1 00:00:00,000 --> 00:00:01,980 - [Instructor] What do vector components mean? 2 00:00:01,980 --> 00:00:04,369 Vector components are a way of breaking any vector 3 00:00:04,369 --> 00:00:06,492 into two perpendicular pieces. 4 00:00:06,492 --> 00:00:08,566 For convenience we typically choose these pieces 5 00:00:08,566 --> 00:00:10,904 to lie along the X and Y directions. 6 00:00:10,904 --> 00:00:12,829 In other words to find the vertical component 7 00:00:12,829 --> 00:00:14,990 of this total vector knowing this angle, 8 00:00:14,990 --> 00:00:17,956 since this vertical component is opposite to this angle, 9 00:00:17,956 --> 00:00:20,708 we could write the vertical component as the magnitude 10 00:00:20,708 --> 00:00:23,245 of the total vector times sine of that angle. 11 00:00:23,245 --> 00:00:24,707 And since this horizontal component 12 00:00:24,707 --> 00:00:26,263 is adjacent to that angle, 13 00:00:26,263 --> 00:00:27,824 we could write the horizontal component 14 00:00:27,824 --> 00:00:29,654 as the magnitude of the total vector 15 00:00:29,654 --> 00:00:31,559 times cosine of that angle. 16 00:00:31,559 --> 00:00:33,484 And if instead we were given this angle 17 00:00:33,484 --> 00:00:34,514 and we wanted to determine 18 00:00:34,514 --> 00:00:36,673 the vertical component of the total vector, 19 00:00:36,673 --> 00:00:40,063 since this vertical component is now adjacent to this angle 20 00:00:40,063 --> 00:00:42,763 we'd write the vertical component as the magnitude 21 00:00:42,763 --> 00:00:45,636 of the total vector times cosine of this angle. 22 00:00:45,636 --> 00:00:47,087 And since the horizontal component 23 00:00:47,087 --> 00:00:48,838 is now opposite to this angle 24 00:00:48,838 --> 00:00:50,202 we'd write the horizontal component 25 00:00:50,202 --> 00:00:51,888 as the magnitude of the total vector 26 00:00:51,888 --> 00:00:53,693 times sine of this angle. 27 00:00:53,693 --> 00:00:56,898 So remember, to find the opposite side you use sine, 28 00:00:56,898 --> 00:00:59,910 and to find the adjacent side, you use cosine. 29 00:00:59,910 --> 00:01:01,228 So what would an example problem 30 00:01:01,228 --> 00:01:03,178 involving vector components look like? 31 00:01:03,178 --> 00:01:05,245 Say you had this question and you wanted to determine 32 00:01:05,245 --> 00:01:07,886 the X and Y components of this velocity vector. 33 00:01:07,886 --> 00:01:10,081 Since the horizontal component is adjacent 34 00:01:10,081 --> 00:01:11,706 to the angle that we're given, 35 00:01:11,706 --> 00:01:13,314 we're gonna write the horizontal component 36 00:01:13,314 --> 00:01:16,385 as the magnitude of the total vector 20 meters per second 37 00:01:16,385 --> 00:01:17,949 times cosine of the angle, 38 00:01:17,949 --> 00:01:19,662 which gives us 10 meters per second. 39 00:01:19,662 --> 00:01:22,739 And since the vertical component is opposite to this angle, 40 00:01:22,739 --> 00:01:25,118 we can write the magnitude of the vertical component 41 00:01:25,118 --> 00:01:28,554 as 20 meters per second times sine of the angle 42 00:01:28,554 --> 00:01:31,876 which gives us 17.3 meters per second. 43 00:01:31,876 --> 00:01:35,040 But since this vertical component is directed downward, 44 00:01:35,040 --> 00:01:36,849 technically this vertical component 45 00:01:36,849 --> 00:01:39,972 would be negative 17.3 meters per second. 46 00:01:39,972 --> 00:01:42,302 So using sine and cosine will give you the magnitude 47 00:01:42,302 --> 00:01:43,765 of the components, but you have to add 48 00:01:43,765 --> 00:01:45,187 the negative signs accordingly. 49 00:01:45,187 --> 00:01:46,626 So if the vector points right, 50 00:01:46,626 --> 00:01:48,425 the horizontal component will be positive. 