July-15-14 10:39 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 • unit conversions (1.3) • dimensional analysis (1.3) • vector and scalar quantities (1.5 − 1.8) • 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) __________________________________________________ Vector and scalar quantities Some of the quantities that we use to describe motion mathematically have magnitudes and directions; these are called vector quantities. Some of the quantities that we use to describe motion mathematically have only magnitudes, with no directionality; these are called scalar quantities. We'll get into the details of vectors and scalars later this week, on Friday. In the meantime, every time a new quantity is introduced, ask yourself if it is a vector or a scalar. Does it have just a magnitude, or does it have a directionality too? In our textbook, vectors are indicated by boldface type and by placing a small arrow above the symbol. (Other books use boldface without the arrow, so watch for this in you linear algebra class or in advanced courses.) I can't do boldface very well in these notes, so I'll just place a small Ch2 Page 1 boldface very well in these notes, so I'll just place a small arrow above a symbol to indicate a vector quantity. Scalar quantities are indicated by symbols in italic type with no arrow above the symbol. The reason we postpone a detailed discussion of vectors until the end of the week is that for motion in a straight line (the main focus of this chapter) there are only two directions, and so we can manage just by using a positive sign for one of the directions and a negative sign for the other direction, without having to use the full power of vector notation. In the next chapter, when we discuss motion in two dimensions, we'll need to use the full power of vector notation, so we'll discuss vectors just before they're needed. _______________________________________________ What is displacement? Example: An object moves along a straight line with initial position 9 m and final position 2 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. If the line of motion is oriented differently (north-south, for example, or in some other direction), then it's up to you to Ch2 Page 2 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 9 m to position 2 m. Then the object moves along the same line another 4 m to the right. (a) Determine the total displacement. (b) Determine the total distance travelled. Solution: How are average velocity and average speed calculated? Example: An object moves along a straight line from position 7 m to position 4 m in 3 s. Then the object moves along the same line back to its starting position in 2 s. Ch2 Page 3 back to its starting position in 2 s. (a) Determine the average velocity for the whole trip. (b) Determine the average speed for the whole trip. Solution: Ch2 Page 4 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: Ch2 Page 5 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 6 This example illustrates that the concepts of motion in one dimension (i.e., straight-line motion) can also be successfully used in certain cases for other types of motion. If the road is curvy, but all we care about is scalar quantities such as distances travelled and speeds, then using concepts of straight-line motion work out fine. And for round-trip motions (even if not in a straight line), the total displacement is zero, so being sloppy and treating the zero vector as if it's the zero scalar is acceptable. (Don't tell your math prof, though!) _________________________________________________ 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 7 _________________________________________________ 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 8 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. __________________________________________________________ Unit conversions We'll take a break from kinematics to discuss how to convert from one unit to another. I assume that you know all about the SI system of units, which is discussed in high school. If you need a refresher, here is a link: http://physics.nist.gov/cuu/Units/ Also, check Section 1.3 in the textbook. Ch2 Page 9 Also, check Section 1.3 in the textbook. _________________________________________________________ Example: Convert 40 km/h to m/s. Solution 1: Note that 1 km is equivalent to 1000 m, and that 1 h is equivalent to 3600 s. Thus, Solution 2: Because 1000 m is equivalent to 1 km, their ratio is exactly 1: Therefore, one can multiply by either of these quotients to convert units. Similarly, Ch2 Page 10 Thus, to convert 40 km/h to m/s, you can proceed as follows: ________________________________________________ Example: Convert 8.7 m/s to km/h. Solution 1: Ch2 Page 11 Solution 2: Dimensional analysis Consider two objects, which have lengths of 5 m and 2 km, respectively. Consider their lengths as physical quantities in their own right, abstracted away from the objects themselves. The two lengths are different, sure, but nevertheless there is something the same about them, because they are both lengths. Ch2 Page 12 something the same about them, because they are both lengths. For example, if the first object has a mass of 20 kg, then its length 5 m and its mass 20 kg are also different, but they are different kinds of quantities. To help us analyze the distinctions discussed in the previous paragraph, we introduce the concept of the dimension of a physical quantity. Although the quantities 5 m and 2 km are different quantities, and they also have different units, we say that they have the same dimension, because they are both lengths. However, the quantities 5 m and 20 kg have different dimensions, because one is a length and one is a mass. The basic dimensions are mass (M), length (L), time (T), and a few others that