1. Motion The basic description of motion is how location (position) changes with time. We call this velocity. Is velocity a vector? (Does it have magnitude and direction?)

2. Motion The basic description of motion is how location (position) changes with time. We call this velocity. Is velocity a vector? (Does it have magnitude and direction?) YES! v =  (x,y) /  t where the  sign means “change in”, or “final minus initial”.

3. Velocity Now the question becomes: how do you divide a vector by a scalar? Since multiplication is simply multiple additions (3*2 means 2+2+2), and since we can add vectors nicely in rectangular form (add the components), we should be able to multiply a vector by a scalar by just multiplying the vector’s rectangular components by the scalar.

4. Velocity And since division is simply the inverse of multiplication, we can divide a vector by a scalar by just dividing the rectangular components of the vector by the scalar. Hence v = (v x, v y ) =  (x,y) /  t, where v x =  x /  t and v y =  y /  t.

5. Velocity Note that the MKS units of velocity are m/s. This definition of velocity indicates that position changes over time. This is really, then, a calculation of an AVERAGE VELOCITY. Is there such a thing as an INSTANTANEOUS VELOCITY?

6. Velocity According to the calculus, in the limit as  t approaches zero [and so  (x,y) also approaches zero], the expression v x =  x /  t becomes v x = dx/dt where x is a function of t. This is the mathematical way of saying we do have a way of finding instantaneous velocity!

7. Average versus Instantaneous v x-average =  x /  t The average deals with discrete data points. We need to know x final and x initial, and t final and t initial, and we get the average speed during the time interval - not necessarily the speed at any particular instant. v x-instantaneous = dx/dt The instantaneous deals with continuous functions. We need to know x as a function of time, x(t); and we get v as a function of time: v(t).

8. Change in velocity But velocity is not the whole story of motion. Sometimes (often) we are interested in how the velocity changes with time! What do we call the change in velocity with respect to time?

9. Acceleration But velocity is not the whole story of motion. Sometimes (often) we are interested in how the velocity changes with time! This leads to the notion of ACCELERATION: a = (a x, a y ) =  (v x,v y ) /  t and a x-average =  v x /  t, or a x-instantaneous = dv x /dt. Note that the units of acceleration are (m/s) / s or more commonly: m/s 2. Question: What is a square second (s 2 )?

10. Where do we stop? Is there a name for the change in acceleration with respect to time? Why haven’t most people heard of it, when most people have heard of velocity and of acceleration?

11. Jerk! To answer the first question, the change in acceleration with respect to time is called Jerk! To answer the second question, the reason most people have not heard of jerk is because it is not normally useful. This is due to reasons we’ll see in Part II of the course.

12. Signs (+ or -) for position Position: Usually we have some reference point that we call zero position. For horizontal positions, plus usually means to the right, and minus means to the left. For vertical positions, plus usually means above (up) and minus means below (down). Warning: these are only the usual conventions; they can be reversed if that is more convenient.

13. Signs (+ or -) for velocity For horizontal motion, moving to the right usually means a positive velocity component, and moving to the left means a negative velocity component. For vertical motion, moving up usually means a positive velocity component, and moving down means a negative velocity component. Warning: if the usual conventions for position are switched, then the sign conventions for the velocity will also be switched. For example, if down is called a positive position, then moving down will be considered a positive velocity. Note: we can have a positive position with either a positive or negative velocity, and we can have a negative position with either a positive or negative velocity.

14. Signs (+ or -) for Acceleration If the velocity is increasing in the positive direction, the acceleration is positive, and if the velocity is decreasing in the positive direction, the acceleration is negative. Warning: the case of negative velocities is more tricky & counter-intuitive! If the velocity is becoming more negative (going faster in the negative direction), the acceleration is negative, and if the velocity is becoming less negative (getting slower in the negative direction), the acceleration is positive.

