Physics For Scientists and Engineers PDF

3/3/2023 Physics For Scientists and Engineers Fifth Edition, Global Edition Chapter 2 Kinematics in One Dimension Copyright © 2023 Pearson Education Ltd. All Rights Reserved....

3/3/2023 Physics For Scientists and Engineers Fifth Edition, Global Edition Chapter 2 Kinematics in One Dimension Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-1 Chapter 2 Kinematics in One Dimension IN THIS CHAPTER, you will learn to solve problems about motion along a straight line. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-2 1  3/3/2023 Chapter 2 Preview (1 of 6) What is kinematics? Kinematics is the mathematical description of motion. We begin with motion along a straight line. Our primary tools will be an object’s position, velocity, and acceleration. ❮❮ LOOKING BACK Sections 1.4–1.6 Velocity, acceleration, and Tactics Box 1.4 about signs Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-3 Chapter 2 Preview (2 of 6) How are graphs used in kinematics? Graphs are a very important visual representation of motion, and learning to ―think graphically‖ is one of our goals. We’ll work with graphs showing how position, velocity, and acceleration change with time. These graphs are related to each other: Velocity is the slope of the position graph. Acceleration is the slope of the velocity graph. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-4 2  3/3/2023 Chapter 2 Preview (3 of 6) How is calculus used in kinematics? Motion is change, and calculus is the mathematical tool for describing a quantity’s rate of change. We’ll find that Velocity is the time derivative of position. Acceleration is the time derivative of velocity. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-5 Chapter 2 Preview (4 of 6) What are models? A model is a simplified description of a situation that focuses on essential features while ignoring many details. Models allow us to make sense of complex situations by seeing them as variations on a common theme, all with the same underlying physics. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-6 3  3/3/2023 Chapter 2 Preview (5 of 6) What is free fall? Free fall is motion under the influence of gravity only. Free fall is not literally ―falling‖ because it also applies to objects thrown straight up and to projectiles. Surprisingly, all objects in free fall, regardless of their mass, have the same acceleration. Motion on a frictionless inclined plane is closely related to free-fall motion. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-7 Chapter 2 Preview (6 of 6) How will I use kinematics? The equations of motion that you learn in this chapter will be used throughout the entire book. In Part I, we’ll see how an object’s motion is related to forces acting on the object. We’ll later apply these kinematic equations to the motion of waves and to the motion of charged particles in electric and magnetic fields. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-8 4  3/3/2023 Reading Question 2.1 The slope at a point on a position-versus-time graph of an object is A. the object’s speed at that point. B. the object’s average velocity at that point. C. the object’s instantaneous velocity at that point. D. the object’s acceleration at that point. E. the distance traveled by the object to that point. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2-9 Reading Question 2.1 Answer The slope at a point on a position-versus-time graph of an object is A. the object’s speed at that point. B. the object’s average velocity at that point. C. the object’s instantaneous velocity at that point. D. the object’s acceleration at that point. E. the distance traveled by the object to that point. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 10 5  3/3/2023 Reading Question 2.2 The area under a velocity-versus-time graph of an object is A. the object’s speed at that point. B. the object’s acceleration at that point. C. the distance traveled by the object. D. the displacement of the object. E. This topic was not covered in this chapter. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 11 Reading Question 2.2 Answer The area under a velocity-versus-time graph of an object is A. the object’s speed at that point. B. the object’s acceleration at that point. C. the distance traveled by the object. D. the displacement of the object. E. This topic was not covered in this chapter. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 12 6  3/3/2023 Reading Question 2.3 The slope at a point on a velocity-versus-time graph of an object is A. the object’s speed at that point. B. the object’s instantaneous acceleration at that point. C. the distance traveled by the object. D. the displacement of the object. E. the object’s instantaneous velocity at that point. