okay so hi welcome to the first project explained physics video for AP Physics C mechanics today we're going to be talking about vectors calculus and kinematics just some brief information before we start the lesson is that this unit in AP Physics C mechanics does use quite a bit of vectors including Vector operations dissolving vectors into components addition and subtraction of vectors dot products and cross products so if you're not familiar with the vectors I would highly suggest you go to the pre-calculus videos in the precalculus stream on the project explained YouTube channel and review that also you need to have a basic understanding of calculus including how to take a derivative and how to take an integral of basic functions including polynomials and trigonometric functions so once again if you're not familiar with that I would encourage you to go onto the project explained YouTube channel and review that before beginning this lesson so we'll do a very brief review of vector operations so the two that I'm going to be talking about are the scalar or dot product and the cross product so at the top of the screen you can see the cross product which is the vector product so the cross product is a vector product between two vectors and it outputs a vector as well so in the cross product U cross V would be represented as the determinant with the first row being the unit vectors i j and k with the subsequent rows being the components of the vectors A b c and d e f so the cross product outputs a vector that is normal to both of the original vectors as well as with magnitude equivalent to the area of the parallelogram spanned by these two vectors the perpendicular and magnitude portions of this are not too important right now they will become pretty useful in the rotation unit as well as in e m but for now just understanding how to calculate the cross product as well as some Basics about the concepts behind it is enough just briefly to review some examples of vector quantities are position velocity acceleration and forces and examples of scalar quantities are speed work and energy so the scalar product or the dot product is the second product on the left side of the screen it's between two vectors again but U dot V is the scalar product so you multiply the respective components of each Vector with each other and then add them together at the bottom over here there's a diagram of head to tail addition of vectors that's the graphical way that we add vectors together with a plus b being the head to tail sum of the original vectors A and B so first we're going to go through some conceptual questions so first of all what angle between two vectors results in the scalar product being a maximum so when Theta the intermediate angle is zero degrees the scalar product is a maximum that's because the scalar product can also be Rewritten in trigonometric form as the magnitude of U Times by the magnitude of V Times by the cosine of theta with cosine of theta being a maximum one theta equals zero so what angle between two vectors results in the magnitude of the vector product being a maximum so this would be Theta is 90 degrees because once again the cross product or the vector product its magnitude can be written as the magnitude of U times the magnitude of V Times by the sine of theta with sine of theta being maximized when Theta is equal to 90 degrees so we already wrote for number three the trigonometric expressions for the magnitudes for both of them and for the magnitude of vector a b once again just for review we just look at Pythagorean theorem so we have a and wave B so the magnitude of this Vector would be the square root of a squared plus b squared just review from Pythagorean theorem so one portion of kinematics that we need to look at is the calculus of vector functions this overlaps with math a little bit but it is something that is important in physics as a lot of times we can represent complex Motion in terms of time using Vector functions so all of these Vector functions are parameterized in terms of time so in addition to the appropriate position velocity or acceleration value being assigned you also have a Time Value which would be the independent variable in this case so the position Vector can be written as X of t i plus y of T J plus Z of t k and the velocity Vector is defined as the first derivative of the position Vector so you just take the derivative of each component separately as you can see in the red box and the acceleration is the derivative of the Velocity vector or the second derivative of the position Vector once again taking the derivative of each component separately so and just to recall remember that the derivative of a constant is zero and the velocity vector v a definition that you should know is that the velocity vector v is always tangent to the curve R so an infinite infinitesimal at least small increment in the curve length is in the direction of the Velocity vector and we're going to use that fact in a second when we do our example problem the other thing that you should know is that the speed is the magnitude of the Velocity Vector so in addition to finding the velocity Vector once we have that information we can calculate the speed which is the magnitude of the Velocity vector so let's go through example number one so we have this Vector function so first we want to find V and a so in order to find V first we have to take the derivative of R each component separately so V the derivative of the first component well the derivative of cosine is negative sine and from chain rule we have to multiply Omega out to the outside so we get Negative Omega sine of Omega T times I hat for the second component the derivative of sine is cosine but once again chain rule we have to multiply the Omega to the outside so Omega cosine of Omega t J not and then lastly the derivative of t with respect to time is just one so plus K hat so