15. Signs (+ and -) for Acceleration Language problems in the vertical If we are going faster in the up direction, we say we are speeding up (and going up). No problem. Is this acceleration positive or negative? If we are going slower in the up direction, we say we are slowing down (but going up). See the language problem? Is this acceleration positive or negative? If we are going faster in the down direction, we say we are speeding up (but going down). See the language problem here? Is this acceleration + or - ? If we are going slower in the down direction, we say we are slowing down (and going down). Is this acceleration + or - ?

16. Signs (+ and -) for Acceleration Language problems in the vertical If we are going faster in the up direction, we say we are speeding up (and going up). No problem. This acceleration is positive. If we are going slower in the up direction, we say we are slowing down (but going up). See the language problem? This acceleration is negative. Warning: these next two are counter-intuitive: If we are going faster in the down direction, we say we are speeding up (but going down). See the language problem here? This acceleration is negative. If we are going slower in the down direction, we say we are slowing down (and going down). This acceleration is positive.

17. Motion We now have two useful definitions (relations) in two different forms: discrete (data points) continuous (functions) v x =  x /  tv x = dx/dt a x =  v x /  t a x = dv x /dt. If we know position and time, we can calculate velocity; if we know velocity and time, we can calculate acceleration.

18. Discrete Case - an example Given the following data, find v x and a x : x (in meters) at t (in seconds) -20 +10.5 +61 +41.5 Can you picture this? 02

19. Discrete case: a picture Note: the time (Δt) for each arrow (which is the change in position, Δx) is 0.5 seconds. -2 m0 sec 1 m0.5 sec 6 m1.0 sec 4 m1.5 sec 0 m2.0 sec -2 0 2 4 6 x

20. Discrete Case - an example Since we know the position at 0 sec and 1 sec, we can find the average velocity in this interval: v x-average =  x /  t v x-avg (between 0 and 1 sec)) = (+1 m - -2 m) / (0.5 sec - 0 sec) = +6 m/s. Since this is the velocity between 0 and 0.5 seconds, we can say that this probably is close to the speed at 0.25 seconds.

21. Discrete Case - an example Can you determine the average v x at other times? x (in meters) at t (in sec.) v x (in m/s) -20 0.25+6 +10.5 0.75 +61 1.25 +41.5 1.75 02

22. Discrete Case - an example Doing similar calculations for the other times: x (in meters) at t (in sec.) v x (in m/s) -20 0.25+6 +10.5 0.75+10 +61 1.25-4 +41.5 1.75-8 02

23. Discrete Case - an example For acceleration we do the same thing: Since we know the approximate velocity at 0.25 sec and 0.75 sec, we can find an approx. average acceleration in this interval: a x-average =  v x /  t a x-avg (between 0.25 and 0.75 sec) = (+10 m/s - +6 m/s) / (.75 sec -.25 sec) = +8 m/s 2. Since this is the acceleration between.25 and.75 seconds, we can say that this probably is close to the acceleration at 0.5 seconds.

24. Discrete Case - an example Doing similar calculations for the other times: x (in meters) at t (in sec.) v x (in m/s) a x (in m/s 2 ) -20 0.25+6 +10.5+8 0.75+10 +61-28 1.25-4 +41.5-8 1.75-8 02

25. Continuous Case - an example Given the following function for position: x(t) = 4 m + (12 m/s)* t + (-5 m/s 3 )* t 3, What are the velocity and acceleration functions, and what is the velocity and acceleration values at t = 2 sec ?

26. Continuous Case - an example x(t) = 4 m + (12 m/s)* t + (-5 m/s 3 )* t 3, v x (t) = dx(t)/dt = d [4 m + (12 m/s)* t + (-5 m/s 3 )* t 3 ] /dt = 0 + 12 m/s - (15 m/s 3 )*t 2 so v x (t=2sec) = 12 m/s - (15 m/s 3 )*(2 sec) 2 = -48 m/s. a x (t) = dv(t)/dt = d(12 m/s - (15 m/s 3 )*t 2 )/dt = 0 - (30 m/s 3 )*t, so a x (t=2 sec) = (-30 m/s 3 )*(2 sec) = -60 m/s 2. Note: when asked for v and a at particular times, you need to take the derivatives BEFORE you substitute in the values for time.

