Interactive Engagement via Thumbs Up. Today s class. Next class. Chapter 2: Motion in 1D Example 2.10 and 2.11 Any Question.

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1 PHYS 01 Interactive Engagement via Thumbs Up 1 Chap.1 Sumamry Today s class SI units Dimensional analysis Scientific notation Errors Vectors Next class Chapter : Motion in 1D Example.10 and.11 Any Question from the previous lecture? Introduction

2 Chap. : Motion in 1D Kinematics is the study of motion. Velocity and acceleration are important physical quantities. To describe straight-line motion in terms of velocity and acceleration To distinguish between average and instantaneous velocity and average and instantaneous acceleration To interpret graphs of position versus time, velocity versus time, and acceleration versus time for straight-line motion To understand straight-line motion with constant acceleration To examine freely falling bodies To analyze straight-line motion when the acceleration is not constant 3 Straight-Line Motion ABC 1. Assume an object as a particle. Be familiar with the terminologies: Displacement: x [ Change in Position ] Velocity: v [ Rate Change in Displacement ] Acceleration: a [ Rate Change in Velocity ] a) Motion with zero acceleration b) Motion with non-zero acceleration 3. Calculate the slopes in x-t, v-t, a-t graphs Review: Sec..1,.,.3,.4 for terminologies 4

3 Anatomy of Exam 1, Problem 5 Exam 1, Problem - Solution 6

4 Average and Instantaneous Velocities In this example, the cheetah s instantaneous velocity increases with time (= accelerating or speeding up ) The figure illustrates how these quantities are related: A particle moving along the x-axis has a coordinate x. The change in the particle s coordinate (displacement) is x = x x 1. Average and instantaneous x-velocity: v av-x = x/ t and v x = dx/dt. Average and instantaneous acceleration: a av-x = v x / t and a x = dv x /dt Instantaneous velocity Average velocity 7 Finding velocity on an x-t graph At any point on an x-t graph, the x-component of instantaneous velocity (v x ) is equal to the slope of the tangent to the curve at that point (P 1 ). The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by v x = dx/dt. The average speed is not the magnitude of the average velocity! 8

5 Finding acceleration on a v x -t graph The v x -t graph may be used to find the instantaneous acceleration (a x ) and the average acceleration. a av-x = v x / t a x = dv x /dt 9 Equations of Motion with Constant Acceleration The four equations shown to the right apply to any straight-line motion with constant acceleration a x. Problem-Solving Strategy 1) Identify the nature of the problem ) Set up the problem 3) Execute the solution. 4) Evaluate the answer. ISEE 10

6 A motorcycle with constant acceleration Problem-Solving Strategy for an accelerating motorcycle. 11 A motorcycle with constant acceleration Problem-Solving Strategy for an accelerating motorcycle. Given! 1

7 KEY: Freely Falling Bodies Free fall is the motion of an object under the influence of only gravity. In the figure, a strobe light flashes with equal time intervals between flashes. The velocity change is the same in each time interval, so the acceleration is constant. Aristotle thought that heavy bodies fall faster than light ones, but Galileo showed that all bodies fall at the same rate. If there is no air resistance, the downward acceleration of any freely falling object is g = 9.8 m/s = 3 ft/s. Follow Example for a coin dropped from the Leaning Tower of Pisa. Freely Falling Bodies Free fall is the motion of an object under the influence of only gravity. In the figure, a strobe light flashes with equal time intervals between flashes. The velocity change is the same in each time interval, so the acceleration is constant. Aristotle thought that heavy bodies fall faster than light ones, but Galileo showed that all bodies fall at the same rate. If there is no air resistance, the downward acceleration of any freely falling object is g = 9.8 m/s = 3 ft/s. Follow Example for a coin dropped from the Leaning Tower of Pisa

8 Problem 1 A -euro coin is dropped from the Leaning Tower of Pisa. It starts from rest at height 50.0 m and falls freely. a) Compute its position after.00 s. b) Find the velocity after.00 s. c) Find the time at which the coin hits the ground. d) Find the velocity just before the coin hits the ground. e) Sketch a y-t graph. 15 ISEE Problem 1 Solution D.A.D. Identify unknowns! y a y = 9.80 m/s y 0 = 50.0 m v y0 = 0.00 m/s y =? v y t =.00 s 50.0 m 0 m unknowns equations if t is given. 16

