Newtonian mechanics: kinematics and dynamics Kinematics: mathematical description of motion (Ch 2, Ch 3) Dynamics: how forces affect motion (Ch 4)

Transcription

1 July :39 AM Chapter 2 Kinematics in One Dimension Newtonian mechanics: kinematics and dynamics Kinematics: mathematical description of motion (Ch 2, Ch 3) Dynamics: how forces affect motion (Ch 4) Summary of Key Concepts position, displacement, distance (2.1) velocity, speed, average velocity, and average speed (2.2) acceleration and average acceleration (2.3) kinematics equations for constant acceleration (2.4, 2.5) freely falling objects (2.6) graphical analysis of one-dimensional kinematics (2.7) What is displacement? Example: An object moves along a straight line with initial position 8 m and final position 3 m. Determine the displacement. Solution: The displacement can be represented by the blue arrow in the figure above. Note that the arrow points to the left, which corresponds to the fact that the displacement is negative. For a positive displacement, the arrow would point to the right. Ch2 Page 1

2 If the line of motion is oriented differently (north-south, for example, or in some other direction), then it's up to you to choose a positive direction along the line; the other direction is therefore negative. How does distance differ from displacement? Example: An object moves along a straight line from position 8 m to position 3 m. Then the object moves along the same line another 2 m to the right. (a) Determine the total displacement. (b) Determine the total distance travelled. Solution: How are average velocity and average speed calculated? Ch2 Page 2

3 Example: An object moves along a straight line from position 8 m to position 3 m in 4 s. Then the object moves along the same line another 2 m to the right in 1 s. (a) Determine the average velocity for each leg of the trip, and for the whole trip. (b) Determine the average speed for each leg of the trip, and for the whole trip. Solution: This way of calculating average speed (i.e., as the length of the average velocity arrow) IS ONLY VALID for a motion in a straight line for which there is no change in direction. For the generally valid method, see below. Ch2 Page 3

4 The average velocity can be interpreted by pretending the actual motion is replaced by the wiggly blue arrow in the diagram above. The wiggly blue arrow represents the total displacement for the entire journey. (At least it would if we replaced it by a straight arrow; I made it wiggly just to distinguish it from the displacement arrows of the actual motion.) The average speed is always the total distance divided by the total time. For a straight line segment of a journey, with no changes of direction, the average speed can also be calculated as the length of the average velocity vector. Example: Alice jogs one complete lap around an oval track, a distance of 200 m, in a time of 40 s. (a) Determine Alice's displacement and average velocity. (b) Determine Alice's distance travelled and her average speed. Solution: (a) Ch2 Page 4

5 Example: Basil drives from St. Catharines to Toronto, a distance of 110 km, and then back again. His average speed on the way to Toronto is 60 km/h, and his average speed on the way back is 90 km/h. Determine Basil's average speed for the whole trip. Solution: Use the basic idea that average speed is distance divided by time: Ch2 Page 5

6 Question: Does the distance between the two cities matter in the previous example? Can you solve the problem if the distance between the two cities is not given? Solution: Yes, we can! Ch2 Page 6

7 Let's check this result using the values of the previous problem; sure enough, if you substitute speeds of 60 km/h and 90 km/h into the result, you reproduce the result of the previous problem. This is comforting. Another way to play with this result is to play with extreme values. If the speed to Toronto is 1 km/h and the speed on the return trip is 100 km/h, we expect that the average speed for the entire trip is a lot closer to 1 km/h than to 100 km/h. Sure enough, if you substitute these values into the previous formula, you obtain about 2 km/h. This is a powerful reminder that to determine the average speed in situations such as this one, you can't just add the two speeds and divide by 2. What is acceleration? How is it calculated? Example: Alice drives at a speed of 50 km/h, then enters a highway, and so increases her speed to 100 km/h in a time of 8 s. Determine her acceleration, in both (km/h)/s and in m/s 2. Solution: Ch2 Page 7

8 Typically, it is easier for us humans to understand units such as (km/h)/s, because of our experience driving cars. However, it is often easier to solve problems using the standard unit of acceleration, which is m/s 2. This means being able to quickly and accurately go from one unit to the other is useful; it's a skill worth practicing. Example: Alice now leaves the highway, and so decreases her speed from 100 km/h to 60 km/h in a time of 10 s. Determine her acceleration, in both (km/h)/s and in m/s 2. Solution: Ch2 Page 8