51 00:01:48,425 --> 00:01:49,802 If the vector points left, 52 00:01:49,802 --> 00:01:51,526 the horizontal component will be negative. 53 00:01:51,526 --> 00:01:52,722 If the vector points up, 54 00:01:52,722 --> 00:01:54,833 the vertical component will be positive. 55 00:01:54,833 --> 00:01:56,199 And if the vector points down, 56 00:01:56,199 --> 00:01:57,996 the vertical component will be negative. 57 00:01:57,996 --> 00:02:01,758 What does tail to tip or head to tail vector addition mean? 58 00:02:01,758 --> 00:02:03,646 This is a graphical way to add 59 00:02:03,646 --> 00:02:05,328 or subtract vectors from each other. 60 00:02:05,328 --> 00:02:07,170 And the way it works is by taking the tail 61 00:02:07,170 --> 00:02:09,638 of the next vector and placing it at the tip 62 00:02:09,638 --> 00:02:11,315 or the head of the previous vector. 63 00:02:11,315 --> 00:02:13,742 And once you're done doing this for all your vectors, 64 00:02:13,742 --> 00:02:16,294 you draw the total vector from the first tail 65 00:02:16,294 --> 00:02:17,463 to the last head. 66 00:02:17,463 --> 00:02:20,359 In other words, if you were adding up vectors A, B, and C, 67 00:02:20,359 --> 00:02:23,748 I would place the tail of vector B to the head of vector A, 68 00:02:23,748 --> 00:02:25,617 and then I'd place the tail of vector C 69 00:02:25,617 --> 00:02:27,095 to the head of vector B. 70 00:02:27,095 --> 00:02:28,786 And once I'm done I would draw the total vector 71 00:02:28,786 --> 00:02:31,385 going from the first tail to the last head. 72 00:02:31,385 --> 00:02:34,243 And that total vector would represent the vector sum 73 00:02:34,243 --> 00:02:35,880 of all three vectors. 74 00:02:35,880 --> 00:02:37,545 And if you had to subtract the vector, 75 00:02:37,545 --> 00:02:39,106 you could still use vector addition. 76 00:02:39,106 --> 00:02:41,825 Simply add the negative of that vector. 77 00:02:41,825 --> 00:02:44,117 In other words, if you had some vector B 78 00:02:44,117 --> 00:02:46,155 and you wanted to subtract vector A, 79 00:02:46,155 --> 00:02:48,597 instead of thinking of it as subtracting vector A, 80 00:02:48,597 --> 00:02:51,438 think of it as adding negative vector A. 81 00:02:51,438 --> 00:02:53,194 And the way you find vector negative A 82 00:02:53,194 --> 00:02:56,463 is by taking vector A and simply placing the arrow head 83 00:02:56,463 --> 00:02:58,258 on the other end of the vector. 84 00:02:58,258 --> 00:03:00,702 So what would an example involving tail to tip 85 00:03:00,702 --> 00:03:02,013 vector addition look like? 86 00:03:02,013 --> 00:03:03,196 Imagine we have this question, 87 00:03:03,196 --> 00:03:04,621 and we have these four vectors, 88 00:03:04,621 --> 00:03:06,978 and we were asked to determine what direction 89 00:03:06,978 --> 00:03:09,547 is the sum off all of those vectors? 90 00:03:09,547 --> 00:03:11,630 So we'll use tail to tip vector addition. 91 00:03:11,630 --> 00:03:14,626 I'll take vector A preserving its direction. 92 00:03:14,626 --> 00:03:16,950 I'm not allowed to rotate it or change its size. 93 00:03:16,950 --> 00:03:18,640 And I'll add to that vector B. 94 00:03:18,640 --> 00:03:21,069 The way I do that is putting the tail of vector B 95 00:03:21,069 --> 00:03:23,736 to the tip of vector A, and we add vector C. 96 00:03:23,736 --> 00:03:26,238 And the way we do that is put the tail of vector C 97 00:03:26,238 --> 00:03:27,729 to the tip of vector B. 98 00:03:27,729 --> 00:03:29,492 And finally we'll add the vector D 99 00:03:29,492 --> 00:03:33,117 by putting the tail of vector D to the tip of vector C. 100 00:03:33,117 --> 00:03:34,836 And now that we've drawn all of our vectors, 101 00:03:34,836 --> 00:03:38,139 our total vector will go from the tail of the first vector, 102 00:03:38,139 --> 00:03:40,748 to the tip or head of the last vector, 103 00:03:40,748 --> 00:03:43,584 which means this is the direction and magnitude 104 00:03:43,584 --> 00:03:47,872 of the total vector, A plus B plus C plus D. 105 00:03:47,872 --> 00:03:50,489 Another more mathematical way of adding vectors 106 00:03:50,489 --> 00:03:53,496 is by simply adding up their individual components. 