Chapter 2. Motion in One Dimension. AIT AP Physics C

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1 Chapter 2 Motion in One Dimension

2 Kinematics Describes motion while ignoring the agents that caused the motion For now, will consider motion in one dimension Along a straight line Will use the particle model A particle is a point-like object, has mass but infinitesimal size

3 Position Defined in terms of a frame of reference One dimensional, so generally the x- or y- axis The object s position is its location with respect to the frame of reference

4 Position-Time Graph The position-time graph shows the motion of the particle (car) The smooth curve is a guess as to what happened between the data points

5 Displacement Defined as the change in position during some time interval Represented as x x = x f - x i SI units are meters (m) x can be positive or negative Different than distance (which is the length of a path followed by a particle)

6 Vectors and Scalars Vector quantities need both magnitude (size or numerical value) and direction to completely describe them Will use + and signs to indicate vector directions Scalar quantities are completely described by magnitude only

7 Average Velocity The average velocity is rate at which the displacement occurs x xf xi vaverage t t The dimensions are length / time [L/T] The SI units are m/s Is also the slope of the line in the position time graph

8 Average Speed Speed is a scalar quantity same units as velocity total distance / total time The average speed is not (necessarily) the magnitude of the average velocity

9 Instantaneous Velocity The limit of the average velocity as the time interval becomes infinitesimally short, or as the time interval approaches zero The instantaneous velocity indicates what is happening at every point of time

10 AP Physics C Mechanics AP Physics C Mechanics Calculus Basics Mechanics

11 AP Physics C Mechanics Calculus Primer Among other things, calculus involves studying analytic geometry (analyzing graphs). The above graph should be familiar to anyone who has studied elementary algebra. The horizontal axis is the 'X' axis and the vertical axis is the 'Y' axis. The primary concern of differential calculus is determining slopes of equations. Since this is a primer, we will review the concept of slope using a linear (straight line) equation. The slope of a line (designated by the letter 'm') is defined as the difference in 'y' divided by the difference in 'x'. (or the "rise over the run" as it is sometimes called). Mathemeticians use the Greek letter delta "D" to represent "difference" and so this equation could be written: Using y=3x + 6 (the red line in the graph above), we take the 2 points (x=2, y=12) and (x=-2, y=0) and calculate the slope: Although the equations are different, if we did slope calculations for the yellow or blue lines, they would have exactly the same slope as the red line (m=3). Mechanics

12 AP Physics C Mechanics Calculus Primer There is an easy method to calculate slope. For equations of the form y = a x + b = 0, the slope equals 'a' (the coefficient of 'x'). This is better than choosing points and calculating differences. Just one more quick example - what is the slope of 5y = -3x + 7? Since the equation has to be of the form y = ax +b then y = -3/5 x + 7/5 and so the slope = -3/5. Mechanics

13 AP Physics C Mechanics Calculus Primer Differential Calculus In the previous section, we learned how to calculate the slope of a linear equation (equations whose exponents = 1). What happens when dealing with quadratic, cubic and higher-power equations? In other word a function that is not a straight line.. The graph at the right is based on a quadratic equation which predicts the distance an object has fallen (the y-axis) in relation to time (the x-axis). You cannot state one specific slope for this equation because the slope is constantly varying. What can be calculated is the slope at any point along the curve. Let's calculate the slope when time = 3 seconds. The formula states the distance (in feet) = ½ g t² (where g = 32 feet/sec²). So, when t = 3 seconds, d= ½ 32 3² = 16 9 = 144 feet. So, we have our first set of values: y=144 and x=3 But what do we do for choosing a second point? We could try using a value of t= 4 seconds, remembering this is not the same slope at t=3. So, we get d= ½ 32 4² = = 256 feet. So, we have our second set of values: y=256 and x=4 Calculating this approximate slope yields: Why don't we choose a closer value of x such as 3.1? When that is the case, the distance equals feet and our slope is: How about choosing a value of x that is even closer to 3 than 3.1? Besides representing a difference, Dx (called 'delta x') also represents the smallest possible quantity greater than zero. D x is less than a millionth, less than a trillionth - it's 1 divided by infinity. So, when x2 = 3 + D x, then y2 equals ½ g t² = 16 * (3 +Dx)² = 16 * (9 + 6 Dx + D x²) = D x + 16 Dx² and the slope at x = 3 can be calculated as: Since Dx is such an incredibly miniscule quantity, we can safely say that at x=3, slope = 96. Incidentally, the process of calculating a slope is called differentiation, the result of these calculations is called the derivative and this branch of mathematics is called differential calculus. The derivative is usually represented by dy/dx or f'(x). Mechanics