27. Motion Often we do NOT know position and time, but rather something else and we wish to predict what the position versus time will be! Can we go backwards as well as forwards in these relations? (That is, knowing acceleration and time, can we figure out what the velocity will be?)

28. Going backwards: the discrete case v x-avg =  x /  t and a x-avg =  v x /  t Since the above definitions involve division, the inverse of division is multiplication. Knowing the AVERAGE velocity and the time, we can find the CHANGE IN position:  x = v x-avg *  t, or x final = x initial + v x-avg *t (where t is really  t ).

29. Going backwards: the discrete case  x = v x-avg *  t, or x final = x initial + v x-avg *t (where t is really  t ) Note that the velocity in this formula is the AVERAGE velocity. If the velocity is constant, then this equation works exactly. However, if the velocity changes, then we need to know the real average velocity. The real average velocity is not necessarily the sum of the initial and final divided by 2! 2 4 avg = 3 avg 3 Using just the endpoints, avg = (2+4)/2 = 3.

30. Going backwards: the discrete case Knowing the AVERAGE acceleration and the time, we can find the CHANGE IN velocity:  v x = a x-avg *  t, or v xfinal = v xinitial + a x-avg *t (where t is really  t) If the acceleration is constant, so that the average acceleration is equal to the acceleration at all times, then this is exact. Otherwise, this is approximate.

31. Example - discrete case x (m)t(sec)v (m/s)a (m/s 2 ) 0 0.5 1+3 1.5 2-1 2.5 3-2 3.5 4

32. Example - discrete case Knowing the acceleration at t=1 sec, we can use the definition of acceleration: a =  v/  t to get:  v = a*  t. Since the accelerations are given in one second intervals, let’s choose  t = 1 sec. This leads to: v(t=1.5 sec) - v(t=0.5 sec) = (3 m/s 2 ) * 1 sec v(t=1.5 sec) = v(t=0.5 sec) + (3 m/s 2 ) * 1 sec However, unless we know one of these v’s, we can’t solve this. Let’s say that we do know the velocity at t=0.5 sec is v(t=0.5 sec) = +5 m/s. v(t=1.5 sec) = +5 m/s + (3 m/s 2 ) * 1 sec = +8 m/s.

33. Example - discrete case x (m)t(sec)v (m/s)a (m/s 2 ) 0 0.5+5 1+3 1.5+8 2-1 2.5 3-2 3.5 4

34. Example - discrete case We now proceed as before to get the next velocities: v(t=2.5 sec) = v(t=1.5 sec) + (-1 m/s 2 ) * 1 sec ; from the previous calculation, we know v(t=1.5 sec) = 8 m/s, so v(t=2.5 sec) = 8 m/s + (-1 m/s 2 ) * 1 sec = 7 m/s. Proceeding: v(t=3.5 sec) = v(t=2.5 sec) + (-2 m/s 2 ) * 1 sec gives v(t=3.5 sec) = 7 m/s + (-2 m/s 2 ) * 1 sec = 5 m/s.

35. Example - discrete case x (m)t(sec)v (m/s)a (m/s 2 ) 0 0.5+5 1+3 1.5+8 2-1 2.5+7 3-2 3.5+5 4

36. Example - discrete case To get position, we now go backwards from velocity: Knowing the velocity at t=0.5 sec, we can use the definition of velocity: v =  x/  t to get:  x = v*  t. Since the velocities are given in one second intervals, let’s choose  t = 1 sec. This leads to: x(t=1 sec) - x(t=0 sec) = (5 m/s) * 1 sec x(t=1 sec) = x(t=0 sec) + (5 m/s) * 1 sec However, unless we know one of these x’s, we can’t solve this. Let’s say that we do know the position at t=0 sec is x(t=0 sec) = -2 m. x(t=1 sec) = -2 m + (5 m/s) * 1 sec = +3 m.