9 y = y 0 + v y0 t + ½ a y t? v y = v y0 + a y t? v y = v y0 + a y (y y 0 )? ISEE Problem 1 Solution (Cont d) a y = 9.80 m/s y 0 = 50.0 m v y0 = 0.00 m/s 17 ISEE Problem 1 Solution (Cont d) y = y 0 + v y0 t + ½ a y t y = ½ ( 9.80) t (1) v y = v y0 + a y t v y = ( 9.80) t () v y = v y0 + a y (y y 0 ) v y = ( 9.80) (y 50.0) (3) (a)eq. 1 (b)eq. (c) Eq. 1 (d)eq. 3 OR Eq. a) Compute its position after.00 s. b) Find the velocity after.00 s. c) Find the time at which the coin hits the ground. d) Find the velocity just before the coin hits the ground. e) Sketch a y-t graph. 18

10 ISEE Problem 1 Solution (Cont d) y = y 0 + v y0 t + ½ a y t y = ½ ( 9.80) t (1) v y = v y0 + a y t v y = ( 9.80) t () v y = v y0 + a y (y y 0 ) v y = ( 9.80) (y 50.0) (3) (a)eq. 1 t =.00 s, thus y = 30.4 m (b)eq. v y = 19.6 m/s (c) Eq. 1 0 = ½ ( 9.80) t, thus, t = 3.19 s (d)eq. 3 y = 0 v y = x m/s or 31.3 m/s OR Eq. v y = ( 9.80)(3.19) = 31.3 m/s y direction] 19 Problem 1 [What If] A -euro coin is dropped from the Leaning Tower of Pisa. It starts from rest at height 50.0 m and falls freely. a) Compute its position after.00 s. b) Find the velocity after.00 s. c) Find the time at which the coin hits the ground. d) Find the velocity just before the coin hits the ground. e) Sketch a y-t graph. See Appendix 0

11 Up-and-down motion in free fall An object is in free fall even when it is moving upward. The vertical velocity, but not the acceleration, is zero at the highest point Follow Example for up-and-down motion. 1 Graphical Check (I)

12 Graphical Check (II) 3 You can find the velocity at t = 4.00 s in the same manner as in the previous example. Problem 4

13 Old Exam 1 for Practice a) 15.8 m/s b) 3.4 m/s c) Do you see the same setting? *) 15 min. *) similar to the HW problems *) a similar figure was used. *) modified to have 3 parts. 5 Today s class Sec..1~.6 Displacement, v, a Change in time slope Motion with constant acceleration Equations of motion Examples Summary of Chap. Next class Chapter 3: Motion in D Example 3.04 and 3.07 Introduction 6

14 Displacement, Time, and Average Velocity The figure illustrates how these quantities are related: A particle moving along the x-axis has a coordinate x. The change in the particle s coordinate is x = x x 1. The average x-velocity of the particle is v av-x = x/ t. x x v av-x 7 Negative Velocity The average x-velocity is negative during a time interval if the particle moves in the negative x-direction for that time interval. The figure illustrates this situation. 8

15 Position-Time Graph A position-time graph (an x-t graph) shows the particle s position x as a function of time t. The figure shows how the average x-velocity (v av-x ) is related to the slope ( x/ t)of an x-t graph. v av-x = x/ t 9 Finding acceleration on a v x -t graph The v x -t graph may be used to find the instantaneous acceleration (a x ) and the average acceleration. a av-x = v x / t a x = dv x /dt 30

16 Average and Instantaneous Velocities In this example, the cheetah s instantaneous velocity increases with time. 31 Average and Instantaneous Velocities In this example, the cheetah s instantaneous velocity increases with time. Average velocity 3

17 Average and Instantaneous Velocities In this example, the cheetah s instantaneous velocity increases with time. Instantaneous velocity Average velocity 33 Motion Diagrams A motion diagram shows the position of a particle at various instants, and arrows represent its velocity at each instant. The figure shows the x-t graph and the motion diagram for a moving particle. 34

18 Motion Diagrams with v x -t graph The figure shows the v x -t graph and the motion diagram for a particle. a x < 0 : negative acceleration 35 Motion with Negative Acceleration For a particle with negative acceleration, the velocity changes throughout the motion to the y direction. a x < 0 : negative acceleration 36

19 Motion with Constant Acceleration For a particle with constant acceleration, the velocity changes at the same rate throughout the motion. 37 Where should the professor be when you release the egg? Problem Egg Drop 38

20 Where should the professor be when you release the egg? Problem Egg Drop What should we do? d =? 0 m 39 ISEE Typical Two 1-D Motion Problem (a) Draw a diagram (b) Motion with constant acceleration (c) bodies, so sets of eqs. Write down kinematic eqs. Solve the equations. 40