107 00:03:53,496 --> 00:03:55,744 So to find the total vector A plus B, 108 00:03:55,744 --> 00:03:57,786 instead of graphing them tail to tip, 109 00:03:57,786 --> 00:03:59,099 you can find the horizontal 110 00:03:59,099 --> 00:04:00,804 and vertical components separately 111 00:04:00,804 --> 00:04:02,970 by adding up the individual components. 112 00:04:02,970 --> 00:04:05,490 In other words, to get the total horizontal component 113 00:04:05,490 --> 00:04:07,488 of the total vector A plus B, 114 00:04:07,488 --> 00:04:09,955 I could just add up the horizontal component of A, 115 00:04:09,955 --> 00:04:12,473 which is negative 20, and the horizontal component of B 116 00:04:12,473 --> 00:04:15,067 which is negative five to get negative 25. 117 00:04:15,067 --> 00:04:16,914 And to find the total vertical component 118 00:04:16,914 --> 00:04:18,731 of the total vector A plus B, 119 00:04:18,731 --> 00:04:21,057 I can simply add up the vertical component of A 120 00:04:21,057 --> 00:04:23,765 which is negative 15 to the vertical component of B 121 00:04:23,765 --> 00:04:25,957 which is 10, to get negative five. 122 00:04:25,957 --> 00:04:27,983 This technique lets you quickly determine 123 00:04:27,983 --> 00:04:31,596 what the individual components are of that total vector. 124 00:04:31,596 --> 00:04:33,813 And again, if you need to subtract a vector 125 00:04:33,813 --> 00:04:35,944 you can still add the components, 126 00:04:35,944 --> 00:04:37,897 except the components of a negative vector 127 00:04:37,897 --> 00:04:40,187 all get multiplied by negative one. 128 00:04:40,187 --> 00:04:43,101 In other words, if vector A has components negative 20 129 00:04:43,101 --> 00:04:45,370 and negative 15, then vector negative A 130 00:04:45,370 --> 00:04:48,659 would have components positive 20 and positive 15. 131 00:04:48,659 --> 00:04:49,786 So what would an example 132 00:04:49,786 --> 00:04:51,940 of vector component addition look like? 133 00:04:51,940 --> 00:04:53,252 Let's say you had this question. 134 00:04:53,252 --> 00:04:55,638 You had vectors A and B with these components, 135 00:04:55,638 --> 00:04:56,944 and you wanted to know the components 136 00:04:56,944 --> 00:04:59,209 of the total vector A plus B. 137 00:04:59,209 --> 00:05:00,782 So to find the horizontal component 138 00:05:00,782 --> 00:05:02,569 of the total vector A plus B, 139 00:05:02,569 --> 00:05:05,772 I can add up the individual components of A and B. 140 00:05:05,772 --> 00:05:07,800 So the horizontal component of A is four, 141 00:05:07,800 --> 00:05:10,709 plus the horizontal component of B is negative one, 142 00:05:10,709 --> 00:05:13,675 gives me a total horizontal component of positive three. 143 00:05:13,675 --> 00:05:15,725 And I could do the same thing for the vertical component. 144 00:05:15,725 --> 00:05:18,627 I could add up the component of A plus the component of B, 145 00:05:18,627 --> 00:05:20,919 which would be five plus negative four, 146 00:05:20,919 --> 00:05:22,505 gives me positive one. 147 00:05:22,505 --> 00:05:24,116 So since my horizontal component 148 00:05:24,116 --> 00:05:25,848 of the total vector is positive, 149 00:05:25,848 --> 00:05:28,438 I know it points to the right three units. 