37. Example - discrete case x (m)t(sec)v (m/s)a (m/s 2 ) -2 0 0.5+5 +3 1+3 1.5+8 2-1 2.5+7 3-2 3.5+5 4

38. Example - discrete case We now proceed as before to get the next positions: x(t=2 sec) = x(t=1 sec) + (+8 m/s) * 1 sec ; from the previous calculation, we know x(t=1 sec) = 3 m, so x(t=2 sec) = 3 m + (+8 m/s) * 1 sec = 11 m. Proceeding: v(t=3 sec) = 11 m + (+7 m/s) * 1 sec = 18 m v(t=4 sec) = 18 m/s + (+5 m/s) * 1 sec = 23 m.

39. Example - discrete case x (m)t(sec)v (m/s)a (m/s 2 ) -2 0 0.5+5 +3 1+3 1.5+8 +11 2-1 2.5+7 +18 3-2 3.5+5 +23 4

40. Example - discrete case Note: In going backwards, we needed to know the acceleration, but we also needed to know where to start, both for the velocity and for the position. These starting points are called “initial conditions”. In going forward, we had no need for such initial conditions.

41. Going backwards: the continuous case From the definition of velocity (in the continuous case): v x = dx(t)/dt ; to go backwards we need the inverse of differentiation, which is integration: dx(t) = v x (t) dt, or on integrating both sides: x(t) - x o = t=0  t v x (t) dt. Likewise for acceleration: from a x = dv x /dt ; on going backwards we get: v x (t) - v xo = t=0  t a x (t) dt.

42. Special Case: Constant Acceleration If the acceleration is constant, then we have (with a x-avg = a x = constant) for both the discrete and the continuous cases: v x-final = v x (t) = v x-initial + a x *t. Since there is acceleration, the velocity does not remain constant and so the discrete way must be in pieces (see previous slides).

43. Special Case: Constant Acceleration However, for the continuous case, we get: x(t) - x o = t=0  t v x (t) dt, which becomes x(t) - x o = t=0  t (v xo + a x t) dt, or x(t) = x o + v xo *t + ½*a x *t 2. Note: In going backwards, as in the discrete case, we do need to know the initial conditions. In this case, the initial conditions are x o and v xo.

44. Falling (without air resistance) In the case of something falling, the acceleration due to gravity near the earth’s surface is approximately constant, if we can also neglect the effects of air resistance. (We’ll talk more about the causes of motion and gravity in Part 2.) In this special case, we can use the equations for constant acceleration.

45. Falling (without air resistance) If we treat up as +y, then we have these two equations: y = y o + v yo *t + ½*g*t 2 and v y = v yo + g*t where g = -9.8 m/s 2. Here, we have simply used y for y final and we have used y o for y initial. The same notation is also used for v.

46. Solving Problems Note that when we have identified a problem as being one of constant acceleration, we have two equations: y = y o + v yo *t + ½*a*t 2 and v y = v yo + a*t. Note that in these two equations we have six quantities: y, y o, v, v o, a, and t. This means we have to identify four of the six in order to use the two equations to solve for the other two quantities.

47. Solving Problems Reading the description of a problem involves several steps: Identify the problem type: does this problem have constant acceleration? If so, we know we have the two equations to work with. Identify what you know: does this problem involve falling under the influence of gravity? If so, we know a = g = -9.8 m/s 2.

48. Solving Problems (list continued from previous slide) We can usually pick out where to start from (if gravity, the ground is usually where y=0 is). This is important for identifying y and y o. Sometimes we are given information about y o, sometimes about y. Special words: The word “stop” or “stationary” means that at this time v=0. This may apply to either v or v o.

49. Solving Problems (list continued from previous slide) Make sure you know what negative signs mean. For y, positive usually means above ground, negative will mean below ground. For v, positive usually means going up (or forward), negative will mean going down (or backwards).

