

Summary 1. Displacement: 2. Average velocity: 4. Average acceleration : 10/19/2011 2. Time interval:

Summary 1. Displacement: 2. Average velocity: 4. Average acceleration : 10/19/2011 2. Time interval: 3. Instantaneou s velocity: 5. nstantaneous acceleration: Norah Ali Almoneef 46