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 13 Reading Question 2.3 Answer The slope at a point on a velocity-versus-time graph of an object is A. the object’s speed at that point. B. the object’s instantaneous acceleration at that point. C. the distance traveled by the object. D. the displacement of the object. E. the object’s instantaneous velocity at that point. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 14 7  3/3/2023 Reading Question 2.4 Suppose we define the y-axis to point vertically upward. When an object is in free fall, it has acceleration in the y- direction A. a y = – g , where g = +9.80 m / s 2 B. a y  g , where g   9.80 m/s2 C. which is negative and increases in magnitude as it falls. D. which is negative and decreases in magnitude as it falls. E. which depends on the mass of the object. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 15 Reading Question 2.4 Answer Suppose we define the y-axis to point vertically upward. When an object is in free fall, it has acceleration in the y- direction A. a y = – g , where g = + 9.80 m / s 2 B. a y  g , where g   9.80 m/s 2 C. which is negative and increases in magnitude as it falls. D. which is negative and decreases in magnitude as it falls. E. which depends on the mass of the object. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 16 8  3/3/2023 Reading Question 2.5 At the turning point of an object, A. the instantaneous velocity is zero. B. the acceleration is zero. C. Both A and B are true. D. Neither A nor B is true. E. This topic was not covered in this chapter. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 17 Reading Question 2.5 Answer At the turning point of an object, A. the instantaneous velocity is zero. B. the acceleration is zero. C. Both A and B are true. D. Neither A nor B is true. E. This topic was not covered in this chapter. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 18 9  3/3/2023 Reading Question 2.6 A 1-pound block and a 100-pound block are placed side by side at the top of a frictionless hill. Each is given a very light tap to begin their race to the bottom of the hill. In the absence of air resistance A. the 1-pound block wins the race. B. the 100-pound block wins the race. C. the two blocks end in a tie. D. There’s not enough information to determine which block wins the race. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 19 Reading Question 2.6 Answer A 1-pound block and a 100-pound block are placed side by side at the top of a frictionless hill. Each is given a very light tap to begin their race to the bottom of the hill. In the absence of air resistance A. the 1-pound block wins the race. B. the 100-pound block wins the race. C. the two blocks end in a tie. D. There’s not enough information to determine which block wins the race. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 20 10  3/3/2023 Chapter 2 Content, Examples, and Quick Check Questions Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 21 Uniform Motion (1 of 2) The simplest possible motion is motion along a straight line at a constant, unvarying speed. We call this uniform motion. An object’s motion is uniform if and only if its position-versus-time graph is a straight line. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 22 11  3/3/2023 Uniform Motion (2 of 2) For one-dimensional motion, the average velocity is simply Δx /Δt (for horizontal motion) or Δy /Δt (for vertical lta e d lta e D lta e d lta e d motion). On a horizontal position-versus-time graph, Δx and Δt are, lta e d lta e d respectively, the ―rise‖ and ―run‖. Because rise over run is the slope of a line, the average velocity is the slope of the position-versus-time graph. The SI units of velocity are meters per second, abbreviated m/s. x y vavg  or  slope of theposition-versus-time graph t t Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 23 Example 2.1 Relating a velocity graph to a position graph (1 of 4) Figure 2.2 on the next page is the position-versus-time graph of a car. 1. Draw the car’s velocity-versus-time graph. 2. Describe the car’s motion. Model Model the car as a particle, with a well-defined position at each instant of time. Visualize Figure 2.2 is the graphical representation. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 24 12  3/3/2023 Example 2.1 Relating a velocity graph to a position graph (2 of 4) Solve a. The car’s position-versus-time graph is a sequence of three straight lines. Each of these straight lines represents uniform motion at a constant velocity. We can determine the car’s velocity during each interval of time by measuring the slope of the line. The position graph starts out sloping downward—a negative slope. Although the car moves a distance of 4.0 m during the first 2.0 s, its displacement is x = xat 2.0 s  xat 0.0 s =  4.0 m  0.0 m =  4.0 m The time interval for this displacement is Δd e lta e q u a ls= 2.0 s, so the velocity during this interval is Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 25 Example 2.1 Relating a velocity graph to a position graph (3 of 4) x 4.0 m νx = = =  2.0 