Motion in 1 Dimension. By Prof. Massimiliano Galeazzi, University of Miami

Motion in 1 Dimension By Prof. Massimiliano Galeazzi, University of Miami When you throw a pebble straight up, how high does it go? How fast is it when it gets back? If you are in your car at a red light and the light gets green, how long does it take you to reach the speed limit? How far have you traveled? These are just some of the questions that we will be able to answer in this chapter. The focus of the chapter is motion in 1-dimension, that is, we will only consider objects moving along a straight line, which could be horizontal, like in the example of the car, vertical, like in the example of the pebble, or have any other orientation. In this chapter we will also neglect the physical shape of the object, for example the fact that the pebble could also rotate, or that the car has many parts moving in different ways. In general, we will refer to all objects a point particles, or simply particles, meaning that their physical geometry is negligible compared to the motion we are describing. --------------------------------------------------------------------------------------------------------------------------------------------------- MATH INSERT The description of this chapter is completely based on calculus and the use of derivatives. Since some of the students reading this chapter are also taking a parallel calculus course, we will start with an overview of the calculus principles that we will need. This is by no means intended as an exhaustive treatment of functions and derivatives, but just a functional description in support of this chapter, mostly for the benefit of students that have not seen it yet, but will cover it, in a much more formal way, in the next few weeks. 2.1 Functions and variables In high school we have learned that or is used to indicate that the variable is a function of the variable (i.e., the values of y depend on the corresponding value of x). The function may have many forms and describes how the variable y changes as a function of the variable x. Before proceedings it is important to remember that y and x are just two of the many symbols that can be used in functions and their preferred choice in many math problems is simply due to the general convention of using x and y as canonical variables for the Cartesian plane, but we can as well write to describe the fact that the variable a is a function of the variable b. In particular, in physics we tend to assign different variables to different quantities to avoid confusion between them, as there are many different quantities that will be introduced in this book alone. So many, in fact, that we will make use of the full Latin and Greek alphabets, both lower and upper case, and we still won t have enough symbols to describe all the physical quantities that we will introduce. From the mathematical point of view, this means that you should get used to see function such as 3, which simply means that the quantity (the Greek letter alpha) depends on the quantity c, more specifically is three times the square of c. For reasons that will be evident later on in this chapter, we will use the quantities x and t for the examples in this insert [ or simply ], but, of course, everything we learn is valid for any combination of variables. 2.2 The First Derivative Sometimes, rather than knowing how x depends on t we may be interested in how x changes when t changes. Let s assume, for example, that we start from and, where and are two specific values of our variables, and the variable t changes by an amount, where (the Greek letter capital delta) is a common symbol used to indicate changes (or intervals) in the value of the variable attached to it and is the new value of the variable t. The new value of t can also be written as. 1

As a consequence of the change in the value of t, the variable x (which depends on t) will change too, and we indicate the change in the value of x as. Since x is a function of t, we can also write it as. It is important to note that the value of depends not only on the change, but also on the initial value of t, in this example. This implies that is, in turn, a function of t (and ), that can be written as [2.1] A particularly interesting case is when the change in the value of the quantity t becomes very small (we will refer to it as infinitesimally small ), that is, the limit of going to zero ( 0). In this limit, we can quantify the rate of the change in the function, that is, how much x changes for a given change in t, by taking the ratio between the two intervals, and. It is common practice to use the symbol to represent discrete intervals (or changes) in the value of a quantity, and the symbol d to represent infinitesimally small intervals. The rate of change is thus expressed as /: lim lim. [2.2] This rate of change is, in turn, a function of the quantity t, (for example, x may change more rapidly for small values of t and less rapidly for higher values) which is called the first derivate of x versus t and the functional dependence is often expressed using the apex:. We can thus use equation [2.2] as an operational definition of the first derivative: lim. [2.3] First derivatives play an important role in physics, but before we begin using them, we will show how we can explicitly calculate them using equation [2.3] for a few simple cases. Example 2.1 The variable x is constant:, where c is a constant lim lim 0 the first derivative of a constant function is zero. Example 2.2 The variable x depends linearly on t:. On a Cartesian plane where t is abscissa and x the ordinate, this is the equation of a straight line with angular coefficient c. lim lim lim lim the first derivative of a straight line is constant and equal to the angular coefficient of the line. Example 2.3 The variable x depends quadratically on t: lim lim lim lim2 2. the first derivative of is proportional to t. 2

