普通物理 General Physics 2 – Straight Line Motion 郭艷光Yen-Kuang Kuo 國立彰化師大物理系暨光電科技研究所 電子郵件: ykuo@cc.ncue.edu.tw 網頁: http://ykuo.ncue.edu.tw

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Outline 2-1 What Is Physics? 2-2 Motion 2-3 Position and Displacement 2-4 Average Velocity and Average Speed 2-5 Instantaneous Velocity and Speed 2-6 Acceleration 2-7 Constant Acceleration: A Special Case 2-8 Another Look at Constant Acceleration 2-9 Free-Fall Acceleration 2-10 Graphical Integration in Motion Analysis 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-1 What Is Physics? NASCAR engineers are fanatical about this aspect of physics as they determine the performance of their cars before and during a race. Geologists use this physics to measure tectonic-plate motion as they attempt to predict earthquakes. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-1 What Is Physics? Medical researchers need this physics to map the blood flow through a patient when diagnosing a partially closed artery, and motorists use it to determine how they might slow sufficiently when their radar detector sounds a warning. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-2 Motion In this chapter, we will study kinematics i.e. how objects move along a straight line. The line may be vertical, horizontal, or slanted, but it must be straight. Kinematics is the part of mechanics that describes the motion of physical objects. We say that an object moves when its position as determined by an observer changes with time. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-2 Motion We will assume that the moving objects are “particles” (by which we mean a point-like object such as an electron) i.e. we restrict our discussion to the motion of objects for which all the points move in the same way. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-3 Position and Displacement Position : To locate an object means to find its position relative to some reference point, often the origin (or zero point) of an axis such as the x axis. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-3 Position and Displacement Position is determined on an axis that is marked in units of length (here meters) and that extends indefinitely in opposite directions. A particle might be located at , which means it is 5 m in the positive direction from the origin. If it were at , it would be just as far from the origin but in the opposite direction. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-3 Position and Displacement Displacement : A change from one position x1 to another position x2 is called a displacement Δx, where , if the particle moves from x1 = 5 m to x2 = 12 m, then Δx = (12 m) - (5 m) = 7 m. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-3 Position and Displacement Displacement involves only the original and final positions. Displacement is a vector quantity, which is a quantity that has both a direction and a magnitude. A displacement of x = -4 m has a magnitude of 4 m and the motion is in the negative direction. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-4 Average Velocity and Average Speed A compact way to describe position is with a graph of position x plotted as a function of time t — a graph of x(t). 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-4 Average Velocity and Average Speed Average Velocity : It is the ratio of the displacement x that occurs during a particular time interval Δt to that interval: The notation means that the position is x1 at time t1 and then x2 at time t2. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-4 Average Velocity and Average Speed The time interval Δt is defined as : Δt = t2 – t1. The unit of vavg is : m/s (the meter per second). is the slope of the straight line that connects two particular points on the x(t) curve. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-4 Average Velocity and Average Speed Graphical determination of vavg Below , and , the corresponding positions are: and 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-4 Average Velocity and Average Speed Average speed : Average speed involves the total distance covered, independent of direction. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (a) You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and stops. Over the next 30 min, you walk another 2.0 km farther along the road to a gasoline station. What is your overall displacement from the beginning of your drive to your arrival at the station? 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (a) Key idea: Your displacement Δx along the x axis is the second position minus the first position. Solutions: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (b) What is the time interval Δt from the beginning of your drive to your arrival at the station? Key idea: This average velocity is the ratio of the displacement for the drive to the time interval for the drive: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (b) Solutions: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (c) What is your average velocity vavg from the beginning of your drive to your arrival at the station? Find it both numerically and graphically. Key idea: vavg for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (c) First solution: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (c) Second solution: 1. we graph the function x(t). 2. vavg is the ratio of the rise (Δx = 10.4 km) to the run (Δt = 0.62 h), which gives us vavg = 16.8 km/h. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (d) Suppose that to pump the gasoline, pay for it, and walk back to the truck takes you another 45 min. What is your average speed from the beginning of your drive to your return to the truck with the gasoline? Key idea: your average speed is the ratio of the total distance you move to the total time interval you take to make that move. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-1 (d) Solutions: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-5 Instantaneous Velocity and Speed Instantaneous Velocity: The velocity at any instant is obtained from the average velocity by shrinking the time interval Δt closer and closer to 0. Speed: We define speed as the magnitude of an object’s velocity vector. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-2 Figure 2-6a is an x(t) plot for an elevator cab that is initially stationary, then moves upward (which we take to be the positive direction of x), and then stops. Plot v(t). 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-2 Solutions: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-3 The position of a particle moving on an x axis is given by with x in meters and t in seconds. What is its velocity at t = 3.5 s? Is the velocity constant, or is it continuously changing? 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-3 Key idea: Velocity is the first derivative (with respect to time) of the position function x(t). Solutions: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-6 Acceleration Average Acceleration: When a particle’s velocity changes, the particle is said to undergo acceleration (or to accelerate). For motion along an axis, the average acceleration over a time interval is 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-6 Acceleration Instantaneous Acceleration: The acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Graphically, the acceleration at any point is the slope of the curve of v(t) at that point. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-6 Acceleration The acceleration of a particle at any instant is the second derivative of its position x(t) with respect to time. The unit of acceleration is: or 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 2-6 Acceleration Figure is a plot of the acceleration of the elevator cab discussed in Example 2-2. Each point on the a(t) curve shows the derivative (slope) of the v(t) curve at the corresponding time. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-4 (a) A particle’s position on the x axis is given by with x in meters and t in seconds. Find the particle’s velocity function v(t) and acceleration function a(t). 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-4 (a) Key idea: (1) To get the velocity function v(t), we differentiate the position function x(t) with respect to time. (2) To get the acceleration function a(t), we differentiate the velocity function v(t) 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-4 (a) Solutions: with v in meters per second. with a in meters per second squared. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-4 (b) Is there ever a time when ? Solutions: Setting v(t) = 0 yields Thus, the velocity is zero both 3s before and 3s after the clock reads 0. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-4 (c) Describe the particle’s motion for . Key idea: To examine the expressions for x(t), v(t), and a(t). Solutions: At t = 0, for 0 < t < 3 s, for t > 3 s, 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-7 Constant Acceleration: A Special Case 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-5 The head of a woodpecker is moving forward at s speed of 7.49 m/s when the beak makes first contact with a tree limb. The beak stops after penetrating the limb by 1.87 mm. Assuming the acceleration to be constant, find the acceleration magnitude in terms of g. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-5 Key idea: We can use the constant-acceleration equations to solve this problem. Solutions: the magnitude is 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-6 Figure 2-9 gives a particle’s velocity v versus its position as it moves along an x axis with constant acceleration. What is its velocity at position x = 0? 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-6 Solutions: We have two such pairs: (1) v = 8 m/s and x = 20 m, and (2) v = 0 and x =70 m. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-7 Two cars approach each other on a straight road. Car A moves at 16 m/s and car B moves at 8 m/s. When they are 45 m apart, both drivers apply their brakes. Car A slows down at 2 m/s2, while car B slows down at 4 m/s2. Where and when do they collide? 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-7 Solution: using We set xA = xB →t = 3 s and 5 s. Stop! Try to find the flaw in the above argument. At t = 3 s, 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-7 You can find vB is strange, because the car reversed its velocity! You can easily verify that B stops at 2 s. B stays at this position until it is hit by A. The collision occurs at 2.8 s and 37 m. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-8 Another Look at Constant Acceleration 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-9 Free-Fall Acceleration When an object either up or down and eliminates the effects of air on its flight, its acceleration is a certain constant rate, called the free-fall acceleration, and its magnitude is represented by g. The free-fall acceleration near Earth’s surface is a = - g = -9.8 m/s2, and the magnitude of the acceleration is g = 9.8 m/s2. Do not substitute -9.8 m/s2 for g. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-9 Free-Fall Acceleration B y 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-8 (a) On September 26, 1993, Dave Munday went over the Canadian edge of Niagara Falls in a steel ball equipped with an air hole and then fell 48 m to the water (and rocks). Assume his initial velocity was zero, and neglect the effect of the air on the ball during the fall. How long did Munday fall to reach the water surface? 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-8 (a) Solutions: Let y = 0 at his starting point and the positive direction up the axis. Then the acceleration is a = along that axis, and the water level is at y = m. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-8 (b) Munday could count off the three seconds of free fall but could not see how far he had fallen with each count. Determine his position at each full second. Solution: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-8 (c) What was Munday’s velocity as he reached the water surface? Solutions: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-8 (d) What was Munday’s velocity at each count of one full second? Was he aware of his increasing speed? Solutions: Once he was in free fall, Munday was unaware of the increasing speed because the acceleration during the fall was always 9.8 m/s2 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-9 A pitcher tosses a baseball up along a y axis, with an initial speed of 12 m/s. (a) How long does the ball take to reach its maximum height? (b) What is the ball’s maximum height above its release point? (c) How long does the ball take to reach a point 5.0 m above its release point? 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-9 Solution: 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-10 Graphical Integration in Motion Analysis A graph of an object’s acceleration versus time, we can integrate on the graph to find the object’s velocity at any given time. Velocity v is defined in terms of the position x as v = dx/dt, then where x0 is the position at time t0 and x1 is the position at time t1. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-10 Graphical Integration in Motion Analysis 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

2-10 Graphical Integration in Motion Analysis 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-10 “Whiplash injury” commonly occurs in a rear-end collision where a front car is hit from behind by a second car. In the 1970s, researchers concluded that the injury was due to the occupant’s head being whipped back over the top of the seat as the car was slammed forward. As a result of this finding, head restraints were build into cars, yet neck injuries in rear-end collisions continued to occur. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-10 In a recent test to study neck injury in rear-end collisions, a volunteer was strapped to a seat that was then moved abruptly to simulate a collision by a rear car moving at 10.5 km/h. Figure gives the accelerations of the volunteer’s torso and head during the collision, which began at time t = 0. The torso acceleration was delayed by 40 ms because during that time interval the seat back had to compress against the volunteer. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-10 The head acceleration was delayed by an additional 70 ms. What was the torso speed when the head began to accelerate? The a(t) curve of the torso and head of a volunteer in a simulation of a rear-end collision. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 Example 2-10 Solution: (b) Breaking up the region between the plotted curve and the time axis to calculate the area. 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授

普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授 End of chapter 2! 2018/11/18 普通物理講義-2 / 國立彰化師範大學物理系/ 郭艷光教授