we won't bother about until future courses. A fundamental principle of physics is that: Each term in a valid physical equation must have the same dimension. There is an every-day slogan that corresponds to this fundamental principle: "You can't add apples and oranges," because it's meaningless to do so. Sure, someone can give you 5 apples and 3 oranges, and sure that means that you have been given 8 items, but in physics 3 kg + 4 m has no meaning. Now, maybe someday someone creative (maybe you?) may figure out a way to make sense of 3 kg + 4 m, but right now this doesn't make sense to us, so we'll stick with the principle that every term of a valid physical equation must have the same dimension. Even though you can't meaningfully add or subtract quantities Ch2 Page 13 Even though you can't meaningfully add or subtract quantities with different dimensions, you can multiply or divide quantities with different dimensions to obtain different meaningful ones. We'll see this throughout the course; for example, a quantity with units kg.m2/s2 might represent energy, or maybe something else. A similar physical principle is that in a valid physical equation the argument of a non-polynomial function must be dimensionless; i.e., a number that has no units. For example, exponents must be dimensionless, angles substituted into sine, cosine, etc., functions must be dimensionless, quantities that you take the logarithm of must be dimensionless, etc. An example of dimensional analysis The fact that each term of a valid physical equation must have the same dimension gives us a tool for (partially) checking physical formulas, and also for guessing them; this tool is called dimensional analysis, and we'll discuss an example shortly. Dimensional analysis provides only a partial check, because errors involving dimensionless numbers are not found using this method, as we'll see. For example, one of the kinematics equations that we'll use later this chapter is or, if you wish, Check the dimensions of each term, and you'll see that they are the same, so this is a partial check on the equation: Ch2 Page 14 Each term in the equation has dimensions of length, so the equation has the possibility of being correct. To be certain that it's correct, we would have to carefully derive it using physical principles, correct mathematics, and correct logic. And of course, the equation would have to then be extensively tested by observations and experiments; in science, experimentation and observation are the ultimate arbiters. The reason that dimensional analysis is only a partial check is that had we written an incorrect version of the above kinematics equation that had incorrect dimensionless coefficients, such as Dimensional analysis could never detect the error. Dimensional analysis only serves to test whether the principle that all terms in a valid physical equation have the same dimension has been satisfied. Dimensional analysis does not verify that the equation is correct! Nevertheless, dimensional analysis is a beloved and useful tool for physicists. It can also be used to guess the form of a physical equation. A classic example is the simple pendulum that all you 1P91 students will be studying in Experiment 1. What factors might affect the period of a simple pendulum? Well, hang a weight on the end of a string and play with it, and you'll soon come up with several factors that seem to play a role. We don't have a string handy, but we might guess by asking ourselves what we could change about the pendulum that might change its period. Guess using dimensional analysis: Ch2 Page 15 Guess using dimensional analysis: Remember, the guess using dimensional analysis is only a guess. To determine the correct formula, one would have to analyze the situation using the relevant dynamical theory (Newton's laws of motion, in this case). Then the formula would have to be extensively tested experimentally. All you 1P91 students will do some of this work in Experiment 1. If you're interested, you can study the development in Section 10.4 of the textbook, where the following correct formula is derived: Ch2 Page 16 Note that dimensional analysis was successful in providing the right functional form, but it is not possible to determine the correct numerical factors without a full analysis based on Newton's laws of motion. But just the functional form is often enough to tell you what you want to know; for example, knowing that the period is proportional to the square root of the length of the pendulum gives us the useful information that increasing the length of the pendulum increases the period. The fact that mass does not appear in the formula tells us that the period is independent of the mass. __________________________________________________ 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 average acceleration, in both (km/h)/s and in m/s2. Solution: Ch2 Page 17 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/s2. 