14 AP Physics C Mechanics Calculus Primer Differential Calculus Although the previous method does work, it has 2 drawbacks - it is rather cumbersome and it only calculates the slope at one particular point. If we wanted to know the slope at x = 2, we would have to go through all those calculations again. Is there an easier way for determining the slope of an equation at any point? Yes Differentiation the Easy Way For a function of the form k x n, the derivative is equal to n k x (n-1) An example is the equation d= ½ g T² (or d = 16 T²) The derivative equals 2 16 T or 32 T Having determined the derivative, we can put it to use by the previous example when we calculated the slope for x=3. When x = 3 (or time = 3 seconds), the slope = 32 3 or 96. What about the slope at 2 seconds? 32 2 equals 64. The derivative of a constant (for example the number 7) is always zero. So, by way of example, the derivative of x2 + 7 is 2 x. Also, if an equation has more than 1 'x' term, simply differentiate each term and then sum those derivatives. Example: What is the derivative of 3x 3-5x 2 + 2x + 13? Answer: 9x 2-10x + 2 One important point to remember is that this method of differentiation works ONLY for equations of the form k x n. Mechanics

15 AP Physics C Mechanics Calculus Primer Integral Calculus As we learned, differential calculus involves calculating slopes and now we'll learn about integral calculus which involves calculating areas. The graph to the right where velocity = g T (or v = 32 T), is based on the derivative of the second graph equation d= ½ g t². Now, if we wanted to determine the distance an object has fallen, we calculate the "area under the curve". If we calculated the sum of the orange, blue and red areas this would equal the distance fallen after 3 seconds. Area of a Right Triangle = ½ (base * height) = ½ (3 seconds * 96 feet per second) = 144 feet Now looking at the previous graph, we see that this is the precise distance after 3 seconds. If we wanted to find the distance fallen between 2 and 3 seconds, we calculate ALL the area from 0 to 3 seconds (144 feet) and then subtract the distance from 0 to 2 seconds: ½ (2 seconds * 64 feet per second) = 64 feet So, the distance fallen between 2 and 3 seconds is = 80 feet. Looking at the previous graph and doing the subtraction, we see the numbers are the same. So how were we calculating these areas? We multiplied the y- axis (which is the quantity g T) by the x-axis (time in seconds or 'T') and we multiplied this by ½. So we calculated the area by the formula ½ (g T T) which equals ½ g T², and this is the precise formula which was used in the previous section!! The process of calculating area is called integration, the resultant formula is called the integral and this branch of mathematics is called integral calculus. (NOTE: the integral is sometimes called the anti-derivative) Mechanics

16 AP Physics C Mechanics Calculus Primer Integral Calculus Integration the Easy Way For a function of the form k x n the integral equals k x (n+1) (n+1) Example 1: What is the integral of 5x 2 + 3x -7? Answer: (5x 3 )/3 + (3x 2 )/2-7x + c where 'c' is a constant - the derivative of a constant is zero and so when calculating an integral we have to allow for a constant. Example 2: Here's an interesting example of integral calculus. The area of a sphere is calculated by the formula 4 p r ²? What is the integral (anti-derivative) of this formula? Answer: (4 p r ³)/3 which happens to be the formula for the volume of a sphere! You now have a basic understanding of calculus (both differential and integral) This was a very quick, "barebones" explanation. However, you should have enough for AP Physics C Mechanics

17 Instantaneous Velocity, equations The general equation for instantaneous velocity is x dx vx t dt lim t 0 The instantaneous velocity can be positive, negative, or zero

18 Instantaneous Velocity, graph The instantaneous velocity is the slope of the line tangent to the x vs. t curve This would be the green line The blue lines show that as t gets smaller, they approach the green line

19 Instantaneous Speed The instantaneous speed is the magnitude of the instantaneous velocity Remember that the average speed is not the magnitude of the average velocity

20 Average Acceleration Acceleration is the rate of change of the velocity v vxf v x xi ax t t Dimensions are L/T 2 SI units are m/s²

21 Instantaneous Acceleration The instantaneous acceleration is the limit of the average acceleration as t approaches 0 a x 2 v dv d x lim x x t 0 t dt dt 2

22 Instantaneous Acceleration -- graph The slope of the velocity vs. time graph is the acceleration The green line represents the instantaneous acceleration The blue line is the average acceleration

23 Acceleration and Velocity, 1 When an object s velocity and acceleration are in the same direction, the object is speeding up When an object s velocity and acceleration are in the opposite direction, the object is slowing down

24 Acceleration and Velocity, 2 The car is moving with constant positive velocity (shown by red arrows maintaining the same size) Acceleration equals zero

25 Acceleration and Velocity, 3 Velocity and acceleration are in the same direction Acceleration is uniform (blue arrows maintain the same length) Velocity is increasing (red arrows are getting longer) This shows positive acceleration and positive velocity

26 Acceleration and Velocity, 4 Acceleration and velocity are in opposite directions Acceleration is uniform (blue arrows maintain the same length) Velocity is decreasing (red arrows are getting shorter) Positive velocity and negative acceleration Stop

27 Kinematic Equations -- summary

28 Kinematic Equations The kinematic equations may be used to solve any problem involving onedimensional motion with a constant acceleration You may need to use two of the equations to solve one problem Many times there is more than one way to solve a problem