Example 2.4 There is a trend common to these example, which can be generalized to all power law functions: the first derivate of is always proportional to. Example 2.5 Finally, I will leave it to you as an exercise to show that when you add and/or subtract different terms (i.e., polynomials), then the first derivative is just the sum and/or difference of the first derivatives of the individual terms. For example, if, where a, b, and c are constants, 2. 2.3 The Antiderivative As we will see later in this chapter, sometimes we may be challenged with the inverse problem, that is, we may know the functional form of the first derivative of a function and need to find what the function itself looks like. The operation is called antiderivative (although it can also be replaced by using the concept of integral). Rather than introducing an operational definition of antiderivative, we will simply use a few examples to see how to use what we have already learned and some additional considerations to invert a derivative. Example 2.6 Given, where a is a constant, find the function. In example 2.3 we have seen that the function whose derivative is proportional to t is a square law, that is, if 2. In our example we could say that 2 and therefore. If you calculate the first derivative of this function you will get, indeed, to first derivative given by the problem. However, the solution is not complete, but requires the addition of a constant factor. The reason is that when you calculate the derivative of a constant factor the result is zero, as seen in example 2.1, therefore the function, where is a constant, has the same derivative as. How do we choose between the two functions or, more in general, how do we choose the value of the constant? The answer to this question depends on the specifics of the problem, in particular, we may notice that is the value of the quantity x when 0, 0, which can be calculated using some additional information given by the problem. To summarize, while given a function we can calculate its first derivative, when we reverse the problem and try to find the function given the first derivative the result will be defined plus/minus a constant factor. The problem must therefore give us some additional information, called initial conditions or boundary conditions to find a unique solution. We can use a simple example to understand why. Let s assume that you are filling a bucket with water using a hose. If you measure the water level as a function of time you can find how fast the level in the bucket rises and therefore the rate at which water comes out of the hose. This is equivalent to finding the first derivative of a function knowing the function itself. However, if you measure the rate of water coming out of the hose you cannot say how much water there is in the bucket at any given 3

time unless you also know how much water is already in the bucket when you start adding water to it with the hose. This is equivalent to finding the function knowing the first derivative. Example 2.7 We can generalize what we have learned in the previous example to any other function and first derivative. Using the result of Example 2.4, for example, we can say that, if Again, we need the problem s initial conditions to find the unique value of the constant that satisfies the specific problem, as shown in the next example. Problem 2.1 Here are two simple problems to see how to practically calculate the antiderivative, using specific numbers (i.e., by replacing the symbols a, b, c, used so far with specific numbers). [a] Find the function knowing that 2 34 and that 0 1. By inverting the first derivative using example 2.7, we obtain 4. By using the initial condition 0 1 we also obtain 1 and thus 41. [b] Find the function knowing that 4 and that 1 2. By inverting the first derivative, we obtain 4. We can again use the initial conditions given above, i.e., when 1 2, or 1 2, to find the value of. In this case we plug the given values of x and t into the equation that we have calculated and solve for : 2 1 41 2 4 The complete function will therefore be: 4. END OF MATH INSERT --------------------------------------------------------------------------------------------------------------------------------------------------- 2.4 Average Speed and Velocity The previous math insert took quite a bit of our time, but it will make the rest of this chapter much easier. Remember that the goal of this chapter is motion in 1 dimension. The problem of describing the motion of a particle can be synthetically expressed as finding a particle position as a function of time. For example, it is possible to describe the motion of an airplane flying from New York to Los Angeles by placing a GPS receiver onboard the airplane to find out where the plane is at any given time. If we limit our work to problems in 1 dimension, then a particle (the airplane in our example) can only move along a straight line. We can therefore define a coordinate system with the x-axis aligned with the particle motion and use the coordinate x to describe its position as a function of time t. The problem thus reduces to finding the function.. 4