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/s2. Solution: Ch2 Page 18 ________________________________________________ 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 v0 a (a) +12 m/s +3.0 m/s2 (b) +12 m/s −3.0 m/s2 (c.) −12 m/s +3.0 m/s2 (d) −12 m/s −3.0 m/s2 Solution: Ch2 Page 19 The conclusion we draw from this example is that (for motion in a 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/s2 east, while cycle B has an average acceleration of 4.0 m/s2 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 20 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 Ch2 Page 21 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: 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. _______________________________________________________ Understanding velocity-time graphs The slope of a velocity-time graph represents acceleration. Ch2 Page 22 ________________________________________________ The area "under" a velocity-time graph represents displacement. The first diagram below illustrates this fact for a motion with constant velocity (i.e., zero acceleration), and the second diagram below illustrates this fact for nonzero acceleration. (In both cases the acceleration is constant.) Ch2 Page 23 __________________________________________________ Arguments based on the diagrams above lead us to basic equations relating the kinematical variables that are valid for constant acceleration. The following four equations are sufficient for solving all the problems in the textbook. (Formal derivations of the equations, for those who are interested, appear later in the notes for this chapter.) __________________________________________________ What are the kinematics equations for constant acceleration? How do you use them? ___________________________________________________ Page 53, Problem 28 (a) Determine the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 8.0 m/s when going down a slope for 5.0 s. (b) Determine how far the skier travels in this time. Solution: Ch2 Page 24 Thus, the magnitude of the average acceleration is 1.6 m/s2. (b) There are several possible methods, all illustrated below: The displacement of the skier is 20 m. __________________________________________________ Ch2 Page 25 Page 53, Problem 29 A jetliner, traveling northward, is landing with a speed of 69 m/s. Once the jet touches down, it has 750 m of runway in which to reduce its speed to 6.1 m/s. Determine the average acceleration (magnitude and direction) of the plane during landing. Solution: One way to solve the problem is to begin with the velocitytime graph. Note that we could calculate the slope of the graph (which is the required acceleration) if we knew the time over which the acceleration takes place. We can do this by writing an expression for the shaded area and equating it to the given displacement, 750 m. Ch2 Page 26 Now that we know the time, we can determine the acceleration. Thus, the magnitude of the acceleration is 3.15 m/s2, and the acceleration is in the negative direction (that is, the acceleration is in the south direction). ______________________________________________________ An alternative solution: _____________________________________________________ Example: Alice starts from rest and accelerates at a constant rate of 2 m/s2 for 5 s. Then her acceleration suddenly increases to a new constant rate of 4 m/s2, which she maintains for an additional 10 s. (a) Determine her average acceleration for the entire time period. (b) Determine her displacement during the entire time period. Solution: Ch2 Page 27 Ch2 Page 28 Ch2 Page 29 ____________________________________________________ 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. Solution 1: Solution 2: Ch2 Page 30 Solution 2: Solution 3: 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 Ch2 Page 31 direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of +5800 m/s, whereas rocket B has an initial velocity of +8600 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/s2. 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: Solution 1: Ch2 Page 32 Now substitute the expression for t on the left into the equation on the right and solve for the acceleration of Rocket B: Solution 2: ________________________________________________ Ch2 Page 33 ________________________________________________ Page 53, Problem 33 A car is travelling at 20.0 m/s, and the driver sees a traffic light turn red. After 0.530 s (the reaction time), the driver applies the brakes and the car decelerates at 7.00 m/s2. Determine the stopping distance of the car, as measured from the point where the driver first sees the red light. Solution: Ch2 Page 34 The car's stopping distance is 39.2 m. ____________________________________________________ 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/s2; 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 Solution 1: Ch2 Page 35 Yes, the cars are side-by-side at t = 0 s, but Car A overtakes Car B after 15.3 s. Solution 2: 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/s2 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 + 25.0 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 Ch2 Page 36 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 vo 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 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. Ch2 Page 37 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. 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. Ch2 Page 38 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, 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 39 What is free fall? https://www.youtube.com/watch?v=E43-CfukEgs Page 54, Problem 43 The greatest height reported for a vertical jump into an air bag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s. To assess the effects of air resistance, determine how fast he would have been travelling on impact had air resistance been absent. Solution: I'll choose downward to be the positive direction. Ch2 Page 40 Solution: I'll choose downward to be the positive direction. In taking the square root, we have to consider both signs. We know the person falls down, in the positive direction, so his impact velocity would have been +44.1 m/s if there were no air resistance. The actual impact velocity is +39 m/s, so you can see that air resistance decreased the impact velocity by about 5 m/s. It's instructive to solve the problem again, but this time choosing upward to be the positive direction. In this case, both the acceleration due to gravity and the displacement are negative, so the result will be Ch2 Page 41 In this case, the impact velocity is −44.1 m/s. In both cases, the impact speed is 44.1 m/s. Thus, the choice of positive direction does not affect the final solution for the physically measurable quantity, the impact speed. ___________________________________________________ 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/s2, 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 42 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 43 speed. 