50. Solving Problems List continued from previous page Note that in the x equation for constant acceleration, there is a t 2 term. That means that, when solving for time, there may be two solutions. Can you identify in the problem what the two solutions would be for?

51. Example of a falling problem To find the height of a tree, a person throws a baseball up so that it just reaches the height of the tree. The person then uses a stopwatch to time the fall of the ball from the highest point (the height of the tree) to the ground. If the time on the stopwatch is 3.4 seconds, how high is the tree?

52. Example of a falling problem Draw a diagram to help define the situation: (highest point) y o = ?, v o = ?, t o = 0 sec. a = ? y = ?, v = ?, t = ?

53. Example of a falling problem We will assume that air resistance is negligible, and that the tree is not too high so we can consider gravity constant. In this case we then have the constant acceleration situation and so can use the two equations: y = y o + v yo *t + ½*a*t 2 and v y = v yo + a*t.

54. Example of a falling problem We need a reference point, so let’s choose a common one: we will say that the ground (which is the final position of the ball) is where y=0. The ball falls, so a = g = -9.8 m/s 2. Then from the statement of the problem, we are looking for y o (which would correspond to the height of the tree), and we know the time for y=0: t=3.4 sec. Thus we know three quantities (y, a, t) and have one unknown so far (y o ):

55. Example of a falling problem y o = ? y = 0 a = -9.8 m/s 2 t = 3.4 seconds That leaves the initial and final velocity. To solve the problem, we need to know four things and can have two unknowns (since we have two equations). That means we need to know either v or v o.

56. Example of a falling problem From the statement of the problem, the ball falls from the highest point, so v o = 0. We do NOT know the final velocity. Note that the ball will HIT the ground, but that does NOT make the final velocity zero - just before it hits it is travelling rather fast! The act of hitting destroys our assumption of constant acceleration due only to gravity.

57. Example of a falling problem We now put the information on our diagram: (highest point) y o = ?, v o = 0, t o = 0 sec. a = g = -9.8 m/s 2 y = 0, v = ?, t = 3.4 seconds

58. Example of a falling problem y = y o + v yo *t + ½*a*t 2 and v y = v yo + a*t. y o = ? y = 0 a = -9.8 m/s 2 t = 3.4 seconds v o = 0 v = ?

59. Example of a falling problem Putting the knowns into the two equations gives: 0 m = y o + (0 m/s)*(3.4 s) + ½*(-9.8 m/s 2 )*(3.4 s) 2 which we see is one equation in one unknown and can be directly solved: y o = 56.64 m. v = 0 m/s + (-9.8 m/s 2 )*(3.4 s) = 33.3 m/s.

60. 2 nd Example: accelerating car A car accelerates (assume constant acceleration) from rest up to a speed of 65 mph in a time of 7 seconds. What is the average acceleration of the car? How far does the car go during the 7 seconds while it is accelerating?

61. Accelerating Car We recognize this as a constant acceleration problem, so we have our two equations: x = x o + v o *t + (1/2)*a*t 2 and v = v o + a*t and six quantities: x o = x = a = t = v o = v = car x o = v o = t o = 0 s x = v = t = a =

62. Accelerating Car From the statement of the problem, we see the following is given: “from rest up to a speed of 65 mph in a time of 7 seconds”. Can we determine which symbols go with which values?

63. Accelerating Car “from rest” means v o = 0; “up to a speed of 65 mph” means v = 65 mph, but mph is not the MKS unit. We need to convert it to m/s (2.24 mph = 1 m/s), so v = 65 mph * (1 m/s / 2.24 mph) = 29 m/s. “in a time of 7 seconds” means t = 7 s. This gives 3 of the six quantities, and we have two equations, so we need to know one more. This one is “hidden” in the problem – since we don’t have a definite starting position, we can assume x o = 0.

