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Physics 1501 Lecture 2 Announcements l l l Lectures available on the web (short Physics 1501 Lecture 2 Announcements l l l Lectures available on the web (short version) For over-enrollment please go to the Physics office P 107 No Labs this week Register on Mastering Physics Options for access numbers: ç If bought a new book at the Coop: included ç If not, exchange it at the Coop (if possible) ç If 1 st ed. , two options: » Buy access code only: without e. Text $56. 30 USD Buy access code only: » Buy access code only: with e. Text $97. 80 USD Buy access code only: Physics 1501: Lecture 2, Pg 1

Physics 1501 Lecture 2 Today’s Topics: • • Units + significant digits (Chapter 1) Physics 1501 Lecture 2 Today’s Topics: • • Units + significant digits (Chapter 1) Motion in One-Dimension (Kinematics) (Chapter 2) ç Displacement, Velocity, Acceleration ç Average vs Instantaneous quantities ç Free Fall Physics 1501: Lecture 2, Pg 2

Working with Numbers l Scientific Notation ç Often times in physics, we deal with Working with Numbers l Scientific Notation ç Often times in physics, we deal with numbers that are really large (distance from the earth to the sun) or numbers that are really small (distance of an electron from the nucleus). ç To handle such numbers, we use scientific notation: » Numbers are expressed as a “reasonable number” multiplied by a factor of 10 ç Examples: » 51985 = 5. 1985 × 104 » 67, 860, 000 = 6. 786 × 1010 » . 0000239 = 2. 39 × 10 -5 Physics 1501: Lecture 2, Pg 3

Working with Numbers l Significant Figures ç The precision of our measurement is implied Working with Numbers l Significant Figures ç The precision of our measurement is implied in the amount of significant figures used (the “reasonable number” mentioned on previous slide). ç In this age of calculators and computers, it is easy to lose sight of which digits are significant and which are not. Physics 1501: Lecture 2, Pg 4

Order of Magnitude Calculations / Estimates Earth’s radius ? l Need to know something Order of Magnitude Calculations / Estimates Earth’s radius ? l Need to know something from your experience: çFlying from NYC to SF one accumulates ~ 3, 500 miles çNYC to SF spans about 1/6 of the Earth’s circumference çSo, the Earth’s circumference L = 3, 500 x 6 ~ 20, 000 mi çSince circumference of a circle is : L = 2 r çEstimate of Earth radius : 3 x 103 mi = 3 x 103 x 1. 61 km ~ 5 x 103 km = 5 x 106 m Physics 1501: Lecture 2, Pg 5

Dimensional Analysis l This is a very important tool to check your work çIt’s Dimensional Analysis l This is a very important tool to check your work çIt’s also very easy! l Example: Doing a problem you get the answer for distance d = v t 2 ( velocity x time 2 ) Quantity on left side = L Quantity on right side = L / T x T 2 = L x T l Left units and right units don’t match, so answer must be wrong !! Physics 1501: Lecture 2, Pg 6

Lecture 2, ACT 1 l There is a famous Einstein's equation connecting energy and Lecture 2, ACT 1 l There is a famous Einstein's equation connecting energy and mass (relativistic). Using dimensional analysis find which is the correct form of this equation : (a) l (b) (c) Note : ç c is speed of light (L/T) ç E is energy (M L 2 / T 2) Solution -> (b) Physics 1501: Lecture 2, Pg 7

Significant Figures l l l The number of digits that matter in a measurement Significant Figures l l l The number of digits that matter in a measurement or calculation. When writing a number, all non-zero digits are significant. Zeros may or may not be significant. ç those used to position the decimal point are not significant. ç those used to position powers of ten ordinals may or may not be significant. l l in scientific notation all digits are significant Examples: ç 2 1 sig fig ç 40 ambiguous, could be 1 or 2 sig figs ç 4. 0 x 101 2 sig figs ç 0. 0031 2 sig figs ç 3. 03 3 sig figs Physics 1501: Lecture 2, Pg 8

Significant Figures l l l When multiplying or dividing, the answer should have the Significant Figures l l l When multiplying or dividing, the answer should have the same number of significant figures as the least accurate of the quantities in the calculation. When adding or subtracting, the number of digits to the right of the decimal point should equal that of the term in the sum or difference that has the smallest number of digits to the right of the decimal point. Examples: ç 2 x 3. 1 = 6 ç 3. 1 + 0. 004 = 3. 1 ç 4. 0 x 101 2. 04 x 102 = 1. 6 X 10 -1 Physics 1501: Lecture 2, Pg 9

Strategies for Learning Physics Most of the ideas and concepts you will learn in Strategies for Learning Physics Most of the ideas and concepts you will learn in this class are not inherently difficult. l The challenges of this class will be in the problem solving: understanding what exactly is being asked of you and figuring out how you can derive the solution. l Your textbook introduces a problem-solving strategy named IDEA (Interpret Develop Evaluate and Assess). ç More later l Physics 1501: Lecture 2, Pg 10