Consider, for example, another simple problem of a car traveling on a straight highway from Miami to Orlando. You could describe the motion of the car by using a coordinate system with the origin in Miami and the positive x-axis in the direction of Orlando. The function x(t) thus describes the distance of the car from Miami as a function of time. There are, however, other quantities related to the motion of the car that you can calculate. For example, if you know the distance L between Miami and Orlando, and the time interval required for the trip, you can calculate the average speed of the car, defined as:. [2.4] Using the SI, the distance is measured in m, the time in s, thus the units of speed are m/s, or, in terms of dimensions, a distance divided by a time. If you know not only the distance traveled, but the initial and final position of the car ( and ), then you can also calculate the average velocity of the car, defined as:, [2.5] Which has the same units (and dimensions) of the speed. At this point it may come natural to ask what is the difference between average speed and average velocity, as their definition seem, at first look, identical. This is because the two definitions are similar (but not identical) and the quantities may have the same numerical value. However, it is important to understand that they are different quantities that represent different physical properties of the particle motion. In particular, the average speed is a scalar quantity, which quantifies how much the car has traveled in a specific time interval. The average velocity, instead, is a vector quantity, which quantifies how the position of the car has changed in a specific time interval. In one dimension the car can move only along the x-axis, thus we only consider the x-component of the average velocity vector, which is still different from the average speed, as shown in the next example (although in many problems their numerical value may be identical). Example 2.8 To get to school you drive 10 km east to pick up your friend, then together you drive west 12 km to school. The total trip takes half an hour. Calculate [a] the average speed and [b] the average velocity. [a] To calculate the average speed we must calculate the distance L traveled by the car and divide it by the time interval for the trip. In this case the total distance traveled is 10 12 22, while the time interval is 0.5.. 44. [b] To calculate the average velocity we need instead the total change in position. First of all, to define x (and therefore ) we must define a coordinate system. We could, for example, define a coordinate system with the positive x-direction pointing East and the origin at the starting point of the trip. The trip can then be described as 10 km in the positive x- direction and 12 km in the negative x-direction, for a total change in position 2. Alternatively, we can say that your initial position is 0 and your final position is 2 km. FIG. 2.1 5

. 4. In this example the average speed and average velocity not only represent different quantities, but also have different numerical values. 2.5 Instant Speed and Velocity In example 2.8 average speed and average velocity are clearly useful quantities that characterize the overall trip, however, they do not tell, for example, whether the car traveled faster in one direction and slower in the other, or how many times it stopped due to traffic. To get additional information it is possible to divide the trip into smaller intervals (for starter each leg of the trip) and calculate the average quantities for each interval. Suppose that we keep dividing the trip into smaller and smaller intervals to get more detailed information, until we get to the point where each time interval becomes infinitesimally small, that is, until 0, then the average quantities become instant ones. This is somewhat equivalent to recording the speedometer of the car at any given instant. Using equation [2.5] and replacing the discrete intervals and with the infinitesimal ones dx and dt, we obtain an expression for the instant velocity (or simply velocity) as: lim, [2.6] that is, the velocity of a particle is the first derivative of its position. This is a very simple and yet very important conclusion, as it allows us to use everything we know about functions and their first derivatives to solve problems involving velocity. We can, for example, find the velocity of a particle at any given time if we know its position as a function of time. Or, given the velocity of a particle as a function of time and its initial position, we can use the antiderivative to find its position at any given time. Note that the quantity dx, and thus the velocity, can be positive or negative, depending on whether the particle moves in the positive or negative direction. In either case, though, the distance traveled is the same and equal to. It follows that we can also define an instant speed (or simply speed) as:, [2.7] That is, in one dimensions the instant speed is equal to the absolute value of the instant velocity. Problem 2.2 A particle, initially at rest in the origin, starts moving at time 0, and its position as a function of time (for positive t) is given by the function 10 5 2. [2.8] Find [a] the particle average velocity between 0 and 1, [b] the particle instant velocity as a function of time, and [c] the time(s) when the particle is at rest. Strategy Equation [2.8] can be used to find the particle position at any given time, including and, and thus to find the average velocity using Eq. [2.5]. We can also calculate its first derivative to find the instant velocity as a function of time and the time at which it is zero. 6