150 00:05:28,438 --> 00:05:30,545 And since the vertical component of the total vector 151 00:05:30,545 --> 00:05:33,325 is positive, I know it points up one unit. 152 00:05:33,325 --> 00:05:35,264 That means my total vector A plus B 153 00:05:35,264 --> 00:05:37,127 points up and to the right. 154 00:05:37,127 --> 00:05:39,872 One unit up, and three units to the right. 155 00:05:39,872 --> 00:05:42,531 How do you deal with 2D kinematics problems? 156 00:05:42,531 --> 00:05:44,787 2D kinemtatics are projectile problems 157 00:05:44,787 --> 00:05:47,760 describe objects flying through the air at angles. 158 00:05:47,760 --> 00:05:49,971 For these objects, if there's nothing acting on them 159 00:05:49,971 --> 00:05:52,765 besides gravity, their vertical acceleration 160 00:05:52,765 --> 00:05:54,391 is gonna be negative 9.8. 161 00:05:54,391 --> 00:05:56,713 And they will have no horizontal acceleration 162 00:05:56,713 --> 00:05:58,905 since gravity doesn't pull sideways. 163 00:05:58,905 --> 00:06:02,819 Also, the X and Y components behave independently. 164 00:06:02,819 --> 00:06:04,561 That means you'll use different equations 165 00:06:04,561 --> 00:06:06,308 to solve for vertical components 166 00:06:06,308 --> 00:06:08,547 than you will for horizontal components. 167 00:06:08,547 --> 00:06:10,817 Since the vertical acceleration is constant 168 00:06:10,817 --> 00:06:12,286 you can use the kinematic formulas 169 00:06:12,286 --> 00:06:14,772 to solve for quantities in the vertical direction. 170 00:06:14,772 --> 00:06:17,419 But you can only plug in vertical quantities 171 00:06:17,419 --> 00:06:18,911 into these equations. 172 00:06:18,911 --> 00:06:22,185 Similarly, since the acceleration is zero in the X direction 173 00:06:22,185 --> 00:06:24,964 you can simply use distance as a rate times time 174 00:06:24,964 --> 00:06:27,317 to relate the quantities in the X direction, 175 00:06:27,317 --> 00:06:29,754 but you should only plug in horizontal components 176 00:06:29,754 --> 00:06:31,107 into this equation. 177 00:06:31,107 --> 00:06:34,026 In other words, as a projectile is flying through the air, 178 00:06:34,026 --> 00:06:36,440 since there's no horizontal acceleration, 179 00:06:36,440 --> 00:06:38,445 the horizontal component of the velocity 180 00:06:38,445 --> 00:06:41,266 is gonna remain the same for the entire trip. 181 00:06:41,266 --> 00:06:43,545 Which means the rate at which this projectile 182 00:06:43,545 --> 00:06:46,014 is moving the X direction never changes. 183 00:06:46,014 --> 00:06:48,663 But since there is acceleration in the vertical direction, 184 00:06:48,663 --> 00:06:50,591 the vertical component of the velocity 185 00:06:50,591 --> 00:06:53,518 will get smaller and smaller until it reaches the top, 186 00:06:53,518 --> 00:06:55,706 and that means the total speed of the projectile 187 00:06:55,706 --> 00:06:58,480 is gonna decrease as well as you approach the top. 188 00:06:58,480 --> 00:07:00,136 And then at the top there is zero 189 00:07:00,136 --> 00:07:01,894 vertical component of velocity 190 00:07:01,894 --> 00:07:05,232 since the projectile's not moving up or down at that moment. 191 00:07:05,232 --> 00:07:06,893 And then on the way down the vertical component 192 00:07:06,893 --> 00:07:09,241 of the velocity gets more and more negative, 193 00:07:09,241 --> 00:07:12,292 which increases the speed of the projectile. 194 00:07:12,292 --> 00:07:14,443 Keep in mind during this entire trip 195 00:07:14,443 --> 00:07:17,969 the vertical acceleration is the same, negative 9.8 196 00:07:17,969 --> 00:07:21,247 on the way up, at the top, and on the way down. 197 00:07:21,247 --> 00:07:23,553 The vertical acceleration never changes. 