1 Zuhairusnizam Md DarusPHYSICS I Zuhairusnizam Md Darus

2 093 B R E A K (18/7 – 25/7) RECOMMENDED TEXT:PHYSICS For Scientists & Engineers With Modern Physics by Giancoli, 4th Edition REFERENCES: Fundamental of Physics by Halliday, Resnick, Walker;6th or 7th Ed., John Wiley &Sons, Inc. Physical Quantities and Units Base Quantities and SI Units Significant Figures Conversion of Units Dimensional Analysis Scalars and Vectors Laws of Thermodyn Heat Capacity of Gases Work and Internal Energy First Law of Thermodynamics Second Law of Thermodynamics 1, 2 15 Ch 1, 3, 7 Ch 19, 20 ASSESSMENT: TESTS – 30% LAB REPORTS – 10% FINAL EXAM – 60 % Mechanics of Motion Motion with Constant Acceleration (1 – D) 3 Ch 2 Gases and Kinetic Theory Gas Laws and Absolute Temp Kinetic Theory of Gases 14 Mechanics of Motion Motion with Constant Acceleration (2 – D) 4 Ch 18 Ch 3 Newton’s Laws and Applications Circular Motion Uniform Circular Motion Centripetal and Angular Accn Centripetal Force Temperature and Heat Temp and Thermal Eqm Thermometers and Temp Scale Thermal Expansion of Solids n Liquids Heat 13 093 5 Ch 17 Ch 4, 5 12 States of Matter Solid – Stress & Strain Young’s Modulus Fluids – Density and Pressure Archimedes’ Principle Bernoulli’s Principle 6 Ch 12, 13 Ch 7, 8 Work and Energy Work by a Varying Force KE n W-KE Theorem PE Conservation of Energy 11 Gravitation Newton’s Law of Gravitation Gravitational Field Strength Gravitational Potential Realtionship bet g and G Satellite Motion in Cicular Orbits Escape velocity 7 Ch 6 9 Ch 9 Ch 10, 11 Rotational Motion Rotational Dynamics Angular Momentum Momentum, Impulse and Collissions 10 8 Ch 12 B R E A K (18/7 – 25/7) Zuhairusnizam Md Darus Phoe: Off : Unit Penerbitan Universiti (UPENA) Statics Equilibrium of Particles Free-body diagram Equilibrium of Rigid Bodies Prepared by Prof Madya Ahmad Abd Hamid, May 2010

3 LECTURES

4 Units of Chapter 1 The Nature of Science Models, Theories, and LawsMeasurement and Uncertainty; Significant Figures Units, Standards, and the SI System Converting Units Order of Magnitude: Rapid Estimating Dimensions and Dimensional Analysis Copyright © 2009 Pearson Education, Inc.

5 1-1 The Nature of Science Observation: important first step toward scientific theory; requires imagination to tell what is important Theories: created to explain observations; will make predictions Observations will tell if the prediction is accurate, and the cycle goes on. No theory can be absolutely verified, although a theory can be proven false. Copyright © 2009 Pearson Education, Inc.

6 1-1 The Nature of Science How does a new theory get accepted?Predictions agree better with data Explains a greater range of phenomena Example: Aristotle believed that objects would return to a state of rest once put in motion. Galileo realized that an object put in motion would stay in motion until some force stopped it. Copyright © 2009 Pearson Education, Inc.

7 1-1 The Nature of Science The principles of physics are used in many practical applications, including construction. Communication between architects and engineers is essential if disaster is to be avoided. Figure 1-1. Caption: (a) This Roman aqueduct was built 2000 years ago and still stands. (b) The Hartford Civic Center collapsed in 1978, just two years after it was built. Copyright © 2009 Pearson Education, Inc.

8 1-2 Models, Theories, and LawsModels are very useful during the process of understanding phenomena. A model creates mental pictures; care must be taken to understand the limits of the model and not take it too seriously. A theory is detailed and can give testable predictions. A law is a brief description of how nature behaves in a broad set of circumstances. A principle is similar to a law, but applies to a narrower range of phenomena. Copyright © 2009 Pearson Education, Inc.

9 1-3 Measurement and Uncertainty; Significant FiguresNo measurement is exact; there is always some uncertainty due to limited instrument accuracy and difficulty reading results. The photograph to the left illustrates this – it would be difficult to measure the width of this board more accurately than ± 1 mm. Figure 1-2. Caption: Measuring the width of a board with a centimeter ruler. The uncertainty is about ±1 mm. Copyright © 2009 Pearson Education, Inc.