Setup In this particular problem the coordinate system is already been defined by the problem itself, so no specific setup is necessary. Solution [a] From Eq. [2.5] we know that, with 0 and 1. We can then use Eq. [2.8] to find the value of and and thus the average velocity: 0 0 ; 1 10 1 5 1 2 1 10 5 2 13; 13. Check units: we are calculating a velocity which has units of m/s, therefore the units of the result are correct. [b] Before we proceed we can rewrite Eq. 2.8 as, [2.9] with 10, 5, and 2. To find the instant velocity we use equation [2.6], i.e., the velocity is the first derivative of the position given by Eq. [2.9]: 23. [2.10] To find a numerical solution, we simply plug the values of,, and into Eq. [2.10]: 10 10 6. Check units: The result is a velocity, with units of m/s, by comparing left and right side of the solution we get, that is, we are adding/subtracting all quantities in units of m/s to get a result in m/s. [c] The particle is at rest when its velocity is zero. In Eq. [2.10] we have already found an expression for the velocity as a function of time, so we simply set 0 and solve the equation for t: 2 3 0. [2.11] The problem has two solutions, but we must verify whether both of them are acceptable. In particular, any solution that gives a negative time is, in general not acceptable. We can do that by plugging numbers directly in Eq. [2.11], which will do in a moment, but we can also do it directly by looking at Eq. [2.11]. We can start from the observation that,, and are all positive, then the expression inside the square root in Eq. [2.11] is the sum of the quantities and (3), which implies that the whole square root term is bigger than the quantity. The solution with the negative sign is therefore negative, thus not acceptable, and we can simply write:. At this point we can plug in the values of,, and to find a numerical solution: 7

. Check units: We are measuring a time and the result is, correctly, in s. Also, throughout the solution, all the terms that have been added/subtracted together have the same units. Problem 2.3 A car moves at constant velocity 4 position of the car as a function of time. toward the positive x direction. At time 2 the car is at the origin. Find the Strategy The problem gives us an expression for the car s velocity, i.e., the first derivative of the position, and an initial condition, i.e., its position at a specific time, although not 0. We thus simply calculate the antiderivative using the initial condition given by the problem to find its position as a function of time. Solution From Eq. [2.6] we can write. [2.12] To find the position as a function of time we must invert Eq. 2.12. In example 2.2 we have seen that the function whose derivative is a constant is proportional to t. In example 2.6 we have also discussed that when we reverse a derivative we must always add a constant that can be determined using the initial conditions given by the problem, therefore. [2.13] The initial conditions to find the value of is given by 2 0, which, when plugged into Eq. [2.13] gives 4 2 0 8. 4 8. [2.14] Check units: the left side of Eq. [2.14] represents a distance, with units of, the units are correct. Problem 2.4 The famous director James Maxwell is shooting the final scene if his romantic masterpiece Found Love. In the final scene, the main character Adam sees his beloved Bernadette at the other end of a square, 50 m away. The two start running toward each other, at constant speed, for a kiss, Adam at 3 m/s and Bernadette at 2 m/s. [a] How long does it take for the two to meet and [b] where do they meet? Strategy In mathematical terms, the fact that Adam and Bernadette meet means that they are in the same place at the same time. To solve the problem we must therefore find the positions of each of them as a function of time, set them equal and solve for t to find when they meet. We then plug the time found into the equation of the position as a function of time to find where they meet. 8

Setup The first thing we must do in this problem is to set up the coordinate system. We can, for example, choose a reference frame with the x-axis coincident with the line going from Adam to Bernadette and the origin at the initial position of Adam. With this choice Adam is initially in the origin and moves at constant speed 3 toward the positive x- direction, while Bernadette is initially on the positive x-axis at 50 and moves at constant speed 2 toward the negative x-direction (see figure). FIG. 2.2 Solution [a] Let s start with the position of Adam, who moves at constant velocity and at time 0 is in the origin, i.e., at 0. Using Eq. [2.6], the problem initial conditions, and reversing the derivative, we get, [2.15] We can repeat the same for Bernadette, who at time 0 is at and moves at constant velocity. Using again Eq. [2.6], the problem initial conditions, and reversing the derivative, we obtain. [2.16] Finally, to find when they meet we set and solve for the time t:. [2.17] Notice that the time is the same necessary for a single particle moving at speed to cover a distance. To find the numerical result we can plug the values of,, and in Eq. [2.17]: 10. Check units: we are calculating a time and the result has the proper units of s. Also, at the denominator we are properly adding two quantities with same units of m/s. [b] To find the position of the encounter we simply plug the time from Eq. [2.17] in either Eq. [2.15] or [2.16]:. [2.18] Numerically: 30. Check units: we are calculating a distance and the result has the proper units of m. 9