9 Is acceleration a vector or a scalar? Page 53, Problem 19 The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since t = 0 s is 2.0 s. Initial Acceleration Velocity v 0 a (a) +12 m/s +3.0 m/s 2 (b) +12 m/s 3.0 m/s 2 (c.) 12 m/s +3.0 m/s 2 (d) 12 m/s 3.0 m/s 2 Solution: The conclusion we draw from this example is that (for motion in a Ch2 Page 9

10 straight line) if the acceleration is in the same direction as the velocity (i.e., they have the same signs) then the object speeds up, whereas if the acceleration opposes the velocity (i.e., they have opposite signs) then the object slows down. In the previous paragraph we have been speaking about acceleration as a vector. In every-day language, we often treat acceleration as a scalar; that is, we use "acceleration" to mean "speeding up" and "deceleration" to mean "slowing down." It would have been so much easier to understand and communicate these concepts if we had two separate words, one for acceleration as a vector and a different one for acceleration as a scalar (just as we use velocity and speed), but we only have one word, so that's life. Page 53, Problem 23 Two motorcycles are travelling east with different speeds. However, four seconds later, they have the same speed. During this four-second interval, cycle A has an average acceleration of 2.0 m/s 2 east, while cycle B has an average acceleration of 4.0 m/s 2 east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster? Discussion: There are two moving objects here, so we should be careful to label the kinematical variables differently for each object. Solution: Ch2 Page 10

11 We are not asked anything about the positions, and we are not given any information about the positions; perhaps we can just ignore positions? If that strategy is to work, we would need to find a way to connect the desired quantities (initial velocities) with the given quantities (accelerations and times). But this is indeed possible: Note that the solution can also be obtained in a relatively simple way by drawing a velocity-time graph. Study the following figure and see if you can understand how to use it to solve the problem: Ch2 Page 11

12 Does the solution make sense? Every second, motorcycle B gains on motorcycle A by 2 m/s. Therefore, after 4 s, motorcycle B gains on motorcycle A by 8 m/s. But after 4 s they have the same velocity. This means that at t = 0, A must have been going faster than B by 8 m/s. What are the kinematics equations for constant acceleration? How do you use them? Page 53, Problem 24 A basketball player starts from rest and sprints to a speed of 6.0 m/s in 1.5 s. Assuming that the player accelerates uniformly, determine the distance she runs. Discussion: We assume that the player moves in a straight line without changing direction along the line. In this case the distance travelled is the same as the magnitude of the displacement. Ch2 Page 12

13 Solution 1: Solution 2: Solution 3: Ch2 Page 13

14 Which solution do you prefer? All are good, so use the method that is best for you. (However, the best problem-solvers are those that are comfortable with many methods, so it s worth practicing a variety of different methods. Solution 4: Page 53, Problem 32 Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of m/s, whereas rocket B has an initial velocity of m/s. After a time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is 15 m/s 2. Determine the acceleration of rocket B. Discussion: Just as in the motorcycle problem solved earlier, there are two moving objects here, so we should be careful to label separate kinematical variables for each object: Ch2 Page 14

15 Solution 1: Now substitute the expression for t on the left into the equation on the right and solve for the acceleration of Rocket B: Ch2 Page 15

16 Solution 2: Page 54, Problem 34 A race driver has made a pit stop to refuel. After refuelling, he starts from rest and leaves the pit area with an acceleration of magnitude 6 m/s 2 ; after 4.0 s, he enters the main speedway. At the same instant, another car on the speedway travelling at a constant speed of 70.0 m/s overtakes and passes the entering car. The entering car maintains its acceleration. Determine the time needed for the entering car to catch the other car. Discussion: There are two moving objects, so use A and B to distinguish their kinematical variables. For the first race car, call its "initial" speed the speed it has after it accelerates for 4 s, which is Ch2 Page 16

17 Solution 1: Yes, the cars are side-by-side at t = 0 s, but Car A overtakes Car B after 15.3 s. Solution 2: Ch2 Page 17

18 Page 54, Problem 40 An airplane has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of the intersection is 25.0 m. The plane decelerates through the intersection at a rate of 5.70 m/s 2 and clears it with a final speed of 45.0 m/s. Determine the time needed for the plane to clear the intersection. Discussion: I interpret "clearing the intersection" to mean that the displacement of the plane is 59.7 m m = 84.7 m. Let t = 0 represent the time at which the plane enters the intersection; thus, we are looking for a positive value of t as the solution of this problem (negative values of t represent times before the plane reaches the intersection, and so are not relevant for this problem). If we knew the value of v o instead of the value of v, we could just use the third kinematics equation to solve the problem. But this would work if we could express the initial velocity in terms of the final velocity, and then substitute the resulting expression into the third kinematics Ch2 Page 18