10 1-3 Measurement and Uncertainty; Significant FiguresEstimated uncertainty is written with a ± sign; for example: 8.8 ± 0.1 cm. Percent uncertainty is the ratio of the uncertainty to the measured value, multiplied by 100: Copyright © 2009 Pearson Education, Inc.

11 1-3 Measurement and Uncertainty; Significant FiguresThe number of significant figures is the number of reliably known digits in a number. It is usually possible to tell the number of significant figures by the way the number is written: 23.21 cm has four significant figures. 0.062 cm has two significant figures (the initial zeroes don’t count). 80 km is ambiguous—it could have one or two significant figures. If it has three, it should be written 80.0 km. Copyright © 2009 Pearson Education, Inc.

12 1-3 Measurement and Uncertainty; Significant FiguresWhen multiplying or dividing numbers, the result has as many significant figures as the number used in the calculation with the fewest significant figures. Example: 11.3 cm x 6.8 cm = 77 cm. When adding or subtracting, the answer is no more accurate than the least accurate number used. The number of significant figures may be off by one; use the percentage uncertainty as a check. Copyright © 2009 Pearson Education, Inc.

13 1-3 Measurement and Uncertainty; Significant FiguresCalculators will not give you the right number of significant figures; they usually give too many but sometimes give too few (especially if there are trailing zeroes after a decimal point). The top calculator shows the result of 2.0/3.0. The bottom calculator shows the result of 2.5 x 3.2. Figure 1-3. Caption: These two calculators show the wrong number of significant figures. In (a), 2.0 was divided by 3.0. The correct final result would be In (b), 2.5 was multiplied by 3.2.The correct result is 8.0. Copyright © 2009 Pearson Education, Inc.

14 1-3 Measurement and Uncertainty; Significant FiguresConceptual Example 1-1: Significant figures. Using a protractor, you measure an angle to be 30°. (a) How many significant figures should you quote in this measurement? (b) Use a calculator to find the cosine of the angle you measured. Figure 1-4. Caption: Example 1–1. A protractor used to measure an angle. Response: (a) If you look at a protractor, you will see that the precision with which you can measure an angle is about one degree (certainly not 0.1°). So you can quote two significant figures, namely, 30° (not 30.0°). (b) If you enter cos 30° in your calculator, you will get a number like However, the angle you entered is known only to two significant figures, so its cosine is correctly given by 0.87; you must round your answer to two significant figures. Copyright © 2009 Pearson Education, Inc.

15 1-3 Measurement and Uncertainty; Significant FiguresScientific notation is commonly used in physics; it allows the number of significant figures to be clearly shown. For example, we cannot tell how many significant figures the number 36,900 has. However, if we write 3.69 x 104, we know it has three; if we write x 104, it has four. Much of physics involves approximations; these can affect the precision of a measurement also. Copyright © 2009 Pearson Education, Inc.

16 1-3 Measurement and Uncertainty; Significant FiguresAccuracy vs. Precision Accuracy is how close a measurement comes to the true value. Precision is the repeatability of the measurement using the same instrument. It is possible to be accurate without being precise and to be precise without being accurate! Copyright © 2009 Pearson Education, Inc.

17 1-4 Units, Standards, and the SI SystemQuantity Unit Standard Length Meter Length of the path traveled by light in 1/299,792,458 second Time Second Time required for 9,192,631,770 periods of radiation emitted by cesium atoms Mass Kilogram Platinum cylinder in International Bureau of Weights and Measures, Paris Copyright © 2009 Pearson Education, Inc.

18 1-4 Units, Standards, and the SI SystemFigure 1-5. Caption: Some lengths: (a) viruses (about 10-7 m long) attacking a cell; (b) Mt. Everest’s height is on the order of 104 m (8850 m, to be precise). Copyright © 2009 Pearson Education, Inc.

19 1-4 Units, Standards, and the SI SystemCopyright © 2009 Pearson Education, Inc.

20 1-4 Units, Standards, and the SI SystemCopyright © 2009 Pearson Education, Inc.

21 1-4 Units, Standards, and the SI SystemThese are the standard SI prefixes for indicating powers of 10. Many are familiar; yotta, zetta, exa, hecto, deka, atto, zepto, and yocto are rarely used. Copyright © 2009 Pearson Education, Inc.