Chap. 2: Mechanics l l l The study of motion ç Kinematics – study Chap. 2: Mechanics l l l The study of motion ç Kinematics – study of motion without regard to its cause. First, let’s learn “proper” physical definitions ç Displacement, velocity, acceleration ç Average vs. instantaneous We start simple – one body moving along a straight line Physics 1501: Lecture 2, Pg 11

Coordinate Systems l l Recall from Chapter 1 that we aim for detailed descriptions Coordinate Systems l l Recall from Chapter 1 that we aim for detailed descriptions in physics – this starts with location Think about when you give directions to someone: ç Always a starting (reference) point ç Directions are based from that starting point A coordinate system gives us a convenient way to describe motion by assigning an artificial reference point (origin) This allows us to describe the exact position of an object relative to this origin Physics 1501: Lecture 2, Pg 12

Coordinate Systems l The most common is the Cartesian coordinate system: x, y, (z) Coordinate Systems l The most common is the Cartesian coordinate system: x, y, (z) l Where we center the origin is completely up to us! Normally, the problem will dictate where to center it ç What is the most convenient for us! ç There is no right or wrong as long as you are consistent! l Physics 1501: Lecture 2, Pg 13

Scalars vs. Vectors l l In chapter 3, we will discuss vectors in detail Scalars vs. Vectors l l In chapter 3, we will discuss vectors in detail A scalar is a mathematical quantity that is described only by its magnitude (size) ç e. g. time, distance traveled, speed A vector is a mathematical quantity that is described by both magnitude and direction ç e. g. displacement, velocity, acceleration Note the differences between distance/displacement and speed/velocity! Physics 1501: Lecture 2, Pg 14

Motion in One-Dimension (Kinematics) Position / Displacement l Position is measured from an origin: Motion in One-Dimension (Kinematics) Position / Displacement l Position is measured from an origin: 10 meters çJoe is 10 feet to the right of the lamp çorigin = lamp çdirection = to the right çposition vector : -x +x 10 meters O Joe Physics 1501: Lecture 2, Pg 15

Position / Displacement is just change in position. ç X = xf - xi Position / Displacement is just change in position. ç X = xf - xi 15 meters 10 meters xi O xf Joe { l x = 5 meters Physics 1501: Lecture 2, Pg 16

Average speed and velocity • l Average velocity = total distance covered per total Average speed and velocity • l Average velocity = total distance covered per total time, Speed is just the magnitude of velocity. ç The “how fast” without the direction. • Instantaneous velocity, velocity at a given instant • Active Figure 1 Physics 1501: Lecture 2, Pg 17

Lecture 2, ACT 2 Average Velocity x (meters) 6 4 2 0 -2 1 Lecture 2, ACT 2 Average Velocity x (meters) 6 4 2 0 -2 1 2 3 4 t (seconds) What is the average velocity over the first 4 seconds ? A) -2 m/s B) 4 m/s C) 1 m/s D) not enough information to decide. Physics 1501: Lecture 2, Pg 18

Lecture 2, ACT 2 Average Velocity x (meters) 6 4 2 -2 1 2 Lecture 2, ACT 2 Average Velocity x (meters) 6 4 2 -2 1 2 3 4 t (seconds) What is the average velocity over the first 4 seconds ? x = 4 m, t = 4 s (C) is correct Physics 1501: Lecture 2, Pg 19

x (meters) Lecture 2, ACT 3 Instantaneous Velocity 6 4 2 -2 1 2 x (meters) Lecture 2, ACT 3 Instantaneous Velocity 6 4 2 -2 1 2 3 4 t (seconds) What is the instantaneous velocity at the fourth second ? A) 4 m/s B) 0 m/s C) 1 m/s D) not enough information to decide. Physics 1501: Lecture 2, Pg 20

x (meters) Lecture 2, ACT 2 Instantaneous Velocity 6 4 2 -2 1 2 x (meters) Lecture 2, ACT 2 Instantaneous Velocity 6 4 2 -2 1 2 3 4 t (seconds) What is the instantaneous velocity at the fourth second ? v = dx/dt = slope of x vs t graph At t = 0 line is flat, slope = 0 v = 0 (B) is correct Physics 1501: Lecture 2, Pg 21

Acceleration l Acceleration is change of velocity per time. ça = v/ t average Acceleration l Acceleration is change of velocity per time. ça = v/ t average ça = dv/dt = d 2 x/dt 2 instantaneous l Similarly, çv = a dt l Also, ç x = v dt displacement Physics 1501: Lecture 2, Pg 22

Recap l If the position x is known as a function of time, then Recap l If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time! x v a t t t Physics 1501: Lecture 2, Pg 23

1 -D Motion with constant acceleration l High-school calculus: l Also recall that l 1 -D Motion with constant acceleration l High-school calculus: l Also recall that l Since a is constant, we can integrate this using the above rule to find: • Similarly, since v = dx/dt we can integrate again to get: Physics 1501: Lecture 2, Pg 24