19 equation. Let`s try this. Solution: This is a quadratic equation with only one unknown, t, and this is the variable that we seek. Thus, use the quadratic formula to solve for t. Remember that the negative solution is not relevant for this problem. Page 54, Problem 42 A train is 92 m long and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant speed. At a time t = 14 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 28 s, the car is again at the rear of the train. Ch2 Page 19

20 Determine the speed of the car and the acceleration of the train. Discussion: I haven't drawn a diagram, nor a velocity-time graph, but it might be helpful to draw either or both if you're having trouble following the solution to this problem. I have constructed a table (below) to record the key information given in the problem. It seems sensible to focus on the three different times of interest, 0 s, 14 s, and 28 s. There are two quantities that we need to determine, the speed of the car and the acceleration of the train. We're not given a lot of information, but the conditions that we are given involve the positions of the car and train. This suggests that we use kinematics equations involving displacement to relate the quantities that we wish to calculate. Solution: In the table below, variables labelled x represent the positions of either the car or train at specific times. My strategy in solving this problem is to write down expressions for the positions of the car at the times 14 s and 28 s, and then write down expressions for the positions of the front of the train at the same times. Then I use the conditions given in the problem to connect the positions of the car and the front of the train. Because the car moves at a constant speed, Ch2 Page 20

21 Because the train's initial speed is zero, its displacement in the first 14 s is The train's displacement from 14 s to 28 s is Ch2 Page 21

22 What is free fall? Page 54, Problem 44 A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0 s later it is rising at a speed of 15 m/s. Assuming air resistance does not exist, calculate the speed of the rock (a) at launch, and (b) 5.0 s after launch. Discussion: Choose "up" to be the positive direction. Then the acceleration due to gravity is 9.8 m/s 2, because the acceleration due to gravity points down, which is the negative direction. Let t = 0 represent the time at which the rock is launched. Thus, we are looking for the speeds of the rock at t = 0 and t = 5 s. Solution: Ch2 Page 22

23 Thus, the initial speed of the rock is 34.6 m/s and the speed of the rock after 5 s is 14.4 m/s. Page 54, Problem 46 A ball is thrown vertically upward, and 8.0 s later the ball returns to its point of release. Determine the ball's initial speed. Discussion: I'll choose "up" to be the positive direction again. Solution 1: We are told that the displacement is zero after 8 s. Substituting this information into the following equation will allow us to calculate the unknown coefficient in the equation, the initial speed. Ch2 Page 23

24 Thus, the initial speed is 39.2 m/s. Solution 2: Page 54, Problem 47 Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground? Ch2 Page 24

25 Discussion: I`ll choose upwards to be the positive direction, as usual. Because there are two moving objects, we shall be careful to distinguish the kinematical variables associated with each. Let h represent the height of the cliff above the ground; thus, let y = 0 represent the vertical position of the ground and let y = h represent the vertical position of the top of the cliff. Solution: We know the displacements of each pellet, we know their initial velocities, we know their accelerations, and we wish to determine information about the times at which they hit the ground. This all suggests that we try working with the third kinematics equation for each pellet: Ch2 Page 25

26 Page 54, Problem 63 While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.20 m, you throw a second stone straight down. What initial speed must you give the second stone if they are both to reach the ground at the same instant? Discussion: I shall take upwards to be the positive direction. I shall let y = 0 represent the position of the ground and y = 15 m represent the position of the bridge. Strategy: Let us calculate the time needed for the first stone to fall 3.2 m, and then the time needed for it to fall from the bridge right to the ground. The difference of these two times is the time needed for the second stone to reach the ground, which will help us determine its initial speed. Remember that we assume no air resistance, so that the acceleration of each stone is the acceleration due to gravity. Ch2 Page 26

27 Solution: Ch2 Page 27

28 Page 54, Problem 64 A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.20 s for the tile to pass her window, which has a height of 1.6 m. Determine the distance between the top of the window and the roof. Discussion: We assume that there is no air resistance, so that the acceleration of the tile is the acceleration due to gravity. As usual, we'll take upwards to be the positive direction. Solution: Ch2 Page 28