22 1-4 Units, Standards, and the SI SystemWe will be working in the SI system, in which the basic units are kilograms, meters, and seconds. Quantities not in the table are derived quantities, expressed in terms of the base units. Other systems: cgs; units are centimeters, grams, and seconds. British engineering system has force instead of mass as one of its basic quantities, which are feet, pounds, and seconds. Copyright © 2009 Pearson Education, Inc.

23 1-5 Converting Units Unit conversions always involve a conversion factor. Example: 1 in. = 2.54 cm. Written another way: 1 = 2.54 cm/in. So if we have measured a length of 21.5 inches, and wish to convert it to centimeters, we use the conversion factor: Copyright © 2009 Pearson Education, Inc.

24 1-5 Converting Units Example 1-2: The 8000-m peaks.The fourteen tallest peaks in the world are referred to as “eight-thousanders,” meaning their summits are over 8000 m above sea level. What is the elevation, in feet, of an elevation of 8000 m? Figure 1-6. Caption: The world’s second highest peak, K2, whose summit is considered the most difficult of the “8000-ers.” K2 is seen here from the north (China). Answer: An elevation of 8000 m is 26,247 ft above sea level. Copyright © 2009 Pearson Education, Inc.

25 1-6 Order of Magnitude: Rapid EstimatingA quick way to estimate a calculated quantity is to round off all numbers to one significant figure and then calculate. Your result should at least be the right order of magnitude; this can be expressed by rounding it off to the nearest power of 10. Diagrams are also very useful in making estimations. Copyright © 2009 Pearson Education, Inc.

26 1-6 Order of Magnitude: Rapid EstimatingExample 1-5: Volume of a lake. Estimate how much water there is in a particular lake, which is roughly circular, about 1 km across, and you guess it has an average depth of about 10 m. Figure 1-7. Caption: Example 1–5. (a) How much water is in this lake? (Photo is of one of the Rae Lakes in the Sierra Nevada of California.) (b) Model of the lake as a cylinder. [We could go one step further and estimate the mass or weight of this lake. We will see later that water has a density 1000 kg/m3, of so this lake has a mass of about (103 kg/m3)(107 m3) ≈ 1010 kg, which is about 10 billion kg or 10 million metric tons. (A metric ton is 1000 kg, about 2200 lbs, slightly larger than a British ton, 2000 lbs.)] Answer: The volume of the lake is about 107 m3. Copyright © 2009 Pearson Education, Inc.

27 1-6 Order of Magnitude: Rapid EstimatingExample 1-6: Thickness of a page. Estimate the thickness of a page of your textbook. (Hint: you don’t need one of these!) Figure 1-8. Caption: Example 1–6. Micrometer used for measuring small thicknesses. Answer: Measure the thickness of 100 pages. You should find that the thickness of a single page is about 6 x 10-2 mm. Copyright © 2009 Pearson Education, Inc.

28 1-6 Order of Magnitude: Rapid EstimatingExample 1-7: Height by triangulation. Estimate the height of the building shown by “triangulation,” with the help of a bus-stop pole and a friend. (See how useful the diagram is!) Figure 1-9. Caption: Example 1–7. Diagrams are really useful! Answer: The building is about 15 m tall. Copyright © 2009 Pearson Education, Inc.

29 1-6 Order of Magnitude: Rapid EstimatingExample 1-8: Estimating the radius of Earth. If you have ever been on the shore of a large lake, you may have noticed that you cannot see the beaches, piers, or rocks at water level across the lake on the opposite shore. The lake seems to bulge out between you and the opposite shore—a good clue that the Earth is round. Suppose you climb a stepladder and discover that when your eyes are 10 ft (3.0 m) above the water, you can just see the rocks at water level on the opposite shore. From a map, you estimate the distance to the opposite shore as d ≈ 6.1 km. Use h = 3.0 m to estimate the radius R of the Earth. Figure Caption: Example 1–8, but not to scale. You can see small rocks at water level on the opposite shore of a lake 6.1 km wide if you stand on a stepladder. Answer: This calculation gives the radius of the Earth as 6200 km (precise measurements give 6380 km). Copyright © 2009 Pearson Education, Inc.