198 00:07:23,553 --> 00:07:24,945 So what would an example problem 199 00:07:24,945 --> 00:07:26,880 involving 2D kinematics look like? 200 00:07:26,880 --> 00:07:29,258 Let's say a meatball rolls horizontally 201 00:07:29,258 --> 00:07:32,452 off of a dinner table of height H, with a speed V. 202 00:07:32,452 --> 00:07:35,063 And we want to know how far horizontally does the meatball 203 00:07:35,063 --> 00:07:37,060 travel before striking the floor. 204 00:07:37,060 --> 00:07:39,325 So the first thing we should do is draw a diagram. 205 00:07:39,325 --> 00:07:41,066 So the height of the table is H. 206 00:07:41,066 --> 00:07:43,724 The initial speed of the meatball is V, 207 00:07:43,724 --> 00:07:46,476 and we want to determine how far horizontally it makes it 208 00:07:46,476 --> 00:07:48,284 from the edge of the table. 209 00:07:48,284 --> 00:07:50,089 But note this problem's symbolic. 210 00:07:50,089 --> 00:07:52,312 We're not given any numbers so we're gonna have to give 211 00:07:52,312 --> 00:07:54,327 our answer in terms of given quantities 212 00:07:54,327 --> 00:07:55,865 and fundamental constants. 213 00:07:55,865 --> 00:07:58,495 The given quantities are things like H and V, 214 00:07:58,495 --> 00:08:01,335 and fundamental constants are things like little G. 215 00:08:01,335 --> 00:08:02,835 So this quantity we want to find 216 00:08:02,835 --> 00:08:04,979 is the displacement in the X direction, 217 00:08:04,979 --> 00:08:07,225 which is gonna be the speed in the X direction 218 00:08:07,225 --> 00:08:08,787 times the time of flight. 219 00:08:08,787 --> 00:08:10,558 We know the speed in the X direction, 220 00:08:10,558 --> 00:08:11,846 it's gonna remain constant. 221 00:08:11,846 --> 00:08:13,686 So this V is gonna be the horizontal speed 222 00:08:13,686 --> 00:08:15,009 for the entire trip. 223 00:08:15,009 --> 00:08:18,010 So we can plug in V for the speed, but I don't know 224 00:08:18,010 --> 00:08:19,480 what the time of flight is gonna be. 225 00:08:19,480 --> 00:08:21,341 To get the time of flight we'll do another equation 226 00:08:21,341 --> 00:08:22,701 for the vertical direction. 227 00:08:22,701 --> 00:08:25,296 The vertical displacement is not gonna be H, 228 00:08:25,296 --> 00:08:28,414 it's gonna be negative H since this meatball fell downward. 229 00:08:28,414 --> 00:08:30,055 And the initial velocity in the Y direction 230 00:08:30,055 --> 00:08:32,950 is not gonna be V, it's gonna be zero, 231 00:08:32,950 --> 00:08:35,582 since this meatball had no vertical velocity 232 00:08:35,582 --> 00:08:37,196 right when it left the table. 233 00:08:37,197 --> 00:08:39,482 It only had horizontal velocity. 234 00:08:39,482 --> 00:08:41,700 The acceleration is negative 9.8, 235 00:08:41,700 --> 00:08:43,852 but we're gonna write that in terms of fundamental constant, 236 00:08:43,852 --> 00:08:45,455 so we'll write that as a negative G. 237 00:08:45,455 --> 00:08:46,757 This lets us solve for T. 238 00:08:46,757 --> 00:08:49,584 We get the square root of two H over G 239 00:08:49,584 --> 00:08:51,405 which we can now bring over to here 240 00:08:51,405 --> 00:08:53,634 to get the horizontal displacement of this meatball 241 00:08:53,634 --> 00:08:55,213 before it hits the ground. 242 00:08:55,213 --> 00:08:56,998 Something else you'll definitely have to know 243 00:08:56,998 --> 00:09:00,118 for the AP exam is how to graph data to a linear fit. 244 00:09:00,118 --> 00:09:02,067 And what I mean by that is that when you graph data 245 00:09:02,067 --> 00:09:03,821 it doesn't always come out linear. 246 00:09:03,821 --> 00:09:05,331 And when you don't get a linear graph 247 00:09:05,331 --> 00:09:07,896 it's hard to find the slope of that curved graph. 