m / s t 2.0 m The car’s position does not change from t = 2s to t = 4 s  Δx  0  , so v x  0. Finally, the displacement between t = 4s and t = 6 s is Δx  10.0 m. Thus the velocity during this interval is 10.0 m νx = = 5.0 m / s 2.0 s These velocities are shown on the velocity-versus-time graph of Figure 2.3 Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 26 13  3/3/2023 Example 2.1 Relating a velocity graph to a position graph (4 of 4) Solve b. The car backs up for 2 s at 2.0 m/s, sits at rest for 2 s, then drives forward at 5.0 m/s for at least 2 s. We can’t tell from the graph what happens for t > 6 s. REVIEW The velocity graph and the position graph look completely different. The value of the velocity graph at any instant of time equals the slope of the position graph. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 27 Tactics: Interpreting Position-versus- Time Graphs Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 28 14  3/3/2023 The Mathematics of Uniform Motion Consider an object in uniform motion along the s-axis, as shown in the graph. The object’s initial position is si at time t i. At a later time t f the object’s final position is sf. The change in time is t  t f  t i. The final position can be found as sf = si + νs Δt (uniform motion) Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 29 The Uniform-Motion Model Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 30 15  3/3/2023 Scalars and Vectors The distance an object travels is a scalar quantity, independent of direction. The displacement of an object is a vector quantity, equal to the final position minus the initial position. An object’s speed v is scalar quantity, independent of direction. Speed is how fast an object is going; it is always positive. Velocity is a vector quantity that includes direction. In one dimension the direction of velocity is specified by the  or  sign. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 31 Quick Check 2.1 An ant zig-zags back and forth on a picnic table as shown. The ant’s distance traveled and displacement are A. 50 cm and 50 cm. B. 30 cm and 50 cm. C. 50 cm and 30 cm. D. 50 cm and –50 cm. E. 50 cm and –30 cm. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 32 16  3/3/2023 Quick Check 2.1 Answer An ant zig-zags back and forth on a picnic table as shown. The ant’s distance traveled and displacement are A. 50 cm and 50 cm. B. 30 cm and 50 cm. C. 50 cm and 30 cm. D. 50 cm and –50 cm. E. 50 cm and –30 cm. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 33 Instantaneous Velocity (1 of 3) Objects rarely travel for long with a constant velocity. Far more common is a velocity that changes with time. If you watch a car’s speedometer, at any instant of time, the speedometer tells you how fast the car is going at that instant. If we include directional information, we can define an object’s instantaneous velocity—speed and direction— as its velocity at a single instant of time. The average velocity vavg  Δs Δt becomes a better and better approximation to the instantaneous velocity as Δt gets smaller and smaller. s ds νs  lim = (instantaneous velocity) t 0 t dt Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 34 17  3/3/2023 Instantaneous Velocity (2 of 3) Motion diagrams and position graphs of an accelerating rocket. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 35 Instantaneous Velocity (3 of 3) As Δt continues to get smaller, the average velocity vavg  Δs Δt reaches a constant or limiting value. The instantaneous velocity at time t is the average velocity during a time interval Δt centered on t, as Δt approaches zero. In calculus, this is called the derivative of s with respect to t. Graphically, Δs Δt is the slope of a straight line. In the limit Δt 0 the straight line is tangent to the curve. The instantaneous velocity at time t is the slope of the line that is tangent to the position-versus-time graph at time t. νs = slope of the position - versus - time graph at time t Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 36 18  3/3/2023 Quick Check 2.2 The slope at a point on a position-versus-time graph of an object is A. the object’s speed at that point. B. the object’s velocity at that point. C. the object’s acceleration at that point. D. the distance traveled by the object to that point. E. I really have no idea. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 37 Quick Check 2.2 Answer The slope at a point on a position-versus-time graph of an object is A. the object’s speed at that point. B. the object’s velocity at that point. C. the object’s acceleration at that point. D. the distance traveled by the object to that point. E. I really have no idea. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 38 19  3/3/2023 Example 2.3 Finding Velocity from Position Graphically (1 of 4) Figure 2.9 shows the position-versus-time graph of an elevator. 1. At which labeled point or points does the elevator have the least velocity? 