30 1-7 Dimensions and Dimensional AnalysisDimensions of a quantity are the base units that make it up; they are generally written using square brackets. Example: Speed = distance/time Dimensions of speed: [L/T] Quantities that are being added or subtracted must have the same dimensions. In addition, a quantity calculated as the solution to a problem should have the correct dimensions. Copyright © 2009 Pearson Education, Inc.

31 1-7 Dimensions and Dimensional AnalysisDimensional analysis is the checking of dimensions of all quantities in an equation to ensure that those which are added, subtracted, or equated have the same dimensions. Example: Is this the correct equation for velocity? Check the dimensions: Wrong! Copyright © 2009 Pearson Education, Inc.

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33 Summary of Chapter 1 Theories are created to explain observations, and then tested based on their predictions. A model is like an analogy; it is not intended to be a true picture, but to provide a familiar way of envisioning a quantity. A theory is much more well developed, and can make testable predictions; a law is a theory that can be explained simply, and that is widely applicable. Dimensional analysis is useful for checking calculations. Copyright © 2009 Pearson Education, Inc.

34 Summary of Chapter 1 Measurements can never be exact; there is always some uncertainty. It is important to write them, as well as other quantities, with the correct number of significant figures. The most common system of units in the world is the SI system. When converting units, check dimensions to see that the conversion has been done properly. Order-of-magnitude estimates can be very helpful. Copyright © 2009 Pearson Education, Inc.

35 Chapter 3

36 Chapter 3 Vectors

37 3-1 Vectors and Scalars A vector has magnitude as well as direction.Some vector quantities: displacement, velocity, force, momentum A scalar has only a magnitude. Some scalar quantities: mass, time, temperature

38 3-2 Addition of Vectors—Graphical MethodsFor vectors in one dimension, simple addition and subtraction are all that is needed. You do need to be careful about the signs, as the figure indicates.

39 3-2 Addition of Vectors—Graphical MethodsIf the motion is in two dimensions, the situation is somewhat more complicated. Here, the actual travel paths are at right angles to one another; we can find the displacement by using the Pythagorean Theorem.

40 3-2 Addition of Vectors—Graphical MethodsAdding the vectors in the opposite order gives the same result:

41 3-2 Addition of Vectors—Graphical MethodsEven if the vectors are not at right angles, they can be added graphically by using the tail-to-tip method.

42 3-2 Addition of Vectors—Graphical MethodsThe parallelogram method may also be used; here again the vectors must be tail-to-tip.

43 3-3 Subtraction of Vectors, and Multiplication of a Vector by a ScalarIn order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction. Then we add the negative vector.

44 3-3 Subtraction of Vectors, and Multiplication of a Vector by a ScalarA vector can be multiplied by a scalar c; the result is a vector c that has the same direction but a magnitude cV. If c is negative, the resultant vector points in the opposite direction.

45 3-4 Adding Vectors by ComponentsAny vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other.

46 3-4 Adding Vectors by ComponentsIf the components are perpendicular, they can be found using trigonometric functions.

47 3-4 Adding Vectors by ComponentsThe components are effectively one-dimensional, so they can be added arithmetically.

48 3-4 Adding Vectors by ComponentsDraw a diagram; add the vectors graphically. Choose x and y axes. Resolve each vector into x and y components. Calculate each component using sines and cosines. Add the components in each direction. To find the length and direction of the vector, use: and .

49 3-4 Adding Vectors by ComponentsExample 3-2: Mail carrier’s displacement. A rural mail carrier leaves the post office and drives 22.0 km in a northerly direction. She then drives in a direction 60.0° south of east for 47.0 km. What is her displacement from the post office?

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51 3-4 Adding Vectors by ComponentsExample 3-3: Three short trips. An airplane trip involves three legs, with two stopovers. The first leg is due east for 620 km; the second leg is southeast (45°) for 440 km; and the third leg is at 53° south of west, for 550 km, as shown. What is the plane’s total displacement?

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53 3-5 Unit Vectors Unit vectors have magnitude 1.Using unit vectors, any vector can be written in terms of its components:

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57 3-6 Vector Kinematics In two or three dimensions, the displacement is a vector:

58 3-6 Vector Kinematics As Δt and Δr become smaller and smaller, the average velocity approaches the instantaneous velocity.

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60 7-2 Scalar Product of Two VectorsDefinition of the scalar, or dot, product: Therefore, we can write: Figure 7-6. Caption: The scalar product, or dot product, of two vectors A and B is A·B = AB cos θ. The scalar product can be interpreted as the magnitude of one vector (B in this case) times the projection of the other vector, A cos θ, onto B.

