Brrring
DISCLAIMER: Once again, I don't own Pokémon or the scientific material being discussed in this story. Nintendo and Shogakukuan own Pokémon...at least Shogakukuan owns the Pokémon anime that this story is largely based off of, and the science is open for everybody to use and learn. So use and learn it. Special thanks go to the Kaplan program for the books that I've used for these chapters, and also my Kaplan instructor Matt Kinney for his diligence in proofreading the chapter for scientific accuracy. In addition, I'd like to thank Angelo Godbey for showing me I still don't have an idea of what I'm talking about when it comes to kinematics in an x and y direction. You use the vf vo + at equation to find the time it takes for the projectile to stop, multiply it by two to get the time it takes to get back to the origin only in a different direction, and the y yo + vot + ½ at 2 equation to find the time if the projectile goes down beneath the origin.
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Ash, Misty, and Brock entered the classroom before the bell rang for their second day of classes. "Another day of toil and trouble," muttered Ash.
"Don't be like that," admonished Misty. "Some of this stuff is pretty interesting."
"And by the way," asked Brock, "did you finish reading the chapter, Ash?"
"How can I? I fell asleep about five minutes into it. The book was SO boring!"
"Well, I must admit it was a bit on the didactic side," said Misty, "but there are far worst ways you could use to teach this material."
Richie ran up to greet his friend. "Hey, Ash, how did studying go last night?"
"Don't even talk about it," Ash groused. "Did you manage to finish the chapter?"
"Yeah, I finished the first chapter and even went on to the second. If this material is proving to be too difficult, I can always go to your place and help you study."
"That would be great," said Ash. "If only we would be able to meet up without this stupid science stuff hanging over our head."
"Even if you do go and help him, you'd just be wasting your time," said Gary.
"Why you…Gary! I'll show you!"
"Oh yeah? Let's see if you can answer this easy question. If I throw a Pokéball 60° degrees in the air with a velocity of four meters per second, and it goes four seconds before it hits the ground, how far away from me is the Pokémon that I caught so gracefully?" Gary's cheerleader cheered behind him.
"Uh, I don't think it's physically possible to have a hang time of four seconds if you only throw the Pokéball four meters per second," said Brock.
"Ah, it doesn't matter. The only thing that matters is whether or not Ash the loser here has the ability to solve this problem."
"I don't think you should try," said Richie, "especially if you didn't read the first chapter."
"Who cares?" cried Ash. "Any idiot would know that it would go sixteen meters, because 4 x 4 is 16!"
"Hah! That's incorrect! I knew you were too stupid to get it!" Gary laughed triumphantly.
"What? How is that possible?" wailed Ash.
"Actually, he's right. The correct answer would only be eight meters," said Misty.
"How can that be?"
"If you read the chapter, you would have known that when the velocity is at an angle, you'd have to consider the horizontal and the vertical components separately."
"That's right, loser!" Gary sneered. "How is it that you have such smart friends and still be so stupid?"
"Don't call Ash stupid, and he's not a loser," Richie cried.
At that moment, the bell rang and Professor Oak walked into the classroom, late but still composed. "All right class," he said, "settle down. We have a lot to cover today."
Team Rocket also scrambled into the room, beat up, dusty, and out of breath.
"What happened to you guys?" asked Tracey.
"We went to the twerp's house to get Pikachu," Jessie grumbled, "and the stupid Mr. Mime his mother has chased us away."
"At least Mr. Mimie can be useful at times," Ash said dryly.
"Ah, so you were doing a bit of running in the morning?" asked Professor Oak. "Perhaps we would be able to use your example in our study this morning."
"Uh, we'd rather not," said James.
"Come now, don't be bashful. Anyways, I hope everybody took the time to read over the first chapter, because this would serve as the foundation for much of what we are going to do in physics: Kinematics. Would anybody like to tell us what the three basic quantities that will be useful in kinematics?"
Todd raised his hand. "The first quantity is displacement. This measures how much you have travelled. The basic units for displacement would be meters. The next quantity is velocity, which measures your change in displacement over a change in time. The basic units for velocity would be meters per second. The final quantity is acceleration. This is the change in velocity over the change in time. The unit for acceleration is meters per seconds squared, since it is change in velocity (m/s) per second."