NOTES: Since in our setup the origin marks the initial position of Adam, we can also say that he travels 30 m while she travels 20 m. Finally, we note that, if to find the position of the encounter we use Eq. [2.16] instead of Eq. [2.15], i.e., the equation for her position, the result still gives the distance traveled by Adam, not Bernadette (you can check that, indeed, it is identical to Eq. [2.18]). The reason is simply that the result always gives the coordinate x of the encounter and in our initial setup we decided to use Adam s initial position as the origin. 2.6 Acceleration Let s assume that we are driving our car at constant velocity and the light in front of us gets red, then we must push on the car breaks and come to a halt, i.e., we change the velocity of the car. How quickly do we come to a halt? That obviously depends on the rate at which the velocity changes. The rate of change of the velocity is called acceleration. Notice, that, somewhat differently from common English language, in physics acceleration represents a change in the velocity, which can be either positive if the velocity increases, or negative if the velocity decreases. In the example of the car and the red light we would therefore say that the car is accelerating with negative acceleration. In Section 2.2 we have seen that the mathematical operator used to quantify a rate of change is the first derivative (e.g., the velocity is the rate of change in the position of a particle). In analogy to what we have done for the velocity, we can thus define the (instant) acceleration of a particle as the first derivative of the velocity: lim. [2.19] Notice that the acceleration, like the velocity, is a vector quantity, but in our 1-dimensional approach we are considering only its x-component. We can also write an expression for the acceleration in terms of the particle position x. To find the particle acceleration given the position we must, in fact, calculate the first derivative of the position to get the velocity, the again the first derivative of the velocity to get the acceleration. The mathematical operation of two first derivatives one after the other is called a second derivative and is expressed as. [2.20] In analogy to Eq. [2.5] we can also define an average acceleration of a particle going from the initial velocity at time to the final velocity at time as:. [2.21] Example 2.9 For the same particle of Problem 2.2, find the time(s) when the particle acceleration is zero. In problem 2.2 we have already found an expression for the velocity as a function of time: 23, [2.22] with 10, 5, and 2. Using Eqs. [2.20] and [2.22] we can therefore calculate an expression for the particle acceleration as a function of time: 10

26. [2.23] To find the time when the acceleration is zero it is therefore sufficient set 0 in Eq. [2.23] and solve for the time t: 2 6 0. [2.24] To find the numerical solution we simply plug in Eq. [2.24] the values of and :. Check units: we are calculating a time and the result is, correctly, in s. Example 2.10 A particle, which at time 0 has velocity 0 and position 0, is accelerated with an acceleration that increases with time as, where is a positive constant with units of m/s 4. Find the position and velocity of the particle as a function of time. Using Eq. [2.19], the initial condition given by the problem, and inverting the derivative (using example 2.4), we get:, [2.25] Repeating the same process using Eq. [2.6]: Problem 2.5. [2.26] A train is moving on a very long straight track and is subject to a negative acceleration proportional to time 12. At time 1 the train is at position 14 moving with velocity 10 /. Find [a] the time(s) when the train passes through the origin and [b] the speed(s) of the train in the origin. Strategy To find when the train passes through the origin, we first need to find an equation for the position of the train as a function of time, set it equal to zero, then solve for t. Once we found the time, we can plug it in the equation for the train velocity as a function of time to find the velocity in the origin. Setup In this problem the coordinate system is already given. What we must do is simply to assign variables to all numerical values (only one in this case). We can write, with 12 /. Solution [a] Using Eq. [2.19] and inverting the derivative we obtain:. [2.27] To find the value of we make use of the initial condition 1 10 : 11

10 12 1 10 6 16. Using the result of Eq. [2.27] with Eq. [2.5] we can then find an expression for the train position:, [2.28] and to find the value of we make use of the initial condition 1 50 : 14 12 1 16 1 14 2 16 0. [2.29], with 12 / and 16 /. [2.30] Equation [2.30] represents the position of the train as a function of time. To find when the train passes through the origin we set the position 0 in the equation and solve for the time t: 0 0,. [2.31] Notice that there is a third solution,, which we have excluded because the time is negative. Also, notice that we could have inferred the value of from Eq. [2.29], which implies that the train is at the origin at time 0. To find numerical solutions we substitute the values of and in Eq. [2.31]: 0, 8 2 2. Check units: we have calculated two times and the results are, correctly, in s. [b] To find the speed when the train passes through the origin we simply plug the values for the times found in Eq. [2.31] into the velocity equation [2.27]: 16, 12 2 2 16 24 16 8, We then make use of the fact that the speed is the absolute value of the velocity to find: 16, 8. Check units: we have calculated two speed and the results are, correctly, in m/s. 2.7 Constant Velocity and Constant Acceleration Although no different from what we have already learned, in this session we will review two examples on the specific cases of constant velocity and constant acceleration. 12