61 7-2 Scalar Product of Two VectorsExample 7-4: Using the dot product. The force shown has magnitude FP = 20 N and makes an angle of 30° to the ground. Calculate the work done by this force, using the dot product, when the wagon is dragged 100 m along the ground. Figure 7-7. Caption: Example 7-4. Work done by a force FP acting at an angle θ to the ground is W = FP·d. Answer: W = 1700 J.

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63 11-2 Vector Cross Product; Torque as a VectorThe vector cross product is defined as: The direction of the cross product is defined by a right-hand rule: Figure The vector C = A x B is perpendicular to the plane containing A and B; its direction is given by the right-hand rule.

64 11-2 Vector Cross Product; Torque as a VectorThe cross product can also be written in determinant form:

65 11-2 Vector Cross Product; Torque as a VectorSome properties of the cross product: Figure The vector B x A equals –A x B.

66 11-2 Vector Cross Product; Torque as a VectorTorque can be defined as the vector product of the force and the vector from the point of action of the force to the axis of rotation: Figure The torque due to the force F (in the plane of the wheel) starts the wheel rotating counterclockwise so ω and α point out of the page.

67 11-2 Vector Cross Product; Torque as a VectorFor a particle, the torque can be defined around a point O: Here, is the position vector from the particle relative to O. Figure τ = r x F, where r is the position vector.

68 11-2 Vector Cross Product; Torque as a VectorExample 11-6: Torque vector. Suppose the vector is in the xz plane, and is given by = (1.2 m) + (1.2 m) Calculate the torque vector if = (150 N) . Solution: This can be done by the determinant method; the answer is τ = (180 m·N)j – it is in the y-direction, as expected.

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70 Describing Motion: Kinematics in One DimensionChapter 2 Describing Motion: Kinematics in One Dimension Chapter Opener. Caption: A high-speed car has released a parachute to reduce its speed quickly. The directions of the car’s velocity and acceleration are shown by the green (v) and gold (a) arrows. Motion is described using the concepts of velocity and acceleration. In the case shown here, the acceleration a is in the opposite direction from the velocity v, which means the object is slowing down. We examine in detail motion with constant acceleration, including the vertical motion of objects falling under gravity.

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72 2-1 Reference Frames and DisplacementAny measurement of position, distance, or speed must be made with respect to a reference frame. For example, if you are sitting on a train and someone walks down the aisle, the person’s speed with respect to the train is a few miles per hour, at most. The person’s speed with respect to the ground is much higher. Figure 2-2. Caption: A person walks toward the front of a train at 5 km/h. The train is moving 80 km/h with respect to the ground, so the walking person’s speed, relative to the ground, is 85 km/h.

73 2-1 Reference Frames and DisplacementWe make a distinction between distance and displacement. Displacement (blue line) is how far the object is from its starting point, regardless of how it got there. Distance traveled (dashed line) is measured along the actual path. Figure 2-4. Caption: A person walks 70m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east.

74 2-1 Reference Frames and DisplacementThe displacement is written: Left: Displacement is positive. Right: Displacement is negative. Figure 2-5. Caption: The arrow represents the displacement x2 – x1. Distances are in meters. Figure 2-6. Caption: For the displacement Δx = x2 – x1 = 10.0 m – 30.0 m, the displacement vector points to the left.

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76 2-2 Average Velocity Speed is how far an object travels in a given time interval: Velocity includes directional information:

77 2-2 Average Velocity Example 2-1: Runner’s average velocity.The position of a runner as a function of time is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, the runner’s position changes from x1 = 50.0 m to x2 = 30.5 m, as shown. What was the runner’s average velocity? Figure 2-7. Caption: Example 2–1. A person runs from x1 = 50.0 m to x2 = 30.5 m. The displacement is –19.5 m. Answer: Divide the displacement by the elapsed time; average velocity is m/s

78 2-2 Average Velocity Example 2-2: Distance a cyclist travels.How far can a cyclist travel in 2.5 h along a straight road if her average velocity is 18 km/h? Answer: distance is average velocity multiplied by time, or 45 km.

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80 2-3 Instantaneous VelocityThe instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short. Ideally, a speedometer would measure instantaneous velocity; in fact, it measures average velocity, but over a very short time interval. Figure 2-8. Caption: Car speedometer showing mi/h in white, and km/h in orange.