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, Ch2 Page 44 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? 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 45 ________________________________________________ Page 54, Problem 49 A hot air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 3.0 m above the ground, the balloonist accidentally drops a compass over the side of the balloon's basket. Determine how much time elapses before the compass hits the ground. Solution: The key point to observe is that at the instant that the compass is dropped, it shares the velocity of the balloon. Think about this. When the compass is in the balloonist's hand, it rises upwards together with the balloon and everything else in it, all at a speed of 2.50 m/s in the upward direction. Thus, when the compass is Ch2 Page 46 in the upward direction. Thus, when the compass is "dropped" it nevertheless continues to rise upwards for a while, relative to the ground. Of course, relative to the balloon, the compass does drop immediately, because the balloon continues to rise upwards at a constant speed of 2.50 m/s, whereas the compass begins to slow down as soon as it is released. Ch2 Page 47 We reject the negative value, as it is not relevant for our problem. Thus, the compass reaches the ground 1.08 s after it is released. __________________________________________________ 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. Solution: Ch2 Page 48 Ch2 Page 49 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: For the motion of the tile across the window, Ch2 Page 50 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 51 Velocity-time graphs for constant, non-zero acceleration Ch2 Page 52 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 53 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 54 Derivations for calculus lovers ONLY: Ch2 Page 55 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 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. Ch2 Page 56 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 = 16.185 m and y = 40.672 m. __________________________________________________ Example: 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 coin is dropped from the balloon. (a) Determine the maximum height above the ground reached by the coin. (b) Determine the time needed for the coin to reach the Ch2 Page 57 (b) Determine the time needed for the coin to reach the maximum height. (c.) Determine the time needed for the coin to reach the ground. Solution: Just as in the previous example, adopting a "functional" approach makes for a smooth solution. That is, think of the vertical position of the coin as a function of time, and think of the vertical velocity of the coin as a function of position. First choose a positive direction; I'll choose "up" as positive, so the acceleration due to gravity is negative, as it is in the downwards direction. One of the things people have trouble wrapping their minds around at first is that when the coin is released its velocity is 7 m/s upwards. We called this the coin's "initial" velocity, imagining that a stopwatch is started just as the coin is released. Of course, the coin immediately begins to gradually slow down on its upward journey, because the acceleration opposes its motion. The coin gradually slows down on its upward journey until it stops for an instant at its peak height. Then the coin begins to accelerate downward, picking up speed gradually and continually as it falls. In class we discussed that the key condition satisfied by the coin's motion when it reaches its peak height is that its velocity is zero then. We can use this condition to write an equation that then allows us to solve parts (a) and (b). Note, however, that the acceleration of the coin is never zero while it is in the air. There is a common misconception that at the peak of the coin's motion its acceleration is momentarily zero. This is false. Our fundamental assumption is that (ignoring air resistance, as we typically do) the acceleration of the coin is constant while the coin is in the air, and therefore is never zero while the coin is in the air. The coin's velocity is zero Ch2 Page 58 never zero while the coin is in the air. The coin's velocity is zero at the peak, not the acceleration. The condition for reaching maximum height is that the velocity is zero. Ch2 Page 59 Ch2 Page 60 Note the symmetry of the graph of the position-time function; the graph is a parabola, and the axis of symmetry corresponds to the time at which the coin Ch2 Page 61 symmetry corresponds to the time at which the coin reaches its peak height (which corresponds to the vertex of the graph). The intercepts of the position-time graph represent the times when the coin is at the ground. The negative solution of the equation is usually rejected, because only the part of the position-time graph for positive times is relevant for this coin's motion. Nevertheless, the negative solution does have a meaning. What is it? (All you physics majors should think about this.) ______________________________________________ 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/s2. 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/s2. 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!") Ch2 Page 62 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 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 63

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