WEBVTT 00:00:00.046 --> 00:00:02.764 This is College Physics Answers with Shaun Dychko. 00:00:03.422 --> 00:00:05.820 A skier descends a 70 meter slope 00:00:06.281 --> 00:00:08.765 which is inclined at an angle of thirty degrees 00:00:09.295 --> 00:00:12.570 and we have to find out what their final speed will be at the bottom, 00:00:12.781 --> 00:00:14.054 assuming there is no friction 00:00:14.336 --> 00:00:17.781 and assuming they have an initial speed of zero for part A. 00:00:18.234 --> 00:00:19.992 Then we do the calculation again 00:00:20.094 --> 00:00:23.602 assuming an initial speed of 2.5 meters per second in part B. 00:00:24.460 --> 00:00:26.414 Then we compare the total time it takes 00:00:26.439 --> 00:00:28.313 to get to the bottom of the slope in each case. 00:00:30.141 --> 00:00:33.813 So we know that the total energy at the end, kinetic plus potential, 00:00:34.281 --> 00:00:36.242 equals the total energy at the beginning. 00:00:37.172 --> 00:00:40.320 So we have no potential energy at the end 00:00:40.345 --> 00:00:43.312 because we'll assume this is our reference level, y equals zero, 00:00:43.337 --> 00:00:45.304 so the final potential energy is zero. 00:00:45.906 --> 00:00:49.078 So we have only kinetic energy when the skier is at the bottom. 00:00:49.171 --> 00:00:51.578 So that's one half mass times final speed squared, 00:00:52.445 --> 00:00:55.294 and in the initial case we have no kinetic energy 00:00:55.319 --> 00:00:57.708 because we assume an initial speed of zero for part A 00:00:58.063 --> 00:01:01.419 and the initial potential energy will be mg times h, 00:01:01.458 --> 00:01:04.089 the vertical height above the ground. 00:01:06.805 --> 00:01:09.128 So we can divide both sides by m 00:01:09.153 --> 00:01:11.098 and then multiply both sides by two 00:01:13.949 --> 00:01:15.550 and also take the square root of both sides. 00:01:15.883 --> 00:01:19.605 We end up with the final speed is the square root of two g times the height. 00:01:20.555 --> 00:01:24.972 Now the height is the opposite leg of this yellow triangle. 00:01:25.375 --> 00:01:29.144 So we go sine theta multiplied by the hypotenuse 00:01:29.339 --> 00:01:31.668 which is d in order to find the height. 00:01:32.121 --> 00:01:34.324 We'll substitute that in for h. 00:01:34.902 --> 00:01:38.297 We have vf equals square root of twogd sine theta. 00:01:39.061 --> 00:01:41.404 So that's the square root of two time 9.8 meters per second squared 00:01:41.430 --> 00:01:45.121 times 70 meters times sine 30 which is 26.2 meters per second. 00:01:46.406 --> 00:01:48.363 So that's the final speed at the bottom of the slope. 00:01:48.719 --> 00:01:50.957 Then the next part of this question is 00:01:51.011 --> 00:01:53.867 what time does it take to get to the bottom of the slope. 00:01:54.847 --> 00:01:57.785 Well, the total displacement along the slope 00:01:58.188 --> 00:02:01.402 is going to equal the average velocity multiplied by the time. 00:02:01.836 --> 00:02:05.828 So we can solve this for t because we know all these other things in the formula. 00:02:08.055 --> 00:02:14.030 So we'll say, let's multiply both sides by two over v i plus vf 00:02:17.922 --> 00:02:19.270 and then switch the sides around. 00:02:19.801 --> 00:02:23.488 We get that t is two d over v i plus vf. 00:02:23.976 --> 00:02:27.113 So that's two times 70 meters divided by zero initial speed, 00:02:27.146 --> 00:02:29.496 plus 26.2 meters per second, final speed, 00:02:29.820 --> 00:02:33.137 giving us 5.35 seconds to descend the slope. 00:02:35.031 --> 00:02:36.152 Now, in part B 00:02:36.567 --> 00:02:39.848 the only difference is that there is an initial kinetic energy. 00:02:39.992 --> 00:02:42.559 So we have this one half m visquared term 00:02:42.863 --> 00:02:47.348 which is the only difference compared to that second line in part A. 00:02:51.547 --> 00:02:55.094 We will multiply both sides by two over m 00:02:57.640 --> 00:03:01.303 and the two and the m cancel in the first term leaving us with vi squared there. 00:03:01.606 --> 00:03:03.209 Then on the second term the m's cancel, 00:03:03.234 --> 00:03:05.586 but we're left with the two behind. So that's two gh there, 00:03:06.133 --> 00:03:08.514 and then we take the square root of both sides to solve for vf. 00:03:09.313 --> 00:03:11.334 So vf equals the square root of vi squared 00:03:11.359 --> 00:03:13.451 plus twog d sine theta 00:03:13.476 --> 00:03:16.248 where we substituted d sine theta in place of the height h. 00:03:17.148 --> 00:03:18.592 Then we substitute in numbers. 00:03:18.873 --> 00:03:21.912 So it's the square root of 2.5 meters per second initial speed squared, 00:03:21.937 --> 00:03:24.444 plus two times 9.8 times 70 times sine 30, 00:03:24.788 --> 00:03:27.959 giving us 26.3 meters per second is the final speed 00:03:28.303 --> 00:03:32.850 which is nearly the same as the final speed we had in part A, 26.2. 00:03:35.310 --> 00:03:37.084 But there is a bit of a -- 00:03:37.382 --> 00:03:40.257 so this difference in speeds is one tenth 00:03:40.844 --> 00:03:44.279 whereas the difference in times is actually going to be more significant. 00:03:45.288 --> 00:03:46.983 It's still a small difference but it's more significant. 00:03:47.008 --> 00:03:49.889 So we have the time is going to be two d over vi 00:03:49.914 --> 00:03:52.100 plus vf for the same reason that it was over here. 00:03:52.125 --> 00:03:54.522 We solved this formula here for t 00:03:56.633 --> 00:04:00.506 and that's two times 70 divided by 2.5 meters per second initial speed, 00:04:00.576 --> 00:04:03.862 plus 26.3106 meters per second final speed. 00:04:04.453 --> 00:04:07.722 That's 4.86 seconds is the total time to get down the slope. 00:04:08.422 --> 00:04:09.948 So the difference in times is small, 00:04:10.430 --> 00:04:14.070 5.35 minus 4.86 is only about half a second. 00:04:14.523 --> 00:04:18.479 We expect that since the initial speed is so much less than the final speed. 00:04:18.984 --> 00:04:20.570 But half a second could make a big difference 00:04:20.595 --> 00:04:22.469 in placement be it first, second, third, fourth 00:04:22.773 --> 00:04:24.242 in a highly competitive event.