"Very good," said Professor Oak. "I'm glad that you did your reading. Todd is correct. Displacement, velocity, and acceleration are the three important quantities in kinematics."
A.J. raised his hand. "Yeah, I was reading the chapter, and it sure seemed like displacement is no different from distance, and velocity is no different from speed, yet they apparently aren't the same thing at all. It really messed me up when I did the questions at the end of the chapter. What is this difference they were talking about anyways?"
"Ah, great question, A.J. Distance and speed are very closely related with displacement and velocity, but there is one difference that separates them. Can anybody explain what this difference is?"
Misty spoke up. "Distance and speed are scalar quantities, meaning that they are independent of direction. If you walk from Viridian City to Pewter City, then from Pewter City to Cerulean City, then the distance you walked is the total distance from Viridian to Pewter, and then from Pewter to Cerulean. However, your displacement is the straight-line difference between Viridian and Pewter. This is because displacement and velocity are vector quantities. Every time you change a direction, even if your speed doesn't change, your velocity would change."
"That is exactly it," said Professor Oak. "This is a good transition into what we were going to cover last class but did not have a chance to do so: the concept of vector addition and breaking a vector down into its components.
"As Misty said, vector quantities are dependent on your direction. If you go north and then go east, then your displacement and velocity would change. It is helpful to draw vectors as lines with an arrow pointing in the desired direction: ––––––––––––
"Whenever you go in one direction for a long time, then the vector quantities are no different from scalar quantities. However, whenever you change directions, the vector quantities would change, and you would have to mathematically determine what the new quantity would be. For example, let's just take Misty's example of walking from Viridian to Pewter, and then to Cerulean. Suppose the distance from Viridian to Pewter is 30,000 meters, while the distance from Pewter to Cerulean is 40,000 meters. Can anybody figure out what your overall displacement is?"
Duplica instantly raised her hand. "Well, 30,000 meters is 3 x 104 while 40,000 meters is 4 x 104. You would draw a straight line that's going north, and another straight line going east. The first line would be labeled 3 x 104 while the second line would be labeled 4 x 10 4. The overall displacement would be the straight line from where the first line begins to where the second line ends. You would use Pythagorean's Theorem, or X 2 + Y 2 is Z 2 to figure out the length. X 2 and Y 2 would be 9 x 10 8 and 16 x 10 8 respectively. Adding those two up would give you 25 x 108, and the square root of that would be 5 x 10 4, so your final answer would be 50,000 meters."
"Very good, Duplica," said Professor Oak. "Now, would you know what the final direction of the displacement would be?" Duplica quietly shook her head.
"It's okay, because this would require a general knowledge of trigonometry, or the study of triangles. There are three trigonometric functions that are dependent only on a particular angle in a right triangle. Here is a sample triangle XYZ, where the angle YXZ is 60° and angle XZY is 30°. We'll consider angle XZY, which we have already established is 30°.
"The first trigonometric function is sine, which is essentially the ratio between the lengths of the opposite side XY with the hypotenuse XZ. No matter how long each of these sides may be, the ratio of the length would be the same if the angle is the same. The cosine of the angle is the adjacent side YZ with the hypotenuse XZ. Similarly, the cosine of an angle remains the same no matter how much you expand or decrease the triangle. Finally, the tangent is the opposite side XY over the adjacent side YZ.
"If I tell you that the sine of 30° is ½, and XZ is 6 meters long, then how long would XY be?"
Meowth raised his hand. "Oh yeah, I know this one. XY would be 3 meters long."
"Very good," said Professor Oak as Jessie whacked Meowth on the head for the audacity of actually participating in class. "Now, let's take this right triangle Duplica described. We'll call it VPC, for Viridian Pewter Cerulean. On one straight edge is 30,000 meters, and on another straight edge you have 40,000 meters. The hypotenuse is 50,000 meters. The angle we are concerned with is this angle PVC, because by figuring out this angle, you would be able to figure out the direction of the vector. Does anybody want to take a crack at solving this?"
Brock took a crack. "Well, we know the length of all of the sides, so we can use any one of the trigonometric functions. I'll use sine, since we've done so much with it. Sine is opposite over hypotenuse, so it would be 40,000 meters over 50,000 meters, which is .8. Our answer would be the angle with a sine of .8."