21 Kinematic Eqs. for Motion with Constant Acceleration Kinematic Eqs. are related by derivatives and integrals. x-t x = x 0 + v 0x t + ½ a x t (1) v x = v 0x + a x t () v x = v 0x + a x (x x 0 ) (3) y-t y = y 0 + v y0 t + ½ a y t (1) v y = v 0y + a y t () v y = v 0y + a y (y y 0 ) (3) [Note] Eq. (3) can be obtained from (1) and () by eliminating t. 41 Where should the professor be when you release the egg? Problem Egg Drop Two 1-D Kinematic Equations for motions of two bodies with constant acceleration y = y 0 + v 0y t + ½ a y t x = x 0 + v 0x t + ½ a x t d =? 0 m Kinematics (D) 4

22 Where should the professor be when you release the egg? ISEE DaD at t =0 a y = 9.80 m/s y 0 = 46.0 m v 0y = 0.00 m/s y 46.0 m a x = 0 m/s x 0 =? m v 0x = 1.0 m/s 1.8 m d =? 0 m x 43 ISEE ISEE 44

23 ISEE ISEE 45 ISEE ISEE 46

24 Two bodies with different accelerations Follow the figure in which the police officer and motorist have different accelerations. Two lines 47 Diagnostic Test To spot misunderstanding 48

25 Motion with Constant Acceleration y v y Diagnostic Test v y -t graph t A B C 49 Motion with Constant Acceleration y v y Diagnostic Test v y -t graph t A B C 50

26 Diagnostic Test Motion with Constant Acceleration y a y a y -t graph t A B C 51 Diagnostic Test Motion with Constant Acceleration y a y a y -t graph t a y = 9.8 m/s A B C 5

27 Diagnostic Test Motion with Constant Acceleration a y a y -t graph t y A B C 53 Diagnostic Test Motion with Constant Acceleration a y a y -t graph a y = +9.8 m/s t y A B C 54

28 Appendix 55 Problem 1 [What If] A -euro coin is dropped from the Leaning Tower of Pisa. It starts from rest at height 50.0 m and falls freely. a) Compute its position after.00 s. b) Find the velocity after.00 s. c) Find the time at which the ball hits the ground. d) Find the velocity just before the ball hits the ground. e) Sketch a y-t graph. See Appendix 56

29 ISEE Problem 1 Solution (Cont d) D.A.D. Identify unknowns! a y = 9.80 m/s y 0 = 0.0 m v y0 = 0.00 m/s 0 m y =? v y t =.00 s 50.0 m y unknowns equations if t is given. 57 ISEE y = y 0 + v y0 t + ½ a y t y = ½ ( 9.80) t (1) v y = v y0 + a y t v y = ( 9.80) t () v y = v y0 + a y (y y 0 ) v y = ( 9.80) (y 0.0) (3) (a)eq. 1 (b)eq. (c) Eq. 1 (d)eq. 3 OR Eq. 58

30 ISEE y = y 0 + v y0 t + ½ a y t y = ½ ( 9.80) t (1) v y = v y0 + a y t v y = ( 9.80) t () v y = v y0 + a y (y y 0 ) v y = ( 9.80) (y 0.0) (3) (a)eq. 1 t =.00 s, thus y = 19.6 m (was 30.4 m) (b)eq. v y = 19.6 m/s (was 19.6 m/s) (c) Eq = ½ ( 9.80) t, thus, t = 3.19 s (d)eq. 3 y = 50.0 v y = 31.3 x m/s or m/s OR Eq. v y = ( 9.80)(3.19) = 31.3 m/s y direction] 59 More Examples 60

31 Problem I-: A person standing at the edge of a cliff throws a ball vertically upward with an initial speed of v 0 = 15.0 m/s from the edge of a cliff that is h = 40.0 m above the ground. The acceleration due to gravity is g = 9.80 m/s pointing down. Ignore air friction. a. (5 pts) How long does it take the ball to reach the ground? b. (5 pts) What is the speed of the ball just before it strikes the ground? c. (10 pts) Sketch y-t, v y -t, and a y -t graphs for the motion. d. (5 pts) If another ball is thrown vertically downward with the same initial speed, the ball to hit the ground with the greater speed is the one initially thrown: (i) upward. (ii) downward. (iii) neither they both hit at the same speed. h = 40.0 m 61 ISEE Hints for Problem I- h = 40.0 m 6