248 00:09:07,896 --> 00:09:10,631 However you can force your data to be linear 249 00:09:10,631 --> 00:09:13,243 if you write down the expression that gives the relationship 250 00:09:13,243 --> 00:09:16,224 between your data in the form of a straight line. 251 00:09:16,224 --> 00:09:19,692 So the form of a straight line is Y equals M X plus B. 252 00:09:19,692 --> 00:09:21,311 Y would be the vertical axis. 253 00:09:21,311 --> 00:09:23,181 X would be the horizontal axis. 254 00:09:23,181 --> 00:09:26,620 M is the slope, and B would be the Y-intercept. 255 00:09:26,620 --> 00:09:28,177 In other words, if you had the expression 256 00:09:28,177 --> 00:09:30,605 P equals 1/2 D squared, 257 00:09:30,605 --> 00:09:33,990 if you just plot P versus D you're gonna get a parabola, 258 00:09:33,990 --> 00:09:35,584 and that means you got problems 259 00:09:35,584 --> 00:09:38,254 'cause finding the slope of a parabola is hard. 260 00:09:38,254 --> 00:09:42,191 But if you instead choose to plot P versus D squared, 261 00:09:42,191 --> 00:09:43,736 you will get a straight line 262 00:09:43,736 --> 00:09:46,918 because now you've required P to be your vertical axis, 263 00:09:46,918 --> 00:09:50,151 you've required D squared to be your horizontal axis, 264 00:09:50,151 --> 00:09:53,292 and the slope that's multiplying what you called X 265 00:09:53,292 --> 00:09:54,934 is just a constant, and that means 266 00:09:54,934 --> 00:09:56,554 your slope's gonna be constant. 267 00:09:56,554 --> 00:09:58,806 So this lets us predict what the slope would be 268 00:09:58,806 --> 00:10:01,650 if I plot P versus D squared since the slope 269 00:10:01,650 --> 00:10:04,495 is always what's multiplying my horizontal axis, 270 00:10:04,495 --> 00:10:07,058 the slope in this case should just be 1/2. 271 00:10:07,058 --> 00:10:09,091 So in other words, if you force your expression 272 00:10:09,091 --> 00:10:11,142 to take the form of a straight line, 273 00:10:11,142 --> 00:10:13,098 now only will you get a linear fit, 274 00:10:13,098 --> 00:10:15,217 but you can predict what the slope is 275 00:10:15,217 --> 00:10:17,742 by looking at everything that's multiplying 276 00:10:17,742 --> 00:10:19,298 what you called X. 277 00:10:19,298 --> 00:10:21,921 But a lot of people find this confusing and strange, 278 00:10:21,921 --> 00:10:23,121 so what would an example problem 279 00:10:23,121 --> 00:10:26,222 where you have to graph data to a linear fit look like? 280 00:10:26,222 --> 00:10:27,523 Say you were given this question. 281 00:10:27,523 --> 00:10:29,944 You repeatedly roll a sphere off a table 282 00:10:29,944 --> 00:10:31,963 with varying speeds V, 283 00:10:31,963 --> 00:10:34,083 and then you measure how far they travel delta X 284 00:10:34,083 --> 00:10:35,405 before they strike the floor. 285 00:10:35,405 --> 00:10:38,286 If the table has a fixed and known height H 286 00:10:38,286 --> 00:10:41,111 what could we plot to determine an experimental value 287 00:10:41,111 --> 00:10:43,777 for the magnitude of the acceleration due to gravity? 288 00:10:43,777 --> 00:10:46,168 So we repeatedly change the speed 289 00:10:46,168 --> 00:10:48,379 and measured how far the ball went. 290 00:10:48,379 --> 00:10:51,463 That lets us know these are the quantities that are varying, 291 00:10:51,463 --> 00:10:54,939 so these are gonna be involved in the X and Y axes. 292 00:10:54,939 --> 00:10:56,813 But to figure out what to plot we need to find 293 00:10:56,813 --> 00:10:59,780 some sort of relationship between these two variables 294 00:10:59,780 --> 00:11:01,335 so that we could put that relationship 295 00:11:01,335 --> 00:11:03,039 in the form of a straight line. 296 00:11:03,039 --> 00:11:05,215 Now in the 2D kinematics section right before this, 297 00:11:05,215 --> 00:11:08,064 we derived a formula for how far a ball goes 298 00:11:08,064 --> 00:11:11,322 rolling off a table in terms of the height of the table 299 00:11:11,322 --> 00:11:13,034 and the acceleration due to gravity. 