Tracey took out his calculator to figure out the answer. "The angle with a sine of .8 is 53.13°, so the direction of our displacement would be 53.13°."
"That is correct," said Professor Oak. "These trigonometric functions would be very useful when it comes to adding or subtracting vectors, which is what you may have to do. Let's suppose you have two vectors 2 meters long, one making a 30° angle to the horizontal, and the other making a 60° angle to the horizontal. If you want to add them together, you would put them end to end. However, this is not enough. In order to find the magnitude and the direction, what you have to do is to find the x, or horizontal, and y, or vertical components.
"Let's just take this first vector. It should be easy to find the x and y components if you make this into a right triangle. You just draw two lines going up or down from the edges of the vector until they meet. Voila! You've figured out the identity of the components, and now all you have to do is use the trigonometric functions to find the magnitudes. Tracey, can you use your calculator to tell us what the x and y components are?"
Tracey worked it out. "The y component would be sine of 30° times two, and that would be 1 meter. The x component would be cosine of 30° times two, and that would be 1.73. For the other vector, the y component would be sine of 60° times two, and that would be 1.73. The x component would be cosine of 60° times two, and that would be 1."
Misty did the math. "The sum of the x components would be 2.73, and the sum of the y components would be 2.73. However, these are only the components, and the final vector would be the hypotenuse of the resulting right triangle."
Tracey plugged numbers into his calculator and came up with an answer. "The final sum would be divided by 3.86 is .707, and so the direction would be 45°."
"That would make sense," said Professor Oak. "The two sides both have the same lengths, so this would be an isosceles right triangle, and the angles in one of those is 90°, 45°, and 45°.
"Most of the sines and cosines you can find with a calculator, but there are a few that you would need to memorize. For example, I already told you that the sine of 30° is ½. The cosine of 30° is √(3) / 2. Similarly, the sine of 60° is √(3) / 2, and the cosine of 60° is ½. The sine and the cosine of 45° are the same, and they are both √(2) / 2. The sine of 0° is zero, and the cosine of 0° is 1. The sine of 90° is 1, and the cosine of 90° is 0. Those are the sines and cosines that you should memorize.
"The last thing you should know about vectors is that it is possible to subtract vectors. To do this, you would just have to flip a vector over, and then add the vectors normally.
"With the introduction to vectors and trigonometric functions out of the way, we can now finally focus on the topic at hand. As Todd said, the important quantities in kinematics are displacement, velocity, and acceleration. All three of these quantities are tied together in some way. Now, Jessie, you said that you were chased away by Mr. Mime. How fast would you say you were going when you started running and at your top speed?"
Jessie didn't say anything, but James said, "Well, I don't think we were moving at all when we started running, but by the time we reached our maximum speed, we were going probably two meters per second."
"And how long did it take you to get to that top speed?"
"Uh, I would guess ten seconds," said James while Jessie made efforts to get him to keep his mouth shut."
"Okay, thank you, James. If we assume constant acceleration, we have enough information to calculate the average velocity, acceleration, and even distance traveled. So, who would want to take a shot at calculating the acceleration?"
"I would," said Lara. "They went from zero m/s to two m/s, so that would be their change in velocity, or delta v. The time it took them to do so is ten seconds. The equation for acceleration is change in velocity over change in time, so it would be 2 m/s / 10 s, and that would be .2 m/s 2."
"That is correct," said Professor Oak. "It is possible to interconvert between displacement, velocity, and acceleration using the equations Todd said at the beginning of class. Now, we know that acceleration usually always changes, but in many of the problems we do on paper, we have to assume constant acceleration. If we make this assumption, then there are four equations that you can use to find displacement, velocity, acceleration, and time even when one of those quantities is missing. Who can tell me what the four equations are?"
"The first equation is vf vi + acceleration change in time," said Gary. "You could use this to find delta v, acceleration, and change in time even if you don't know the displacement."
"The second equation is the square of the initial velocity plus two times the acceleration multiplied the change in distance is equivalent to the square of the final velocity," said Samurai. "You may use this equation to find the change in velocity, the acceleration, or the change in displacement even though you do not know the change in time."
"The third equation is △d vit + ½ at 2," said Richie. "You can use this to find the change in displacement, the initial velocity, the acceleration, or the time even if you do not know the final velocity."
"These are all correct," said Professor Oak. "And what is the final equation?"