acceleration is the derivative of velocity so now we need to take our velocity vector and differentiate each component of it in order to find the acceleration vector so once again the derivative of sine is cosine and once again chain rule multiply the Omega out to the outside so you get Negative Omega squared cosine Omega T times I hat for the second component we get that the derivative of cosine is negative sine once again multiply the Omega to the outside by Chain rule so you get Negative Omega squared sine of Omega T J and the derivative of a constant which in this case would be the vector constant K is just zero so it's negligible so one thing to note here is at least in the X and Y directions we can see that a is equal to negative Omega squared r and that's a result that we're going to become more familiar with as we look at uniform circular motion and especially during the unit of simple harmonic motion so that's something that is a bit of a preview for the future so Part B find the magnitudes of them so for the magnitude of the Velocity Vector it would be the square root of each component so it'd be Omega squared sine squared Omega t plus Omega squared cosine squared Omega t plus 1 squared right and then this portion over here just becomes Omega squared because sine squared plus cosine squared by python Pythagorean identity is just one so then for the magnitude we get the square root of Omega squared plus one is of course the 1 squared is just one and for the magnitude of a we have dot a magnitude is equal to Omega to the fourth cosine squared Omega t plus Omega to the fourth sine squared Omega t plus 0 squared so we get the square root of Omega to the fourth once again using Pythagorean identity and then you get the square root of Omega to the fourth which is equal to Omega squared so for the last part we have to show that a is normal to the curve R so this is using the result that I mentioned earlier which is an infinitesimal increment in the length of the curve R so basically the arc length is in the same direction as the velocity Vector so in order to show that a is normal to the curve R we need to show that a DOT V basically is equal to zero because based on the trigonometric form of the scalar product that we looked at earlier and we saw that it was the magnitude of U magnitude of V cosine of theta so cosine of theta is zero when Theta is 90 degrees so if the dot product is zero then we can show that they're perpendicular to each other and just looking at the components which we derived you would get negative Omega sine Omega T negative Omega squared cosine and Omega T so that would be a positive and you would have the same magnitude of the opposite direction for this component because it would be negative so this All wipes out to zero so it is verified so here's a short multiple choice question quiz so use these to test your knowledge of vector functions and the calculus that relates them to each other based on what we just studied okay I'm hoping you're back and you knew how to solve the problems so for question number one we are asked to find the X and Y components of the instantaneous velocity so we have the position function so we need to differentiate each component to find the velocity Vector so this one is a little easier because it's just polynomial so the derivative of eight t squared is 16 T derivative of 4 negative four T is negative 4 times I and then for the J component the derivative of two t cubed that'd be six t squared minus derivative of three t squared would be negative 6t because we have the negative sign there plus six derivative of 6t is just six times J so then we have to plug in the v of two since we're being asked for numerical value and get 32 minus 4 I hat plus 24 minus 12 plus 6 j-hot which simplifies down to 28 I hat plus 18 J hot meters per second so the correct answer should be B so for question number two in another scenario the acceleration of a vector of a particle is given as the equation on the screen what is the magnitude of the particle's Velocity so for this one we don't actually have enough information to choose because they haven't mentioned what the initial velocity of the particle was so when we carry out that integral we only know what the change in velocity was over the time period but we don't know what the particle's actual velocity is because of that plus C so we don't know what the initial velocity was so there's not enough information to choose an answer so when acceleration is uniform so when it's a constant value there are some equations that we can use which have derivations on the screen so we know that just from the definition of velocity and acceleration and the relationship between them that V of T is equal to the integral of a DT which would be equal to a t plus a v naught V naught being the constant of integration here which is the initial velocity using that equation then we know that the position as a function of time is the integral of velocity so that would be the integral of V naught plus a t DT and then once again we'd get V naught t plus one half a t squared just from the integrational polynomials and then the constant of an integration would be X naught which in this case is the initial position um another really important constant acceleration equation that we want to use is the one that relates the velocity acceleration and displacement so the derivation of it is right over here so we know that the displacement is equal to when acceleration is constant is equal to the average velocity Times by time so we know that the final velocity plus the initial velocity over 2 is that average velocity Times by the time which in this case would be the final velocity so basically the change in velocity over the acceleration you can check and make sure that the dimensions work out for yourself but you should get meters per second over meters per second squared so you get seconds so