300 00:11:13,034 --> 00:11:14,718 So this is the expression that relates 301 00:11:14,718 --> 00:11:17,371 how fast it was going to how far it went. 302 00:11:17,371 --> 00:11:19,680 And we need to put this in the form of a straight line 303 00:11:19,680 --> 00:11:22,014 so it's easiest to just make this left-hand side 304 00:11:22,014 --> 00:11:24,082 since it's already solved for Y. 305 00:11:24,082 --> 00:11:27,913 So our Y quantity on our vertical axis would just be delta X 306 00:11:27,913 --> 00:11:29,712 and that's okay 'cause that's one of the quantities 307 00:11:29,712 --> 00:11:31,634 that we're varying over here. 308 00:11:31,634 --> 00:11:34,728 Similarly, the other quantity that's varying is V, 309 00:11:34,728 --> 00:11:36,539 so I'll just call V X. 310 00:11:36,539 --> 00:11:39,146 That means the horizontal axis is gonna be V. 311 00:11:39,146 --> 00:11:41,526 I was able to do that since this was just V. 312 00:11:41,526 --> 00:11:44,620 If this had been V squared, I would've had to have plotted 313 00:11:44,620 --> 00:11:46,667 V squared on the horizontal axis. 314 00:11:46,667 --> 00:11:48,660 And if this was square root of V, 315 00:11:48,660 --> 00:11:51,667 I would have to plot square root of V on the X axis. 316 00:11:51,667 --> 00:11:54,783 But since it was just V, I can get away with just plotting V 317 00:11:54,783 --> 00:11:56,949 on my horizontal axis and now we can figure out 318 00:11:56,949 --> 00:11:58,139 what our slope would represent. 319 00:11:58,139 --> 00:12:01,077 We've got Y equals M times X. 320 00:12:01,077 --> 00:12:03,100 Everything that multiplies what we called X 321 00:12:03,100 --> 00:12:04,107 is gonna be our slope. 322 00:12:04,107 --> 00:12:05,036 The way it's written here 323 00:12:05,036 --> 00:12:07,148 the M is multiplied on the right-hand side, 324 00:12:07,148 --> 00:12:09,027 but it doesn't matter 'cause M times X 325 00:12:09,027 --> 00:12:11,236 is the same as X times M. 326 00:12:11,236 --> 00:12:15,083 That means this entire term, root two H over G 327 00:12:15,083 --> 00:12:17,099 is the slope of this graph. 328 00:12:17,099 --> 00:12:19,093 In other words, we're gonna get a linear fit, 329 00:12:19,093 --> 00:12:21,231 and the number we find for the slope 330 00:12:21,231 --> 00:12:25,562 is gonna equal the square root of two H over G. 331 00:12:25,562 --> 00:12:27,815 And since there's no added B term, 332 00:12:27,815 --> 00:12:30,809 there's no Y-intercept to worry about over here. 333 00:12:30,809 --> 00:12:32,157 So how do we actually determine 334 00:12:32,157 --> 00:12:33,840 the experimental value for G? 335 00:12:33,840 --> 00:12:35,756 We take our data, we plot them, 336 00:12:35,756 --> 00:12:37,950 we draw a best fit line through the data. 337 00:12:37,950 --> 00:12:39,726 We would use points on our line 338 00:12:39,726 --> 00:12:41,612 to determine the slope of this line 339 00:12:41,612 --> 00:12:43,981 by taking the rise over the run. 340 00:12:43,981 --> 00:12:46,986 That rise over run would be equal to the slope, 341 00:12:46,986 --> 00:12:49,152 and we know that that number's gonna equal 342 00:12:49,152 --> 00:12:51,940 the square root of two H over G. 343 00:12:51,940 --> 00:12:53,560 So if we know this number for the slope, 344 00:12:53,560 --> 00:12:56,532 and the table has a known and fixed value H, 345 00:12:56,532 --> 00:00:00,000 the only unknown is G which we can now solve for.