Duplica spoke up. "The final equation is △d ½ (vi + vf) x △t. This equation would allow you to find the displacement, the change in velocity, or the change in time even if you do not know the acceleration."
"That's very good. These are the basic equations you would have to remember for one-dimensional kinematics."
Ash was aghast. "What? Do you mean there's two-dimensional kinematics?"
"That is correct," said Professor Oak, "but before we go into two-dimensional kinematics, we'd have to discuss a special case of one-dimensional kinematics: free fall.
"I am sure you all remember Team Rocket's spirited entrance yesterday. That is a perfect example of free fall, since they were falling with a constant acceleration."
"The acceleration due to gravity," said Misty.
"That is correct. The Earth has gravity, and it has the same pull on anybody no matter where on Earth you are, as long as you're still on the surface and not miles underground in a Diglett's tunnel, that is. The acceleration g is equal to -9.8 m/s2, although it is sometimes acceptable to round that to 10 m/s2. Using these kinematic equations, we can calculate anything about the free fall, from the final velocity to the time it takes for something to fall. There is one thing you must remember, and that is that the acceleration due to gravity is often negative when you use it in the equations."
"Why is that?" asked A.J.
"As we said, it is because these vector quantities are dependent on direction. If we are doing one-dimensional kinematics, it is important to signify one direction as positive, and so the other direction would be negative. Take acceleration for example. When James was speeding up to get away from the Mr. Mime this morning, we would say that his acceleration is positive. Once he gets far enough away from it, he would slow down."
"And when he slows down, his acceleration would be negative," said Brock.
"That is right. In addition, you can do it for velocity and displacement as well. Let's suppose Ash is going north, from Viridian City to Pewter City. After that, he goes from Pewter City to Viridian City. What would his net displacement be?"
"The net displacement would actually be zero, since he went back to where he started," said Lara.
"Ah, that's because we're assuming the northern direction to have a positive displacement, and the southern direction to have a negative displacement," said A.J. "The two would cancel each other out of they're opposite but equal. I get it now."
"That's terrific," said Professor Oak. "It's the same if you're doing a free fall problem. Our positive for acceleration is always in the upwards direction, which doesn't make any sense because it is physically impossible to toss something like a Pokéball in the air and have it accelerate upwards. A Pokéball would slow down and eventually falls if we throw it up in the air. This is our negative acceleration.
"So let's think back to when Team Rocket fell from the ceiling yesterday. We'll say that it is 3 meters from the floor to the ceiling, and the group is starting from rest. How fast would they be going when they hit the floor, and how long would it take them?"
With the kinematic equations in hand, the class began tackling this deceptively simple problem. Misty was the first to put her pencil down, but she waited until everybody was finished before saying her answers.
"Well, we know the initial velocity, the displacement, and the acceleration. We can find the final velocity and the time with the vf 2 is vi 2 + 2a△d and the △ d vit + ½at 2 equations respectively. In this situation, it is essential that we assume the negative direction to be positive, because the initial velocity is zero, and the quantity we want to calculate is a square."
"Ah, that's a very good point," said Professor Oak. "While it is customary to use the upwards direction as the positive, it may sometimes be essential to make the downwards direction positive, mostly if you're starting from rest. I hope most of you caught this." Most of the students nodded, but a few of the students were abashed. "Please continue."
"Okay," said Misty, "for the first equation, the initial velocity is zero, the acceleration is 9.8 m/s2, and the displacement is 3 meters. Plugging all these into the equation, it would be 2 x 9.8 x 3, which is 58.8 m2/s2. Taking the square root of that would tell you that their final velocity was 7.68 m/s. Now that we know final velocity, we can actually use any equation, but using the △ d vit + ½at2 equation, we rearrange it to find out that 2△ d / a t 2. Plugging everything in, we get that t 2 is around .6, so the square root of that is around .78 seconds."
"It sure seemed longer than that as we were falling," said James.
"That is correct," said Professor Oak. This is the basics of free fall. We can use the same equations if we consider a situation where we throw an object in the air, except this time the initial velocity would be something other than zero, and we cannot get away with using a positive acceleration. This throw can be straight up in the air, or to the side. In the latter situation, the kinematics is known as projectile motion.