64. Accelerating Car x = x o + v o *t + ½*a*t 2 and v = v o + a*t and six quantities: x o = 0 m x = x o + v o *t + ½*a*t 2 x = ? x = 0 m + 0 m/s*(7 s) + ½ a*(7 s) 2 a = ? t = 7 s v = v o + a*t v o = 0 m/s29 m/s = 0 m/s + a*(7 s) v = 29 m/s

65. Accelerating Car The first (x) equation has two unknowns (x and a). However, the second (v) equation has only one unknown (a). Therefore, we can solve the second equation for a, and then use the first equation to solve for x. 29 m/s = 0 m/s + a*(7 s) or a = 29 m/s / 7s = 4.14 m/s 2 and then x = 0 m + 0 m/s*(7 s) + ½ (4.14 m/s 2 )*(7 s) 2 = 101.5 m. Note that v avg =  x  t, but x = v avg *t ≠ (29 m/s)*(7 s) = 203 m because v is not a constant 29 m/s over the 7 second time interval. However, because the acceleration is constant, we can use v avg = ½*(v o +v f ) = ½*(0 m/s + 29 m/s) = 14.5 m/s, so that x = (14.5 m/s)*(7 s) = 101.5 m. Note that v avg = ½*(v o +v f ) does not always work – it only works if a = constant, which it is here.

66. Graphical Representations From the definition of velocity (working in rectangular components): v x-avg =  x /  t or v x = dx/dt we can recognize that the slope of the x vs t curve at any time = value of the velocity at that time. Note that this means the VALUE of x has NOTHING to do with v. It is only how x CHANGES that affects the velocity.

67. Graphical Representations Given the graph of x vs t, can you figure out the graph of v vs t? xvt

68. Graphical Representations At t=0, the SLOPE of x is a small positive amount, so the VALUE of v is a small positive? xvt

69. Graphical Representations A little before t=0, the x vs t curve is flat so SLOPE=0, so at this time the VALUE of v=0. xvt

70. Graphical Representations At an even earlier time, the SLOPE of x is slightly negative, so the VALUE of v is also slightly negative at that earlier time. xvt

71. Graphical Representations At a later time, the SLOPE of x is slightly more positive than it was at t=0, so the VALUE of v is also slightly more positive. xvt

72. Graphical Representations At the latest time, the SLOPE of x is about the same, so the VALUE of v is also about the same. xvt

73. Graphical Representations Now we just connect the dots to get a graph of v vs t based on the graph of x vs t. xvt

74. Graphical Representations Since the definition of acceleration is a x =  v x /  t we see that the slope of v gives the value of a. Thus we can use the same procedure to get the graph of a vs t from the graph of v vs t as we did to get the graph of v vs t from the graph of x vs t.

75. Graphical Representations To go in reverse, that is, knowing the graph of v and trying to find the graph of x, work with the idea that the VALUE of v gives the SLOPE of x. One thing to note: since the value of v gives no information about the value of x, only about the slope of x, we need to be given one value of x to begin. This is usually x o.

76. Graphical Representations Given this curve for v(t), can you sketch (roughly) the curve for x(t)? x t v t

77. Graphical Representations For the most negative time on the graph, the velocity is zero, which means the slope of the x curve at this time is zero. But this doesn’t tell us where to draw a flat curve. We need to know where to start. Let’s assume that x is negative at this time. x t v t 1 1

78. Graphical Representations Now a little later, the velocity is a small positive value, so the slope of the x curve should be a little positive and getting a little more positive. x t v t 2 3 2 3

79. Graphical Representations At t=0, the velocity value is still positive, but not as large as it was a little before that. This means the slope of the x curve is still positive, but not as steep as it was a little earlier. A little after t=0 the velocity is zero so the x curve is flat at that point. x t v t 4 5 4 5

80. Graphical Representations for t>0, the velocity value turns negative. This means the slope of the x curve will be negative, first getting steeper and then getting flatter until it is flat when the velocity reaches zero. x t v t 6 6 7 7

81. Graphical Representations Computer homework programs Vol.1-#3 (on Motion Graphs) and Vol.1-#4 (on Acceleration Due to Gravity) provide some information that you can get by graphing the information.