Example 2.11 A particle which, at time t=0 is located at, moves with constant velocity. Find [a] the acceleration of the particle and [b] its position as a function of time. [a] From Eq. [2.19] 0. [2.32] When the velocity is constant its rate of change, i.e., the acceleration, is zero. [b] We have actually already answered this question in problem 2.4. Using Eq. [2.6] and the initial condition 0 we get:. [2.33] When the velocity is constant the position changes linearly with time. Example 2.12 A particle which, at time t=0 is located at and has velocity, moves with constant acceleration. Find [a] the velocity of the particle and [b] its position as a function of time. [a] Using Eq. [2.19] and the initial condition 0 we obtain:. [2.34] [b] Using equations [2.6] and [2.34], and the initial condition 0, we obtain: 2.8 Gravity. [2.35] Four centuries ago Galileo showed that all objects fall toward the center of the Earth with constant acceleration called gravitational acceleration. The magnitude of the gravitation acceleration is usually represented by the symbol g and has value 9.81 /. Problems involving gravity are, in practice, variations of what has been shown in Example 2.12, with a bit of attention to the problem setup. Problem 2.6 A small ball is released from a height h above the floor. How fast does it move when in reaches the floor? Strategy To find the speed at the floor we find the equations for the velocity and position of the ball. We use the position equation to find the time at which the ball reaches the floor and plug it into the velocity equation to find the velocity. Setup #1 As a useful exercise we will solve this problem using two different setups, to show that we can solve any problem in multiple ways, depending on the setups used, but, as long as we are consistent in our calculations, we will get the same result. In the first setup we use the y-axis aligned vertically, with the positive axis pointing up and the origin at the floor 13

level. Thus the ball is subject to a constant acceleration in the y-direction (pointing down), has initial zero velocity, and initial position. Obviously, with this setup our position is defined by the variable y instead of the variable x that we used so far, but this should not worry you! FIG. 2.3 Solution #1 Using Eq. [2.19] and the initial condition 0 0 we obtain:. [2.36] Using equations [2.6] and [2.36], and the initial condition 0, we get:. [2.37] Equations [2.36] and [2.37] represent the equations for the velocity and position of the ball as a function of time. In our setup the floor corresponds to 0, therefore to find the time at which the ball hits the floor we set 0 in Eq. [2.37] and solve for t: 0. [2.38] Only the positive solution of Eq. [2.38] is physically acceptable, therefore. [2.39] To find the velocity when the ball hits the floor we then plug the solution from Eq. [2.39] into Eq. [2.36]: 2. [2.40] The final velocity of the ball has magnitude 2, pointing down (negative y-direction in our setup). Check units:, the units of the left are the same as the units of the right side of Eq. [2.40]. 14

Setup #2 We could solve the same problem using a different choice of setup. For example, we could choose our axes with the x- axis vertical, the positive x-axis pointing down, and the origin at the initial position of the ball. Then the ball moves with constant acceleration (pointing down), has initial zero velocity, and initial position 0. FIG. 2.4 Solution #2 Using Eq. [2.19] and the initial condition 0 0 we obtain:. [2.41] Using equations [2.6] and [2.41], and the initial condition 0 0, we get:. [2.42] Equations [2.41] and [2.42] represent the equations for the velocity and position of the ball as a function of time. The equations look different from Eqs. [2.36] and [2.37], even if they represent the same quantities. The reason is just the different initial setup that we have chosen. It is thus very important, when we solve a problem, to clearly specify the setup used! A problem solution, even if it looks perfect, does not make any sense if the setup is not specified. In this second setup the floor corresponds to, therefore to find the time at which the ball hits the floor we set in Eq. [2.42] and solve for t:. [2.43] Again, only the positive solution of Eq. [2.43] is physically acceptable, therefore. [2.44] Notice that Eq. [2.44] is identical to Eq. [2.39]. While the equations for the motion and velocity may look different, depending on the setup, the time it takes the ball to reach the floor is always the same and thus the result MUST be independent on the setup used. To find the velocity when the ball hits the ground we then plug the solution from Eq. [2.44] into Eq. [2.41]: 15