2. At which point or points does the elevator have maximum velocity? 3. Sketch an approximate velocity-versus-time graph for the elevator. Figure 2.9 Position-versus-time graph. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 39 Example 2.3 Finding Velocity from Position Graphically (2 of 4) Model Model the elevator as a particle. Visualize Figure 2.9 is the graphical representation. Solve a. At any instant, an object’s velocity is the slope of its position graph. Figure 2.10a shows that the elevator has the least velocity—no velocity at all!—at points 1 and 3 where the slope is zero. At point 1, the velocity is only instantaneously zero. At point 3, the elevator has actually stopped and remains at rest. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 40 20  3/3/2023 Example 2.3 Finding Velocity from Position Graphically (3 of 4) b. The elevator has maximum velocity at 2, the point of steepest slope. c. Although we cannot find an exact velocity-versus-time graph, we can see that the slope, and hence v y is initially negative, becomes zero at point 1, rises to a maximum value at point 2, decreases back to zero a little before point 3, then remains at zero thereafter. Thus Figure 2.10b shows, at least approximately, the elevator’s velocity-versus-time graph. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 41 Example 2.3 Finding Velocity from Position Graphically (4 of 4) REVIEW Once again, the shape of the velocity graph bears no resemblance to the shape of the position graph. You must transfer slope information from the position graph to value information on the velocity graph. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 42 21  3/3/2023 Quick Check 2.3 Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? A. B. C. D. E. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 43 Quick Check 2.3 Answer Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? A. B. C. D. E. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 44 22  3/3/2023 Quick Check 2.4 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? E. None of the above. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 45 Quick Check 2.4 Answer Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? E. None of the above. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 46 23  3/3/2023 Quick Check 2.5 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 47 Quick Check 2.5 Answer Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 48 24  3/3/2023 A Little Calculus: Derivatives ds/dt is called the derivative of s with respect to t. ds/dt is the slope of the line that is tangent to the position- versus-time graph. Consider the function u(t) = ct n , where c and n are constants: du The derivative of u = ct n is = nct n 1 dt The derivative of a constant is zero: du = 0if u = c = constant dt The derivative of a sum is the sum of the derivatives. If u and w are two separate functions of time, then d du dw (u + w) =  dt dt dt Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 49 Derivative Example Suppose the position of a particle as a function of time is s(t) = 2t 2 m where t is in s. What is the particle’s velocity? Velocity is the derivative of s with respect to t: ds νs   2  2t 21 = 4t dt The figure shows the particle’s position and velocity graphs. The value of the velocity graph at any instant of time is the slope of the position graph at that same time. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 50 25  3/3/2023 Quick Check 2.6 Here is a position graph of an object: At t = 1.5 s, the object’s velocity is A. 40 m/s B. 20 m/s C. 10 m/s D. –10 m/s E. None of the above. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 51 Quick Check 2.6 Answer Here is a position graph of an object: At t = 1.5 s, the object’s velocity is A. 40 m/s B. 20 m/s C. 10 m/s D. –10 m/s E. None of the above. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 52 26  3/3/2023 Quick Check 2.7 Here is a position graph of an object: At t = 3.0 s, the object’s velocity is A. 40 m/s B. 20 m/s C. 10 m/s D. –10 m/s E. None of the above. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 53 Quick Check 2.7 Answer Here is a position graph of an object: At t = 3.0 s, the object’s velocity is A. 40 m/s B. 20 m/s C. 10 m/s D. –10 m/s E. None of the above. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 54 27  3/3/2023 Quick Check 2.8 When do objects 1 and 2 have the same velocity? A. At some instant before time t0. B. At time t0. C. At some instant after time t0. D. Both A and B. E. Never. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 55 Quick Check 2.8 Answer When do objects 1 and 2 have the same velocity? A. At some instant before time t0. B. At time t0. C. At some instant after time t0. D. Both A and B. Same slope at this time E. Never. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 56 28  3/3/2023 Finding Position from Velocity (1 of 2) Suppose we know an object’s position to be si at an initial time ti. We also know the velocity as a function of time between ti and some later time tf. Even if the velocity is not constant, we can divide the motion into N steps in which it is approximately constant, and compute the final position as N sf = si  lim  (vs ) k Δt = si +  νs dt tf t 0 ti k 1 The curlicue symbol is called an integral. The expression on the right is read ―the integral of νs dt from ti to tf." Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 57 Finding Position from Velocity (2 of 2) The integral may be interpreted graphically as the total area enclosed between the t-axis and the velocity curve. The total displacement s is called the ―area under the curve.‖ sf = si + area under the velocity curve νs between ti and tf Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 58 29  3/3/2023 Quick Check 2.9 Here is the velocity graph of an object that is at the origin (x = 0 m) at t = 0 s. At t = 4.0 s, the object’s position is A. 20 m. B. 16 m. C. 12 m. D. 8 m. E. 4 m. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 59 Quick Check 2.9 Answer Here is the velocity graph of an object that is at the origin (x = 0 m) at t = 0 s. At t = 4.0 s, the object’s position is A. 20 m. B. 16 m. C. 12 m. Displacement = area under the curve D. 8 m. E. 4 m. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 60 30  3/3/2023 Example 2.5 The Displacement During a Drag Race (1 of 3) Figure 2.16 shows the velocity-versus-time graph of a drag racer. How far does the racer move during the first 3.0 s? Figure 2.16 Velocity-versus-time graph for Example 2.5. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 61 Example 2.5 The Displacement During a Drag Race (2 of 3) Model Model the drag racer as a particle with a well-defined position at all times. Visualize Figure 2.16 is the graphical representation. SOLVE The question ―How far?‖ indicates that we need to find a displacement x rather than a position x. According to Equation 2.12, the car’s displacement x  x f  xi between t = 0 s and t = 3 s is the area under the curve from t = 0 s to t = 3 s. The curve in this case is an angled line, so the area is that of a triangle: Δx = areaof trianglebetween t = 0s and t = 3s = 1 2 × base× height = 1 2 × 3s× 12m/s = 18m The drag racer moves 18 m during the first 3 seconds. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 62 31  3/3/2023 Example 2.5 The Displacement During a Drag Race (3 of 3) REVIEW The ―area‖ is a product of s with m/s, so x has the proper units of m. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 63 A Little More Calculus: Integrals Taking the derivative of a function is equivalent to finding the slope of a graph of the function. Similarly, evaluating an integral is equivalent to finding the area under a graph of the function. n For the important function u(t) = ct , the essential result from calculus is that tf tf tf ct n1 ct n1 ct n1  udt   ct dt   f  i (n  1) n ti ti n1 t n1 n1 i The vertical bar in the third step means the integral evaluated at tf minus the integral evaluated at ti. The integral of a sum is the sum of the integrals. If u and w are two separate functions of time, then: tf tf tf ti (u + w) dt =  u dt +  wdt ti ti Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 64 32  3/3/2023 Quick Check 2.10 Which velocity-versus-time graph goes with this position graph? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 65 Quick Check 2.10 Answer Which velocity-versus-time graph goes with this position graph? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 66 33  3/3/2023 Motion with Constant Acceleration (1 of 2) Imagine a competition between a Table 2.1 Velocities of a Porsche Volkswagen Beetle and a and a Volkswagen Beetle Porsche to see which can t (s) vPorsche (m/s) vVW (m/s) achieve a velocity of 30 m/s in 0.0 0.0 0.0 the shortest time. 0.1 0.5 0.2 The table shows the velocity of 0.2 1.0 0.4 each car, and the figure shows 0.3 1.5 0.6 the velocity-versus-time graphs. ⁝ ⁝ ⁝ Both cars achieved every velocity between 0 and 30 m/s, so neither is faster. But for the Porsche, the rate at which the velocity changed was νs 30m/s rate of velocity change =   5.0(m/s)/s t 6.0s Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 67 Motion with Constant Acceleration (2 of 2) The SI units of acceleration are (m/s)/s, or m s2. It is the rate of change of velocity and measures how quickly or slowly an object’s velocity changes. The average acceleration during a time interval Δt is νs aavg  (average acceleration) t Graphically, aavg is the slope of a straight-line velocity- versus-time graph. If acceleration is constant, the acceleration as is the same as aavg. Acceleration, like velocity, is a vector quantity and has both magnitude and direction. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 68 34  3/3/2023 Quick Check 2.11 A cart slows down while moving away from the origin. What do the position and velocity graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 69 Quick Check 2.11 Answer A cart slows down while moving away from the origin. What do the position and velocity graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 70 35  3/3/2023 Quick Check 2.12 A cart speeds up toward the origin. What do the position and velocity graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 71 Quick Check 2.12 Answer A cart speeds up toward the origin. What do the position and velocity graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 72 36  3/3/2023 Quick Check 2.13 Here is a motion diagram of a car speeding up on a straight road: The sign of the acceleration ax is A. positive. B. negative. C. zero. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 73 Quick Check 2.13 Answer Here is a motion diagram of a car speeding up on a straight road: The sign of the acceleration ax is A. positive. B. negative. Speeding up means vx and ax have the same sign. b u s b u s C. zero. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 74 37  3/3/2023 Example 2.9 Running the Court (1 of 4) A basketball player starts at the left end of the court and moves with the velocity shown in Figure 2.20. Draw a motion diagram and an acceleration-versus-time graph for the basketball player Figure 2.20 Velocity-versus-time graph for the basketball player of Example 2.9. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 75 Example 2.9 Running the Court (2 of 4) VISUALIZE The velocity is positive (motion to the right) and increasing for the first 6 s, so the velocity arrows in the motion diagram are to the right and getting longer. From t = 6 s to 9 s the motion is still to the right (vx is still positive), but the arrows are getting shorter because vx is decreasing. There’s a turning point at t = 9 s, when vx = 0 m/s, and after that the motion is to the left (vx is negative) and getting faster. The motion diagram of Figure 2.21a shows the velocity and the acceleration vectors. Figure 2.21 Motion diagram and acceleration graph for Example 2.9. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 76 38  3/3/2023 Example 2.9 Running the Court (3 of 4) Solve Acceleration is the slope of the velocity graph. For the first 6 s, the slope has the constant value Δνx 6.0m / s ax = = = 1.0m / s 2 Δt 6.0s The velocity then decreases by 12 m/s during the 6 s interval from t = 6 s to t = 12 s, so Δνx 12m / s ax = = = 2.0m / s 2 Δt 6.0s The acceleration graph for these 12 s is shown in Figure 2.21b. Notice that there is no change in the acceleration at t = 9 s, the turning point. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 77 Example 2.9 Running the Court (4 of 4) Review The sign of a x does not tell us whether the object is sub speeding up or slowing down. The basketball player is slowing down from t = 6 s to t = 9 s, then speeding up from t = 9 s to t = 12 s. Nonetheless, his acceleration is negative during this entire interval because his acceleration vector, as seen in the motion diagram, always points to the left. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 78 39  3/3/2023 Quick Check 2.14 A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 79 Quick Check 2.14 Answer A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 80 40  3/3/2023 Quick Check 2.15 A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 81 Quick Check 2.15 Answer A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 82 41  3/3/2023 Quick Check 2.16 A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 83 Quick Check 2.16 Answer A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 84 42  3/3/2023 The Kinematic Equations of Constant Acceleration (1 of 4) Suppose we know an object’s velocity to be νis at an initial time ti. We also know the object has a constant acceleration of as over the time interval t = tf  ti. We can then find the object’s velocity at the later time tf as νfs = νis + as Δt Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 85 The Kinematic Equations of Constant Acceleration (2 of 4) Suppose we know an object’s position to be si at an initial time ti. It’s constant acceleration as is shown in graph (a). The velocity-versus-time graph is shown in graph (b). The final position sf is si plus the area under the curve of νs between ti and tf : sf = si + νis Δt + 12 as (Δt )2 Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 86 43  3/3/2023 The Kinematic Equations of Constant Acceleration (3 of 4) Suppose we know an object’s velocity to be νis at an initial position si. We also know the object has a constant acceleration of as while it travels a total displacement of s = sf  si. We can then find the object’s velocity at the final position sf : νfs 2 = νis 2 + 2as Δs Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 87 The Constant-Acceleration Model Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 88 44  3/3/2023 The Kinematic Equations of Constant Acceleration (4 of 4) Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 89 Example 2.11 A Two-Car Race (1 of 5) Fred is driving his Volkswagen Beetle at a steady 20 m/s when he passes Betty sitting at rest in her Porsche. Betty instantly begins accelerating at 5.0 m/s2. How far does Betty have to drive to overtake Fred? Model Model the VW as a particle in uniform motion and the Porsche as a particle with constant acceleration. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 90 45  3/3/2023 Example 2.11 A Two-Car Race (2 of 5) Visualize Figure 2.24 is the pictorial representation. Fred’s motion diagram is one of uniform motion, while Betty’s shows uniform acceleration. Fred is ahead in frames 1, 2, and 3, but Betty catches up with him in frame 4. The coordinate system shows the cars with the same position at the start and at the end—but with the important difference that Betty’s Porsche has an acceleration while Fred’s VW does not. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 91 Example 2.11 A Two-Car Race (3 of 5) SOLVE This problem is similar to Example 2.2, in which Bob and Susan met for lunch. As we did there, we want to find Betty’s position (x1)B at the instant t1 when (x1)B = (x1)F. We know, from the models of uniform motion and uniform acceleration, that Fred’s position graph is a straight line but Betty’s is a parabola. The position graphs in Figure 2.24 show that we’re solving for the intersection point of the line and the parabola. Fred’s and Betty’s positions at t1 are ( x1 ) F  ( x0 ) F  (v0 x ) F (t1  t0 )  (v0 x ) F t1 1 1 ( x1 ) B  ( x0 ) B  (v0 x ) B (t1  t0 )  (a0 x ) B (t1  t0 ) 2  ( a0 x ) B t12 2 2 Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 92 46  3/3/2023 Example 2.11 A Two-Car Race (4 of 5) By equating these, (ν0x )F t1  12 (a0x )B t12 we can solve for the time when Betty passes Fred: t1  12 (a0x )B t1  (ν0x )F   0 t1   0s 2( ν0x )F / ( a0x )B =8.0 s Interestingly, there are two solutions. That’s not surprising, when you think about it, because the line and the parabola of the position graphs have two intersection points: when Fred first passes Betty, and 8.0 s later when Betty passes Fred. We’re interested in only the second of these points. We can now use either of the distance equations to find ( x1 ) B  ( x1 ) F  160m. Betty has to drive 160 m to overtake Fred. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 93 Example 2.11 A Two-Car Race (5 of 5) Review 160 m ≈ 160 yards. Because Betty starts from rest while Fred is moving at 20 m/s ≈ 40 mph, needing 160 yards to catch him seems reasonable. Note The purpose of the Review step is not to prove that an answer must be right but to rule out answers that, with a little thought, are clearly wrong. Figure 2.24 Pictorial representation for Example 2.11. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 94 47  3/3/2023 Quick Check 2.17 Which velocity-versus-time graph goes with this acceleration graph? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 95 Quick Check 2.17 Answer Which velocity-versus-time graph goes with this acceleration graph? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 96 48  3/3/2023 Free Fall (1 of 2) The motion of an object moving under the influence of gravity only, and no other forces, is called free fall. Two objects dropped from the same height will, if air resistance can be neglected, hit the ground at the same time and with the same speed. Consequently, any two objects in free fall, regardless of their mass, have the same acceleration: r In a vacuum, the apple and 2 afree fall = (9.80 m s , vertically downward) feather fall at the same rate and hit the ground at the same time. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 97 Free Fall (2 of 2) Figure (a) shows the motion diagram of an object that was released from rest and falls freely. Figure (b) shows the object’s velocity graph. The velocity graph is a straight line with a slope: a y = afreefall =  g Where g is a positive number which is equal to 9.80 m/s2 on the surface of the earth. Other planets have different values of g. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 98 49  3/3/2023 Quick Check 2.18 A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration a y is sub A. positive. B. negative. C. zero. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 99 Quick Check 2.18 Answer A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration a y is sub A. positive. B. negative. C. zero. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 100 50  3/3/2023 Motion on an Inclined Plane Figure (a) shows the motion diagram of an object sliding down a straight, frictionless inclined plane. Figure (b) shows the free-fall r acceleration afree fall the object would have if the incline suddenly vanished. This vector can be broken into two r r pieces aP and a. The surface somehow ―blocks‖ a  , so the one-dimensional acceleration along the incline is as = ± g sin θ The correct sign depends on the direction the ramp is tilted. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 101 Quick Check 2.19 The ball rolls up the ramp, then back down. Which is the correct acceleration graph? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 102 51  3/3/2023 Quick Check 2.19 Answer The ball rolls up the ramp, then back down. Which is the correct acceleration graph? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 103 Advanced Topic: Instantaneous Acceleration (1 of 2) Figure (a) shows a realistic velocity- versus-time graph for a car leaving a stop sign. The graph is not a straight line, so this is not motion with a constant acceleration. Figure (b) shows the car’s acceleration graph. The instantaneous acceleration as is the slope of the line that is tangent to the velocity-versus-time curve at time t: dνs as  = slope of the velocity - versus - time graph at time t dt Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 104 52  3/3/2023 Advanced Topic: Instantaneous Acceleration (2 of 2) Suppose we know an object’s velocity to be v is at an sub initial time t i. sub We also know the acceleration as a function of time between t i and some later time t f. sub sub Even if the acceleration is not constant, we can divide the motion into N steps of length ∆ t in which it is delta approximately constant. In the limit Δt 0 we can compute the final velocity as delta t gives 0 tf νfs = νis   as dt ti The graphical interpretation of this equation is νfs = νis  area under the acceleration curve as between ti and tf Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 105 Example 2.17 Finding Velocity from Acceleration (1 of 2) Figure 2.37 shows the acceleration graph for a particle with an initial velocity of 10 m/s. What is the particle’s velocity at t = 8 s? Figure 2.37 Acceleration graph for Example 2.17. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 106 53  3/3/2023 Example 2.17 Finding Velocity from Acceleration (2 of 2) MODEL We’re told this is the motion of a particle. VISUALIZE Figure 2.37 is a graphical representation of the motion. SOLVE The change in velocity is found as the area under the acceleration curve: νfs = νis  area under the acceleration curve as between ti and tf The area under the curve between ti = 0s and tf = 8s can be subdivided into a rectangle 0s  t  4 s  and a triangle  4s  t  8 s . These areas are easily computed. Thus νs (at t = 8s)  10m / s  (4m / s2 )(4 s)  12 (4m / s2 )(4 s)  34m / s Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 107 Chapter 2 Summary Slides Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 108 54  3/3/2023 General Principles (1 of 2) Kinematics describes motion in terms of position, velocity, and acceleration. General kinematic relationships are given mathematically by: Instantaneous velocity νs  ds / dt = slope of position graph Instantaneous acceleration as  dνs / dt = slope of velocity graph  tf Final position sf = si +  νs dt = si  area under the velocity curve from ti to tf ti  tf Final velocity νfs = νis +  a s dt = νis  area under the acceleration curve from ti to tf ti Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 109 General Principles (2 of 2) Solving Kinematics Problems MODEL Uniform motion or constant acceleration. VISUALIZE Draw a pictorial representation. SOLVE Uniform motion sf = si + νs  t Constant acceleration νfs = νi s + a s  t sf = si + νis Δt + 12 as (Δt )2 νf2s = νi2s + 2as Δs REVIEW Is the result reasonable? Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 110 55  3/3/2023 Important Concepts (1 of 2) Position, velocity, and acceleration are related graphically. The slope of the position-versus-time graph is the value on the velocity graph. The slope of the velocity graph is the value on the acceleration graph. s is a maximum or minimum at a turning point, and v s = 0. sub Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 111 Important Concepts (2 of 2) Displacement is the area under the velocity curve. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 112 56  3/3/2023 Applications (1 of 2) The sign of vs indicates the direction of motion. u v s > 0 is motion to the right or up. sub v s < 0 is motion to the left or down. sub The sign of a s indicates which way a points, not whether the sub object is speeding up or slowing down. a s > 0 if a points to the right or up. sub a s < 0 if a points to the left or down. sub The direction of a is found with a motion diagram. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 113 Applications (2 of 2) An object is speeding up if and only if vs and as have the same sign. An object is slowing down if and only if vs and as have opposite signs Free fall is constant-acceleration motion with 2 a y   g = 9.80 m s Motion on an inclined plane has as = ±g sinθ. The sign depends The sign depends on the direction of the tilt. Copyright © 2023 Pearson Education Ltd. All Rights Reserved. 2 - 114 57