everything checks out there so you get v f squared minus phenol squared over 2A and then just rearranging the equation multiplying 2A to both sides you get VF squared minus V naught squared is equal to 2A Delta X so that's another one of the important equations that we're going to use and it's really important that you understand especially in AP Physics C because there will be situations where the acceleration is variable and time dependent these equations only apply when a is constant which includes both magnitude and direction okay so let's look at example number two so a runner is moving in the positive X Direction and passes the origin at time T is equal to zero between T is equal to zero and T is equal to two seconds the runner has a constant speed of 8 meters per second um and at time T is equal to two it'd be good they begin accelerating at a constant rate of 4 meters per second squared and the negative X Direction and we want to find the runner's location at T is equal to five seconds so it's for a situation like this where there's a section where there is an acceleration and is not an acceleration or the magnitude of it is changing between two time intervals but within those time intervals it's still constant let's break it up into two intervals and then we can just add the displacements together so for the first interval it's super easy to find we know that Delta X in that case there's no acceleration so just be V naught um into T which in this case is just eight times two which is equal to 16 meters in the second interval Delta X2 so then we do have an acceleration so we have the same initial velocity that would be eight times three with eight being the initial velocity times T minus one half minus because the acceleration is in the opposite direction um towards a negative x-axis minus one-half the acceleration was four meters per second squared times by t squared which would be 3 squared in this case so that'd be 24 minus one half times 36 which would be equal to 6 meters so then adding them together the Delta X total would be 16 plus 6 is equal to twenty two meters so in addition to One D kinematics a big part of kinematics is 2D kinematics and the most um common form of 2D kinematics and AP Physics C is projectiles so one thing that you should always remember whenever you have any multi-dimensional kinematics problem is that the X Y and if applicable Z components of motion are independent of each other so motion one dimension doesn't impact Motion in another dimension so looking at the diagram on the screen we can see the path of an ideal projectile so it's launched with initial velocity of vector v with a certain magnitude and Direction so the direction is indicated by the angle Theta so in order to find the equations of motion for each of the dimensions we want to find the initial velocity and then identify if there was an acceleration in that direction so starting in the X Direction it's really easy the velocity initial in the X direction is just v0 cosine of theta just using the trigonometric relationships that we have we know that we have Theta and we have V so this v x um the sine the cosine of theta would be equal to of VX over the initial velocity so um v x is equal to the initial velocity Times by the cosine of theta so also in the X direction we obviously don't have any acceleration we're neglecting air resistance in this case so there's no acceleration so it's very simple just the initial velocity Times by time the y direction similarly we can find the initial velocity of the V 0 sine Theta because the sine of theta would be v y over V naught so um v y would be equal to V naught sine Theta and then we multiply that by T for the first part of our kinematics equation but we do have an acceleration in this case so minus one half a t squared minus because the acceleration which is the gravitational acceleration is down whereas the initial velocity is upward so we're establishing upward is positive and downward as negative so we have minus one half GT squared where G is the gravitational acceleration and free fall on Earth which is equal to 9.8 meters per second squared so just some conceptual questions to evaluate what are the X and Y components of the Velocity at 0.8 so at Point a we would consider that it's still moving with horizontal velocity which would be V naught X which is equal to V 0 cosine of theta but at a because of what in the y direction is changing directions v y is equal to zero so it only has an X component a velocity at that point and how much time does it take for the projectile to reach point B versus point a you could verify this mathematically but for an ideal projectile the path traveled for the first half and the second half is completely symmetric so the magnitude of the Velocity at the initial point and the magnitude of velocity at the final point is the same the speed is the same and the path is the same so the time it takes to get up here is the same amount of time as it takes to go down here for an ideal projectile so the amount of time it takes to get to point B is twice not it takes to get to point a so here are some useful ideal projectile results so first of all just remember that you can only use these um results when the projectile is not being launched from an initial height if it is being launched from an initial height then you just have to go through and solve the quadratic equations manually but if it's an ideal projectile result we can use these formulas to get information such as the range of the projectile or the maximum horizontal distance that it travels and the maximum height really quickly so for the range formula what we want to do is find uh the time at which the projectile hits the ground so we set y of t equal to zero solving for that we get T is equal to 2 V naught sine Theta by G and then we're going to plug in that time value into the X component of position or in order to find