32 ISEE ISEE ISEE ISEE ISEE 63 Problem I-3: A person standing at the edge of a cliff throws a ball vertically upward with an initial speed of v 0 = 15.0 m/s from the edge of a cliff that is h = 40.0 m above the ground. The acceleration due to gravity is g = 9.80 m/s pointing down. Ignore air friction. a. (5 pts) How long does it take the ball to reach the ground? b. (5 pts) What is the speed of the ball just before it strikes the ground? c. (10 pts) Sketch y-t, v y -t, and a y -t graphs for the motion. d. (5 pts) If another ball is thrown vertically downward with the same initial speed, the ball to hit the ground with the greater speed is the one initially thrown: (i) upward. (ii) downward. (iii) neither they both hit at the same speed. h = 40.0 m y y = h y = 0 64

33 Problem I-5: Using v x = v 0x + a x (x x 0 ) t 0 v 0x Motion with constant acc. (negative : deacceleration) v x t x x 0 x Q: Braking Distance? 65 Problem I-6 An antelope moving with constant acceleration covers the distance between two points that are 80.0 m apart in 7.00s. Its speed as it passes the second point is 15.0 m/s. 0) Have a diagram a) What is the speed at the first point? b) What is the acceleration? 66

34 Problem I-6 Solution D.A.D. Identify unknowns! t=0 a x = constant =? t=7.00 s v x0 =? m/s 80.0 m v x =15.0 m/s unknowns equations v x = v 0x + a x t 15.0 = v 0x a x x = x 0 + v 0x t + ½ a x t 80.0 = v 0x x (7.00) + ½ a x (7.00) Solve equations simultaneously! Eliminating a x, v 0x = 7.86 m/s Eliminating v 0x, a x = 1.0 m/s 67 Problem I-7: How Many g s? A sports car is advertised to be able to stop in a distance of 55.0 m from a speed of 100 km/h. a) What is its acceleration in m/s? b) How many g s is this? 1) Conversion: 100 km/h?? m/s ) Use v x v 0x = a x (x x 0 ) to find a x. 3) a x /g where g = 9.80 m/s?? 68

35 Problem I-7: How Many g s? A sports car is advertised to be able to stop in a distance of 55.0 m from a speed of 100 km/h. a) What is its acceleration in m/s? b) How many g s is this? 1) Conversion: 100 km/h?? m/s ) Use v x v 0x = a x (x x 0 ) to find a x. 3) a x /g where g = 9.80 m/s 7.8 m/s 7.03 m/s Problem II-1 (-body problem): A police car at rest, passed by a speeder traveling at a constant 100 km/hr, takes off in hot pursuit. The police officer catches up to the speeder in 800 m, maintaining a constant acceleration. a) How long did it take the police officer to overtake the speeder? b) What was the acceleration of the police car? c) What was the speed of the police car at the overtaking point? ISEE v xs0 = 7.8 m/s D.A.D d = 800 m Hint: Kinematic eqs. for each body. 70

36 ISEE Problem II-1 Solution x = x 0 + v x0 t + ½ a x t d = ½ a p t (1) v x = v x0 + a x t v xp = a p t () v x = v x0 + a x (x x 0 ) v xp = a p d (3) 71 ISEE Problem II-1 Solution x = x 0 + v x0 t + ½ a x t d = ½ a p t (1) v x = v x0 + a x t v xp = a p t () v x = v x0 + a x (x x 0 ) v xp = a p d (3) d = v xs0 t (4) v xs = v xs0 (5) v xs = v xs0 (6) 7

37 ISEE Problem II-1 Solution x = x 0 + v x0 t + ½ a x t d = ½ a p t (1) v x = v x0 + a x t v xp = a p t () v x = v x0 + a x (x x 0 ) v xp = a p d (3) d = v xs0 t (4) v xs = v xs0 (5) v xs = v xs0 (6) Eq. (4) t = d / v xs0 = 8.8 s (7) Eq. (7) Eq. (1) : a p = d / ( d / v xs0 ) = ( v xs0 ) / d = 1.93 m/s (8) 73 Diagnostic Test some 74

38 Diagnostic Test Ready? Quick Quiz: Which car s slope (f) increases in magnitude? (g) decreases in magnitude? some C B A B C Ready for Quick Quiz? 75 Straight-line Parabola Diagnostic Test Describe the motion from the graph: Look at slope Speed Zero? Constant? Increase? Decrease? v-t graph 76

39 Student A s Graph How do you grade this (0-10 scale)? arbitrary scale v t 77 Student B s Graph How do you grade this (0-10 scale)? 78

40 Student C s Graph How do you grade this (0-10 scale)? 79 arbitrary scale x v t 80

41 v a t t 81 Problem I-G1 8

42 arbitrary scale 83 Problem I-G 84