"Gary already touched on the important thing about projectile motion when he quizzed Ash before class. The important thing to know is that you must take the x and the y components independently. The x component is the easy one, because the horizontal acceleration of a projectile is zero if we assume no external forces like air resistance – friction by the air – is in play. This is related to Newton's First Law of Motion, which we will discuss in the next class. The y component is just like what we talked about with free fall. The only thing to keep in mind is that at the top of the projectile's parabolic path, the velocity is zero. So Gary, can you repeat what you told Ash before class?"
"Of course I can," Gary said confidently. "If I throw a Pokéball 60° degrees in the air with a velocity of four meters per second, and it goes four seconds before it hits the ground, how far away from me is the Pokémon that I caught so gracefully?"
"Thank you, Gary. As Brock said, the time Gary said can be discounted in this case, but it's understandable why Gary would mention it because he was testing Ash about the fact that horizontal motion has zero acceleration. However, we would have a new time and a new horizontal displacement if we want to make this Pokémon capture physically acceptable. Who would be willing to take a shot at figuring out the new location of Gary's new Pokémon."
"I'll do it," said Richie. "We'll make the assumption that the Pokémon is at the same height as when Gary releases the ball. A projectile takes a parabolic motion that is symmetrical as long as the initial height is the same as the final height. To find the time, all we would have to do is to find the time when the Pokéball is at the apex of its path, and double that. At the peak, the Pokéball has a vertical velocity of zero, so we can use the equation vf vi + a△t to find the time. The final velocity is zero, so we'd be able to tell that the initial velocity is equal and opposite of the acceleration times the change in time. If we rearrange the equation so that the time is on one side, we get △t vi / a. We would need the overall velocity times the sine of 60°, and that is 3.46 m/s. Divide that by the acceleration 9.8, and we get .35 seconds. Double that and you get .70 seconds.
"To find the horizontal displacement, you would just multiply the horizontal velocity, four times the cosine of 60°, by the time. This is the equation △x 1/2(vf + vi) △t, only the initial and final velocity are the same. The cosine of 60° is ½, so the horizontal velocity is 2 m/s. The Pokémon would only be 1.4 meters away."
"Hah!" cried Ash. "That's only like the distance from me to Brock! That's not very impressive at all!"
"Yeah? Well I'll bet you'd miss the Pokémon even when it's that close," Gary countered.
"All right, that's enough from you two," said Professor Oak. "That is correct, Richie. If you want to figure out the time if the Pokémon is lower than the point of release, then you would just use the △d vit + ½ a t2 equation with the negative initial velocity as the starting velocity and add the time to the .7 seconds you got. If you want to figure out the time if the Pokémon is higher than the point of release, then you would use that equation with zero as your initial velocity and add it to the .35 seconds you got at the top.
"Well, our time is almost up. Your homework for tonight is to do the homework problems for Chapter 1 if you haven't done so already, and also to read Chapter 2. In addition, I'll throw in an extra credit problem: what was the maximum height of Gary's Pokéball? See you all tomorrow."
The bell rang, and the students slowly streamed out of the room.
"Oh, that Gary," Ash muttered. "I'll get the best of him, if it's the last thing I do!"
"Well, if it's in physics, I don't think that'll ever happen," said Misty.
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AUTHOR'S NOTE: I think the personalities for each of the characters are coming out with this chapter. There are the people that will obviously be good students, people that would obviously be bad students, and people that just don't give a darn (Team Rocket.) Then there are the people like Joe who fade away into obscurity. I think I completely forgot about him in a few of the early chapters, but how could you not be anonymous with a name like Joe Smith. And then there is the rivalry between Ash and Gary, which is the primary plot point outside of all of the science junk. Gary's not too bad in these early chapters, but I've definitely exaggerated his pompous behavior and descending attitude towards Ash and everybody else. You'll see what I mean.
Anyways, as I said in the disclaimer, I made an egregious error in the calculations of Gary's Pokéball example near the end of the chapter. I think I used the y yo + vot + ½ at 2 equation and set y equal to zero. Thankfully I checked it over with somebody first. The rest should be okay. The only major integration of Pokémon I could think of in this chapter is the Pokéball example and the three major Pokémon cities in the discussion about vector addition. The lack of superscripts, subscripts, and equal signs are annoying, just because so many equations have these values. Why, ? WHY?