1 00:00:01,199 --> 00:00:23,570 Since now we have the basic concepts of kinematics, let's try now to apply them. 2 00:00:23,570 --> 00:00:30,109 In this video we are dealing with some very simple problems just to practice. 3 00:00:30,109 --> 00:00:36,289 For this is the kind of problems I will ask you to solve in the exam. 4 00:00:36,289 --> 00:00:41,609 We can start with changing the units of velocities. 5 00:00:41,609 --> 00:00:48,729 For example, from kilometers per hour into meters per second. 6 00:00:48,729 --> 00:01:02,590 We multiply 1000 times and we divide 3600, because one hour has 3600 seconds. 7 00:01:02,590 --> 00:01:11,099 So 36 kilometers per hour is the same as 10 meters per second. 8 00:01:11,099 --> 00:01:24,519 Conversely, to pass from meters per second to kilometers per hour, we just multiply 3600 9 00:01:24,519 --> 00:01:35,859 and divide by 1000, so in fact we multiply by 3.6, thus 20 meters per second is the same 10 00:01:35,859 --> 00:01:41,780 as 72 kilometers per hour. 11 00:01:41,780 --> 00:01:49,140 should practice this for the exam. Another typical exercise about velocity 12 00:01:49,140 --> 00:01:57,859 is finding the point where two mobiles meet. To do this, we are just dealing with 13 00:01:57,859 --> 00:02:07,400 mean velocities. In this case, two trains, one leaving Madrid, the other one 14 00:02:07,400 --> 00:02:14,400 Zaragoza, both at the same time, the one from Madrid at a speed of 200 km per hour 15 00:02:14,400 --> 00:02:22,400 and the one from Zaragoza at 150 km per hour, must meet at a certain point 16 00:02:22,400 --> 00:02:28,400 and that point is just the purpose of this exercise, finding it. 17 00:02:28,400 --> 00:02:36,400 Let's just remember that speed is the relationship between space and time. 18 00:02:36,400 --> 00:02:44,400 On the other hand, we know that when the two trains meet, the time will be the same for both. 19 00:02:44,400 --> 00:02:53,400 So we can derive the formula for time and then make both times the same number. 20 00:02:53,400 --> 00:03:03,400 And so we have obtained the key equation where S1 is the space between Madrid and the meeting point, 21 00:03:03,400 --> 00:03:13,960 s2, the distance from Zaragoza up to the meeting point, and v1, v2, their respective velocities. 22 00:03:13,960 --> 00:03:21,710 We can solve this, provided we know the total distance between Madrid and Zaragoza. 23 00:03:21,710 --> 00:03:29,969 In order to do this, we remove one variable using the given value of the total distance. 24 00:03:29,969 --> 00:03:34,259 Thus, we can write. 25 00:03:34,259 --> 00:03:40,860 this equation we have only one variable that we can derive just by multiplying 26 00:03:40,860 --> 00:03:55,330 in cross. Please try to follow this by writing on a paper by yourself. Now we 27 00:03:55,330 --> 00:04:03,430 can group the terms having S1 on the left side of the equation and so we can 28 00:04:03,430 --> 00:04:15,120 derive the value for S1. This means that the trains meet at 200 kilometers away from Madrid. 29 00:04:17,189 --> 00:04:28,069 Of course we can also calculate at what time and since the velocity for train 1 is 200 kilometers 30 00:04:28,069 --> 00:04:38,290 per hour, it's clear that it has spent just one hour. Well, that's all for the moment. 31 00:04:39,250 --> 00:04:47,329 Try to practice, write me your hands on it up, and watch the Padlet where I'm uploading 32 00:04:47,329 --> 00:04:51,810 an exercise that I expect you to send me back before Wednesday.