81 2-3 Instantaneous VelocityThe instantaneous speed always equals the magnitude of the instantaneous velocity; it only equals the average velocity if the velocity is constant. Figure 2-9. Caption: Velocity of a car as a function of time: (a) at constant velocity; (b) with varying velocity.

82 2-3 Instantaneous VelocityOn a graph of a particle’s position vs. time, the instantaneous velocity is the tangent to the curve at any point. Figure Caption: Graph of a particle’s position x vs. time .The slope of the straight line P1P2 represents the average velocity of the particle during the time interval Δt = t2 – t1. Figure Caption: Same position vs. time curve as in Fig. 2–10, but note that the average velocity over the time interval ti – t1 (which is the slope of P1Pi) is less than the average velocity over the time interval t2 – t1. The slope of the thin line tangent to the curve at point P1 equals the instantaneous velocity at time t1.

83 2-3 Instantaneous VelocityExample 2-3: Given x as a function of t. A jet engine moves along an experimental track (which we call the x axis) as shown. We will treat the engine as if it were a particle. Its position as a function of time is given by the equation x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. (a) Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s. (b) Determine the average velocity during this time interval. (c) Determine the magnitude of the instantaneous velocity at t = 500 s. Figure Caption: Example 2–3. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B. Solution: a. Using the equation given, at 3.00 s, the engine is at 21.7 m; at 5.00 s it is at 55.3 m, so the displacement is 33.6 m. b. The average velocity is the displacement divided by the time: 16.8 m/s. c. Take the derivative: v = dx/dt = 2At = 21.0 m/s.

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86 2-4 Acceleration Acceleration is the rate of change of velocity. Example 2-4: Average acceleration. A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration? Figure Caption: Example 2–4.The car is shown at the start with v1 = 0 at t1 = 0. The car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of our time interval, t2 = 5.0 s. We assume the acceleration is constant and equals 5.0 m/s2. The green arrows represent the velocity vectors; the length of each arrow represents the magnitude of the velocity at that moment. The acceleration vector is the orange arrow. Distances are not to scale. Solution: The average acceleration is the change in speed divided by the time, 5.0 m/s2.

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88 2-4 Acceleration Conceptual Example 2-5: Velocity and acceleration.If the velocity of an object is zero, does it mean that the acceleration is zero? (b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples. Solution: a. No; if this were true nothing could ever change from a velocity of zero! b. No, but it does mean the velocity is constant.

89 2-4 Acceleration Example 2-6: Car slowing down.An automobile is moving to the right along a straight highway, which we choose to be the positive x axis. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration? Figure Caption: Example 2–6, showing the position of the car at times t1 and t2, as well as the car’s velocity represented by the green arrows. The acceleration vector (orange) points to the left as the car slows down while moving to the right. Solution: The average acceleration is the change in speed divided by the time; it is negative because it is in the negative x direction, and the car is slowing down: a = -2.0 m/s2

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91 2-4 Acceleration There is a difference between negative acceleration and deceleration: Negative acceleration is acceleration in the negative direction as defined by the coordinate system. Deceleration occurs when the acceleration is opposite in direction to the velocity. Figure Caption: The car of Example 2–6, now moving to the left and decelerating. The acceleration is +2.0 m/s.

92 2-4 Acceleration The instantaneous acceleration is the average acceleration in the limit as the time interval becomes infinitesimally short. Figure Caption: A graph of velocity v vs. time t. The average acceleration over a time interval Δt = t2 – t1 is the slope of the straight line P1P2: aav = Δv/ Δt. The instantaneous acceleration at time is t1 the slope of the v vs. t curve at that instant.

93 2-4 Acceleration Example 2-7: Acceleration given x(t).A particle is moving in a straight line so that its position is given by the relation x = (2.10 m/s2)t2 + (2.80 m). Calculate (a) its average acceleration during the time interval from t1 = 3.00 s to t2 = 5.00 s, and (b) its instantaneous acceleration as a function of time. Figure Caption: Example 2–7. Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B. Note that increases linearly with and that the acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve. Solution: The velocity at time t is the derivative of x; v = (4.20 m/s2)t. a. Solve for v at the two times; a = 4.20 m/s2. b. Take the derivative of v: a = 4.20 m/s2.