the maximum horizontal distance it travels so X of 2 V naught sine Theta by G would be equal to V naught squared sine two Theta by G so that's the range of an ideal projectile for the maximal height of an ideal projectile what we want to do is find the time value at which the projectiles Y velocity turns to zero which indicates that it's changing direction which indicates that it's at its maximum height so we take the derivative of the Y component so v y which would be V naught sine Theta minus GT we set it equal to zero we get our time value which is V naught sine Theta by G and then we plug that in to the Y expression to find that maximum height what we end up getting is V naught sine squared theta by 2G so that's the maximum height so example number three is asking us to find the angle for which the range of an ideal projectile is equal to the maximum height so in order to do this we need to set the range equal to the maximal height so we get that V 0 squared sine squared theta by 2G which is the maximal height is equal to V naught squared sine of two Theta by G so V naught squared obviously cancels everywhere sine squared theta we're going to leave that how it is and we're going to expand sine of two Theta using double angle identity into 2 sine Theta cosine Theta divided by G so we're going to multiply both sides by 2G and in that case we're going to get that sine squared theta is equal to 4 sine Theta cosine Theta dividing both sides by the sine of theta we get that sine Theta is equal to 4 cosine of theta and bringing that back up here that we divide both sides by cosine then we get that the tangent of theta is equal to four at this point we need to use calculator and find that Theta is equal to the arctangent of four in which case we get the Theta is approximately 75.96 degrees so this is a very common type of question that you could be asked where it's very useful to have these ideal projectile results ready because otherwise it would take a long time to get the answer to this kind of a problem but using our results we got the answer super fast so there are a few more complex types of problems which I want to briefly discuss so the first is an incline plane problem so you're going to look at inclined plans a lot when you get into um forces and Dynamics but you do need to know a little bit about them during kinematics so this is an incline plane and it's frictionless and the coefficient of restitution is one so anything that's on it it's not going to lose any energy through collisions with the incline it's inclined at an angle of theta and we have ball or something that's being launched off of it with initial velocity V so with these kinds of incline plane problems they're very similar to General projectile problems there's just one modification that we need to make we're going to tilt our axes so we're going to consider that the plane itself so the parallel to the incline is the x-axis and then perpendicular to the incline is the y-axis so then from there we need to dissolve the gravitational acceleration so the component of gravitational acceleration which we would now consider in the y direction would be um G cosine of theta and the gravitational acceleration component in the X Direction which now we do have an acceleration in the X Direction because we tilted our axes would be G times the sine of theta so once again from here you can apply your kinematics equations in the same way that you would for a general projectile problem the only difference is now the X component of position and velocity isn't going to be as simple because you have an acceleration now this G sine Theta but other than that the procedures are very similar the only difference is now because this isn't an ideal projectile you have to do all the work solving the um quadratic equations on your own um you can't use the um Nifty range in maximal height formulas that I mentioned earlier some other facts that you should just be aware of um is that the time between each balance is the same because as the um ball for example bounces down the incline um even though the horizontal speed is increasing the vertical speed at each balance is remaining the same so the actual y portion y of T the path every single time vertically for each bounce is the same that's not the same for the X component though which is why the distance between each bounce horizontally across the plane increases so that's just something to be aware of the other very common type of problem that you'll have is a running problem so you will have um a projectile being launched usually it's an ideal projectile to some distance and you'll have a person who's trying to catch that projectile right as it hits the ground so for those kinds of problems what you want to do is find the range of the projectile find based on finding the range of the projectile figure out how far that is going to be away from the person right and then based on that you can find what speed the person should be traveling with in order to reach this point exactly when the ball hits the ground so that's another very common setup that you have for these kinds of problems which not only uses the knowledge of projectiles but then also you have to do additional um calculations and applications um for One D kinematics with the person running towards that point on a 1D kind of situation so that's pretty much it for today so thank you so much for watching this intro video for AP Physics C mechanics um so feel free to look in the comments section there will be a practice multiple choice quiz attach in order to test your understanding of the chapter and I would highly recommend re-watching this video if there were any parts of it that you felt first time were a little unclear um re-watching and slowing down can really help digest these physics Concepts more and the next video which will be on forces and Newton's Laws of Motion will be coming soon thank