2. [2.45] The final velocity of the ball has magnitude 2, pointing down (positive x-direction in this setup), which is exactly the same result obtained from Eq. [2.40]. Problem 2.7 A projectile is launched straight up with initial velocity 10.0 /, find how high the projectile goes. Strategy The maximum height of the projectile is the point where the projectile changes direction, from moving up to moving back down, i.e., it is the point where the projectile velocity changes sign. At that point the velocity is thus zero. To solve the problem we must therefore find the equations for the velocity and position of the projectile, find the time at which the velocity is zero, and finally find the projectile position at that time. Setup To solve this problem we choose the y-axis vertical with the positive axis pointing up and the origin at the initial position of the projectile. The projectile is thus subject to constant acceleration 9.81 /, and has initial velocity 10 / and initial position 0. FIG. 2.5 Solution Using Eq. [2.19] and the initial condition 0 we obtain:. [2.46] Using equations [2.6] and [2.46], and the initial condition 0 0, we get:. [2.47] To find the time at which the particle reaches the maximum height we use Eq. [2.46], set 0 and solve for t: 0. [2.48] 16

To find the maximum height we then plug the time from Eq. [2.48] into Eq. [2.47]:. [2.49] To get a numerical solution we finally replace the symbols and with their numerical values:.. 5.10. Check units: we are calculating a length and the result is, correctly, in m. 2.9 Graphical Representation We conclude this chapter with another example that links the physics of motion in one dimension with the mathematical properties of first derivatives. In particular, as we have discussed in session 2.2, the fact that the first derivative of a function represents its slope on a Cartesian plane. Problem 2.8 A new locomotive is tested on a straight track and its velocity as a function of time is shown in Fig. 2.6. Assuming that the train is at the station ( 0 at time 0, find the position of the train as a function of time. FIG. 2.6 Strategy We can calculate the slope of the graphed curve to get the acceleration as a function of time and from there obtain its velocity and position. It is also necessary to divide the time into three intervals: 02, 26, and 6. Setup In this particular problem the axes setup is already defined by the problem itself, so no additional setup is necessary (other than dividing the time into three intervals as already specified in the strategy). Solution [a] 0 2 In this interval the velocity is constant and equal to 2/. Using the initial condition that the train is at the origin (the station) at 0, we get: 17

2. [2.50] [b] 26 In this interval the velocity increases linearly with time, i.e., its slope (and thus acceleration) is constant. More specifically, the velocity changes from 2 / at 2 to 4 / at 6. Since the acceleration is constant in this interval its average value will be equal to its instant value, thus:. [2.51] From the acceleration we can get the velocity and then the position as a function of time, but we need the proper initial conditions. From the graph we find that at 2 the velocity is equal to 2 /. For the position, instead, we can use the result from the first time interval. In particular, from Eq. [2.50] we obtain the train position at 2: 2 2 2 4. [2.52] Using Eq. [2.51] we then get:, [2.53] And using the initial conditions: 2 2 5. [2.54] Note that it would have been wrong to assume that 2 /, i.e., the actual train velocity at 0. The value of, in fact, represent the velocity the train would have had at 0 if it moved at constant acceleration as in this time interval for the whole time. At this point we can use Eqs. [2.53] to find the train position:. [2.55] And using the initial condition [2.52]: 4 2 5 2 4 3 10 3. [2.56] Combining Eqs. [2.55] and [2.56] we finally get: [c] 6 5 3. [2.57] In this time interval the velocity is, again, constant and equal to 4, therefore. [2.58] As for the initial conditions, from Eq. [2.57] we can calculate the train position at 6: 18

6 6 5 6 3 27 30 3 0, [2.59] therefore: 0 4 6 24. [2.60] Combining Eqs. [2.58] and [2.60] we finally obtain: 4 24. [2.61] Combining Eqs. [2.50], [2,57], and [2.61] we finally obtain the train position at any given time: 2, for 02 5 3 for 26 4 24, for 6. Equation Sheet [2.4] [2.5] [2.6] [2.7] [2.19] [2.21] 19