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95 2-4 Acceleration Conceptual Example 2-8: Analyzing with graphs.This figure shows the velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s. Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars. Figure 2-19. Solution: a. Average acceleration is the same; both have the same change in speed over the same time. b. Car A accelerates faster than B at the beginning but then slower than B towards the end (look at the slope of the lines). c. Car A is always going faster than car B, so it will travel farther.

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97 2-5 Motion at Constant AccelerationThe average velocity of an object during a time interval t is The acceleration, assumed constant, is

98 2-5 Motion at Constant AccelerationIn addition, as the velocity is increasing at a constant rate, we know that Combining these last three equations, we find:

99 2-5 Motion at Constant AccelerationWe can also combine these equations so as to eliminate t: We now have all the equations we need to solve constant-acceleration problems.

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102 2-6 Solving Problems Read the whole problem and make sure you understand it. Then read it again. Decide on the objects under study and what the time interval is. Draw a diagram and choose coordinate axes. Write down the known (given) quantities, and then the unknown ones that you need to find. What physics applies here? Plan an approach to a solution.

103 2-6 Solving Problems 6. Which equations relate the known and unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, and check that your result is sensible (correct dimensions). 7. Calculate the solution and round it to the appropriate number of significant figures. 8. Look at the result—is it reasonable? Does it agree with a rough estimate? 9. Check the units again.

104 2-6 Solving Problems Example 2-10: Acceleration of a car.How long does it take a car to cross a 30.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m/s2? Figure 2-20. Solution: We are given the acceleration, the initial speed, and the distance, and are asked for the time. Substituting in the appropriate equation gives t = 5.48 s.

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106 2-6 Solving Problems Example 2-11: Air bags.Suppose you want to design an air bag system that can protect the driver at a speed of 100 km/h (60 mph) if the car hits a brick wall. Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver? Figure Caption: Example An air bag deploying on impact. Solution: Assume the acceleration is constant; the car goes from 100 km/h to zero in a distance of about 1 m (the crumple zone). This takes a time t = 0.07 s, so the air bag has to inflate faster than this. The seat belt keeps the driver in position, and also assures that the driver decelerates with the car, rather than by hitting the dashboard.

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108 2-6 Solving Problems Example 2-12: Braking distances. Estimate the minimum stopping distance for a car. The problem is best dealt with in two parts, two separate time intervals. (1) The first time interval begins when the driver decides to hit the brakes, and ends when the foot touches the brake pedal. This is the “reaction time,” about 0.50 s, during which the speed is constant, so a = 0. Figure Caption: Example 2–12: stopping distance for a braking car. Solution: In this first part, the speed is a constant 14 m/s, so in 0.50 s the car travels 7.0 m.

109 2-6 Solving Problems Example 2-12: Braking distances. (2) The second time interval is the actual braking period when the vehicle slows down (a ≠ 0) and comes to a stop. The stopping distance depends on the reaction time of the driver, the initial speed of the car (the final speed is zero), and the acceleration of the car. Calculate the total stopping distance for an initial velocity of 50 km/h (= 14 m/s ≈ 31 mi/h) and assume the acceleration of the car is -6.0 m/s2 (the minus sign appears because the velocity is taken to be in the positive x direction and its magnitude is decreasing). Figure Caption: Example 2–12. Graphs of (a) v vs. t and (b) x vs. t. Solution: We are given the initial and final speeds and the acceleration, and are asked for the distance. Substituting in the appropriate equation gives 16.0 m; adding the 7.0 m the car traveled before the driver hit the brakes gives a total distance of 23.0 m.

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112 2-6 Solving Problems Example 2-13: Two moving objects: Police and speeder. A car speeding at 150 km/h passes a still police car which immediately takes off in hot pursuit. Using simple assumptions, such as that the speeder continues at constant speed, estimate how long it takes the police car to overtake the speeder. Then estimate the police car’s speed at that moment and decide if the assumptions were reasonable. Figure 2-24. Solution: First, assume the speeder continues at a constant speed, and that the police car’s acceleration is constant. If a car can go from 0 to 100 km/h in 5 sec, this is an acceleration of 5.6 m/s2. Using this, we find that the police car’s speed when it catches up to the speeder is 84 m/s, about 300 km/h. Not a good idea. Probably the acceleration of the police car is not constant, and maybe the speeder slows down.

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114 2-7 Freely Falling ObjectsNear the surface of the Earth, all objects experience approximately the same acceleration due to gravity. This is one of the most common examples of motion with constant acceleration. Figure Caption: Multiflash photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.