Recap l So for constant acceleration we find: x v a t t t Recap l So for constant acceleration we find: x v a t t t Physics 1501: Lecture 2, Pg 25

Derivation: l Solving for t: l Plugging in for t: Physics 1501: Lecture 2, Derivation: l Solving for t: l Plugging in for t: Physics 1501: Lecture 2, Pg 26

Average Velocity l Remember that v v vav v 0 t t Physics 1501: Average Velocity l Remember that v v vav v 0 t t Physics 1501: Lecture 2, Pg 27

Recap: l For constant acceleration: l From which we know: Active Figure 2 Physics Recap: l For constant acceleration: l From which we know: Active Figure 2 Physics 1501: Lecture 2, Pg 28

Lecture 2, Act 4 Motion in One Dimension l When throwing a ball straight Lecture 2, Act 4 Motion in One Dimension l When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path? (a) Both v = 0 and a = 0. (b) v 0, but a = 0. y (c) v = 0, but a 0. Physics 1501: Lecture 2, Pg 29

Lecture 2, Act 4 Solution l Going up the ball has positive velocity, while Lecture 2, Act 4 Solution l Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is momentarily zero. x l Since the velocity is continually changing there must t v be some acceleration. çIn fact the acceleration is caused t by gravity (g = 9. 81 m/s 2). ç(more on gravity in a few lectures) a The answer is (c) v = 0, but a 0. t Physics 1501: Lecture 2, Pg 30

Free Fall l When any object is let go it falls toward the ground Free Fall l When any object is let go it falls toward the ground !! The force that causes the objects to fall is called gravity. l The acceleration caused by gravity is typically written as g l Any object, be it a baseball or an elephant, experiences the same acceleration (g) when it is dropped, thrown, spit, or hurled, i. e. g is a constant. Physics 1501: Lecture 2, Pg 31

Gravity facts: l g does not depend on the nature of the material! çGalileo Gravity facts: l g does not depend on the nature of the material! çGalileo (1564 -1642) figured this out without fancy clocks & rulers! l demo - feather & penny in vacuum l Nominally, g = 9. 81 m/s 2 çAt the equator g = 9. 78 m/s 2 çAt the North pole g = 9. 83 m/s 2 l More on gravity in a few lectures! Physics 1501: Lecture 2, Pg 32

Problem: l On a bright sunny day you are walking around the campus watching Problem: l On a bright sunny day you are walking around the campus watching one of the many construction sites. To lift a bunch of bricks from a central area, they have brought in a helicopter. As the pilot is leaving, she accidentally releases the bricks when they are 1000 m above the ground. The worker below is getting ready to walk away in 10 seconds. Does he live? Physics 1501: Lecture 2, Pg 33

Problem Solution Method: Five Steps: (Simimilar to IDEA) 1) Focus the Problem - 2) Problem Solution Method: Five Steps: (Simimilar to IDEA) 1) Focus the Problem - 2) Describe the physics - 3) What are the relevant physics equations Execute the plan - 5) What physics ideas are applicable What are the relevant variables known and unknown Plan the solution - 4) Draw a picture – what are we asking for? Solve in terms of variables Solve in terms of numbers Evaluate the answer - Are the dimensions and units correct? Do the numbers make sense? Physics 1501: Lecture 2, Pg 34

Problem: 1. We need to find the time it takes for the brick to Problem: 1. We need to find the time it takes for the brick to hit the ground. 1000 m Physics 1501: Lecture 2, Pg 35

Problem: 2. Free Fall Constant acceleration at value g l Choose coordinate system 1000 Problem: 2. Free Fall Constant acceleration at value g l Choose coordinate system 1000 m l Variables: h – height of helicopter = 1000 m g – acceleration due to gravity = 9. 81 m/s 2 t – time to drop – goal y m – mass of brick – unknown vo – initial velocity = 0 why? y=0 Physics 1501: Lecture 2, Pg 36

Problem: 3. Next write position for constant acceleration: 1000 m Realize that v 0 Problem: 3. Next write position for constant acceleration: 1000 m Realize that v 0 y = 0 Solve for when y=0 • Why ? ? y y=0 Physics 1501: Lecture 2, Pg 37

Problem: 4. Substitute y=0 and vo=0 Solve for t in symbols: y 0 = Problem: 4. Substitute y=0 and vo=0 Solve for t in symbols: y 0 = 1000 m Solve for t in numbers: y y=0 Physics 1501: Lecture 2, Pg 38

Problem: 5. Does it make sense? ü Units worked out to be ‘s’ That’s Problem: 5. Does it make sense? ü Units worked out to be ‘s’ That’s correct for time ü y 0 = 1000 m It takes 14. 3 s. That means that the man escapes !!! y y=0 Physics 1501: Lecture 2, Pg 39

Tips: l Read ! çBefore you start work on a problem, read the problem Tips: l Read ! çBefore you start work on a problem, read the problem statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem. l Watch your units ! çAlways check the units of your answer, and carry the units along with your numbers during the calculation. Physics 1501: Lecture 2, Pg 40