29 For the motion of the tile across the window, Now that we know the velocity of the tile at the top of the window, we can calculate the initial position by focussing on the motion above the window (i.e., from the initial position to the top of the window): How can we describe motion graphically? Position-time graphs for zero acceleration (i.e., constant velocity) Recall the key facts about the relations between position-time graphs and velocity-time graphs: the slope of a position-time graph is the velocity the slope of a velocity-time graph is the acceleration the area under a velocity-time graph is the displacement Here are some basic position-time graphs and velocity-time graphs: Ch2 Page 29

30 Velocity-time graphs for constant, non-zero acceleration Ch2 Page 30

31 Position-time graphs for constant, non-zero acceleration For each of the seven position-time graphs displayed in the past few pages, draw the corresponding velocity-time graph. Here's one done for you to give you a sense for what I am asking you to do: Ch2 Page 31

32 Do you understand which of the four curved position-time graphs given above matches with which of the four velocitytime graphs given above? Note that we typically assume constant acceleration in this course, which means that position-time graphs are either straight lines or pieces of parabolas, and therefore velocitytime graphs are straight. If you know calculus, you can play with position-time graphs that are more general. Derivations of the kinematics equations for constant acceleration Ch2 Page 32

33 Ch2 Page 33

34 Derivations for calculus lovers ONLY: Derivations for equations (3) and (4) are as above. Page 56, Problem 86 A hot-air balloon is rising straight up at a constant speed of 7.0 m/s. When the balloon is 12.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above the ground are these places? Discussion: As usual, I'll take the upwards direction to be positive, and we'll pretend that there is no air resistance. I'll label the ground as the zero-level for height. There are several ways to solve this problem, as with virtually all problems. I'll adopt a "functional" approach; I encourage all you advanced students (including, and especially, physics majors) to Ch2 Page 34

35 think about this. Most of the time when we solve kinematics problems we think in terms of displacement, but in this problem I want to shift the focus to thinking in terms of positions; in particular, I want to highlight the fact that each moving object has a position function. You can write the position of each moving object as a function of time. Because this problem asks specifically for positions, it makes sense to think in terms of position functions. So, write a position function for each moving object. The position function of an object moving vertically has the form Solution: Remember that the balloon moves at a constant speed, and so its acceleration is zero. The pellet is a projectile, in free fall, so its acceleration is the acceleration due to gravity. Thus, the position functions of the moving objects are We are asked about when the vertical positions of the bullet and pellet are the same; to determine the corresponding times, set the two position functions equal and solve for t. Substituting these times into either of the position functions, we obtain that the balloon and pellet have the same height above the ground at the same time at the heights y = s Ch2 Page 35

36 and y = m. Page 56, Problem 88 A football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of 1.5 s. Then, during a negligible amount of time, he changes the magnitude of his acceleration to 1.1 m/s 2. With this acceleration, he continues in the same direction for another 1.2 s, until he reaches a speed of 3.4 m/s. Determine the value of his acceleration (assumed to be constant) during the initial 1.5-s period. Discussion: Try sketching a velocity-time graph to help you out on this one. My result is 1.39 m/s 2. Page 56, Problem 81 A woman and her dog are out for a morning run to the river, which is 4.0 km away. The woman runs at 2.5 m/s in a straight line. Her dog is unleashed and runs back and forth at 4.5 m/s between his owner and the river, until the woman reaches the river. Determine the distance travelled by the dog. Solution: This is a neat little problem. The dog runs in a complicated back-and-forth motion, and if we assume that the dog has zero width, and makes turns instantly, then there are an infinite number of legs to the dog's journey. ("Balto, a dog cannot make this journey alone; but maybe, a wolf of zero width can!") For those of you who will take MATH 1P02 or 1P06 next semester, you'll learn about infinite series, so you'll have a shot at solving the problem using an infinite series, but there has to be a faster and more way to solve the problem see what you can do with this one. The problem is reminiscent of a famous story involving mathematician John von Neumann, who was famous for being able to do complex calculations mentally. At a party, another mathematician gave him a similar problem; von Neumann Ch2 Page 36

37 thought for a few seconds, and then gave the correct answer. "Most people try to sum the infinite series," commented the other mathematician, assuming that because von Neumann answered so quickly he must have used the quick, insightful method. Nonplussed, von Neumann replied, "Yes, that's what I did." Yes, he was a very smart guy. Ch2 Page 37