115 2-7 Freely Falling ObjectsIn the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance. Figure Caption: (a) A ball and a light piece of paper are dropped at the same time. (b) Repeated, with the paper wadded up.

116 2-7 Freely Falling ObjectsThe acceleration due to gravity at the Earth’s surface is approximately 9.80 m/s2. At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration. Figure Caption: A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.

117 2-7 Freely Falling ObjectsExample 2-14: Falling from a tower. Suppose that a ball is dropped (v0 = 0) from a tower 70.0 m high. How far will it have fallen after a time t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s? Ignore air resistance. Figure Caption: Example 2–14. (a) An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. (See also Fig. 2–26.) (b) Graph of y vs. t. Solution: We are given the acceleration, the initial speed, and the time; we need to find the distance. Substituting gives t1 = 4.90 m, t2 = 19.6 m, and t3 = 44.1 m.

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119 2-7 Freely Falling ObjectsExample 2-15: Thrown down from a tower. Suppose a ball is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped. (a) What then would be its position after 1.00 s and 2.00 s? (b) What would its speed be after 1.00 s and 2.00 s? Compare with the speeds of a dropped ball. Solution: This is the same as Example 2-14, except that the initial speed is not zero. At t = 1.00 s, y = 7.90 m. At t = 2.00 s, y = 25.6 m. At t = 1.00 s, v = 12.8 m/s. At t = 2.00 s, v = 22.6 m/s. The speed is always 3.00 m/s faster than a dropped ball.

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121 2-7 Freely Falling ObjectsExample 2-16: Ball thrown upward, I. A person throws a ball upward into the air with an initial velocity of 15.0 m/s. Calculate (a) how high it goes, and (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance. Figure Caption: An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2–16, 2–17, 2–18, and 2–19. Solution: a. At the highest position, the speed is zero, so we know the acceleration, the initial and final speeds, and are asked for the distance. Substituting gives y = 11.5 m. b. Now we want the time; t = 3.06 s.

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123 2-7 Freely Falling ObjectsConceptual Example 2-17: Two possible misconceptions. Give examples to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction, and (2) that an object thrown upward has zero acceleration at the highest point. If acceleration and velocity were always in the same direction, nothing could ever slow down! At its highest point, the speed of thrown object is zero. If its acceleration were also zero, it would just stay at that point.

124 2-7 Freely Falling ObjectsExample 2-18: Ball thrown upward, II. Let us consider again a ball thrown upward, and make more calculations. Calculate (a) how much time it takes for the ball to reach the maximum height, and (b) the velocity of the ball when it returns to the thrower’s hand (point C). The time is 1.53 s, half the time for a round trip (since we are ignoring air resistance). v = m/s

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126 2-7 Freely Falling ObjectsExample 2-19: Ball thrown upward, III; the quadratic formula. For a ball thrown upward at an initial speed of 15.0 m/s, calculate at what time t the ball passes a point 8.00 m above the person’s hand. Figure Caption: Graphs of (a) y vs. t, (b) v vs. t for a ball thrown upward, Examples 2–16, 2–18, and 2–19. Solution: We are given the initial and final position, the initial speed, and the acceleration, and want to find the time. This is a quadratic equation; there are two solutions: t = 0.69 s and t = 2.37 s. The first is the ball going up and the second is the ball coming back down.

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129 2-7 Freely Falling ObjectsExample 2-20: Ball thrown upward at edge of cliff. Suppose that a ball is thrown upward at a speed of 15.0 m/s by a person standing on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. (a) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation). Figure Caption: Example 2–20. The person in Fig. 2–30 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below. Solution: a. We use the same quadratic formula as before, we find t = 5.07 s (the negative solution is physically meaningless). b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total distance of 73.0 m.

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132 Summary of Chapter 2 Kinematics is the description of how objects move with respect to a defined reference frame. Displacement is the change in position of an object. Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time. Instantaneous velocity is the average velocity in the limit as the time becomes infinitesimally short.

133 Summary of Chapter 2 Average acceleration is the change in velocity divided by the time. Instantaneous acceleration is the average acceleration in the limit as the time interval becomes infinitesimally small. The equations of motion for constant acceleration are given in the text; there are four, each one of which requires a different set of quantities. Objects falling (or having been projected) near the surface of the Earth experience a gravitational acceleration of 9.80 m/s2.