29 Kinematic Equations, specific For constant a, Can determine an object s velocity at any time t when we know its initial velocity and its acceleration Does not give any information about displacement v v at xf xi

30 Kinematic Equations, specific For constant acceleration, v x v xi 2 v xf The average velocity can be expressed as the arithmetic mean of the initial and final velocities

31 Kinematic Equations, specific For constant acceleration, 1 x f xi ( vxi vx f ) t 2 This gives you the position of the particle in terms of time and velocities Doesn t give you the acceleration

32 Kinematic Equations, specific For constant acceleration, 1 x f xi vxit axt 2 Gives final position in terms of velocity and acceleration Doesn t tell you about final velocity 2

33 Kinematic Equations, specific For constant a, v v 2 a ( x x ) 2 2 xf xi x f i Gives final velocity in terms of acceleration and displacement Does not give any information about the time

34 Graphical Look at Motion displacement time curve The slope of the curve is the velocity The curved line indicates the velocity is changing Therefore, there is an acceleration

35 Graphical Look at Motion velocity time curve The slope gives the acceleration The straight line indicates a constant acceleration

36 Graphical Look at Motion acceleration time curve The zero slope indicates a constant acceleration

37 Freely Falling Objects A freely falling object is any object moving freely under the influence of gravity alone. It does not depend upon the initial motion of the object Dropped released from rest Thrown downward Thrown upward

38 Acceleration of Freely Falling Object The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g = 9.80 m/s 2 g decreases with increasing altitude g varies with latitude 9.80 m/s 2 is the average at the Earth s surface

39 Acceleration of Free Fall, cont. We will neglect air resistance Free fall motion is constantly accelerated motion in one dimension Let upward be positive Use the kinematic equations with a y = g = m/s 2

40 Free Fall Example Initial velocity at A is upward (+) and acceleration is g (-9.8 m/s 2 ) At B, the velocity is 0 and the acceleration is g (-9.8 m/s 2 ) At C, the velocity has the same magnitude as at A, but is in the opposite direction The displacement is 50.0 m (it ends up 50.0 m below its starting point)

41 Motion Equations from Calculus Displacement equals the area under the velocity time curve f lim v t v ( t) dt t n 0 n xn n x t The limit of the sum is a definite integral t i

42 Kinematic Equations General Calculus Form a x x xf xi x 0 x dv dt v v a dt v dx dt x x v dt f i x 0 t t

43 Kinematic Equations Calculus Form with Constant Acceleration The integration form of v f v i gives v v a t xf xi x The integration form of x f x i gives 1 x f xi vxit a xt 2 2

44 Kinematic Equations Derivations: Calculus Form with Constant Acceleration acceleration Integration can also be from t ii to t f velocity dv a so a dt dv dt 0 x x x x x t 0 x t v v a dt a ( t 0) a t t dv a dt xf xi x x x 0 dx v and vdt dx dt vdt dx v dt x x x f i if v v v a t then x xf xi x t t t x x ( v a t) dt v dt a tdt f i xi x xi x t 1 x f xi vxi ( t 0) ax[ 0] vxit axt Integrate both sides If acceleration is constant then Integrate both sides Put velocity as a function of time This should look familiar

45 General Problem Solving Strategy Conceptualize Categorize Analyze Finalize

46 Problem Solving Conceptualize Think about and understand the situation Make a quick drawing of the situation Gather the numerical information Include algebraic meanings of phrases Focus on the expected result Think about units Think about what a reasonable answer should be

47 Problem Solving Categorize Simplify the problem Can you ignore air resistance? Model objects as particles Classify the type of problem Try to identify similar problems you have already solved

48 Problem Solving Analyze Select the relevant equation(s) to apply Solve for the unknown variable Substitute appropriate numbers Calculate the results Include units Round the result to the appropriate number of significant figures

49 Problem Solving Finalize Check your result Does it have the correct units? Does it agree with your conceptualized ideas? Look at limiting situations to be sure the results are reasonable Compare the result with those of similar problems

50 Problem Solving Some Final Ideas When solving complex problems, you may need to identify sub-problems and apply the problem-solving strategy to each sub-part These steps can be a guide for solving problems in this course

51 Problem Solving Steps 1. Don t Panic! Every problem has a solution. Well, at least the problems this year in Physics 2. READ the problem, READ the problem, READ the problem! 3. Construct an informative diagram of the physical situation (sketch). 4. Identify and list the given information in variable form (make a table of values) 5. Identify and list the unknown information in variable form. 6. READ the problem again to make sure you still understand what it is that you are solving for 7. Identify and list the equation which will be used to determine the unknown information from the known variables. 8. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown or solve for the unknown variable first and then substitute known values (this is generally preferable and easier) 9. Check your answer to ensure that it is reasonable and mathematically correct (include UNITS in you solution and final answer)

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