Transcript for:
AP Physics 1: Kinematics Review

hello i'm brian brown coming to you from central new jersey i'm really glad that you're joining me on on the program tonight where we review ap physics 1 topics you've accepted the challenge to take the course in your high school and now we're just going to make sure you're as prepared as possible to do very well on the exam in this show we're going to talk about kinematics and what kinematics is all about kinematics answers three questions about the motion of an object where is the object that's its position how fast an object's position is changing that's its velocity and how fast is the object's velocity changing that's its acceleration we could keep going and ask about how fast an object's acceleration is changing but we don't really need to get into that because we stopped with the definition of acceleration in this course so what we're going to learn tonight is talk about the key topics of kinematics from a couple different lenses we'll be defining the terms in more detail we'll be looking at the equations and i'll let you know which ones are on the ap physics 1 equation sheet and which ones aren't we'll be looking at graphs and all the information you can get from different graphs and kinematics and really importantly when you can apply these concepts and when you can't apply them then we'll take a look at some practice questions both from a multiple choice and free response format so let's begin our review so the first thing is the position of the object right we're trying to describe where the cart is located so the first thing is the cart has some size to it we've kind of have to pin down a point on the cart that we concern or that we're going to concern ourselves with so it could be the front of the card it could be the back it could be the middle it doesn't matter as long as that point doesn't change throughout our problem but i can't just say well the cart's there we can only describe the position of the object after we define a frame of reference that we're going to use to make all of our measurements or observations for linear motion choose a single axis in which it's moving x for horizontal y for vertical but if it's moving down a ramp then use a tilted axis that would be appropriate for 2d motion we're definitely going to want to choose an x-y coordinate system to help define its position in any case it's important to note where the zero position or origin is and which direction is considered positive these are choices that you get to make there is no right or wrong answer but we again we need to define it so that it's clear and we can all agree that these are the proper measurements in our reference frame so if i draw a x-axis and define the right to be positive i would say that this position the position of the object here is at the positive 5 meter mark it's important to say positive 5 as opposed to negative 5 because if it was at an x equals negative 5 meter position it would be to the left of the origin in this case a term that goes along with position is the displacement the displacement which is given the symbol delta x is the change of an object's position whenever you see delta in front of any symbol it's always final minus initial so here you our expression for delta x is x minus x naught x naught is the object's initial position so i've marked two positions on the coordinate axis one is at two meters and one is at eight meters if a cart moves from two meters to eight meters we would say the displacement is a positive six meters because it's moved to the right if the cart moves from eight meters to two meters then the displacement is opposite it would be a negative six meters because now the carts move to the left notice the position could be positive in both cases but the displacement could be positive or negative depending on which way the object's moving now we can start looking at velocity it's about how the position is changing so it is related to the displacement in fact its technical definition is the rate that the position is changing so at this point in time the cart could be moving to the right or left based on the reference frame we've defined if it's moving to the right in other words the displacement is positive then the velocity is positive if the cart were moving to the left because its displacement was negative then its velocity would also be negative just as positions measured relative to another point the velocity can only be described clearly when the frame of reference is defined the next term would be acceleration how is the velocity changing now we have to start getting into some different diagrams these are called motion diagrams because i want to know how the velocity is changing so i can adjust address how the rate of that velocity is changing in this case i've labeled three positions of a of an object and it has three different velocities shown by the length of the velocity vectors so the object starting at v1 then going to v2 then going to v3 and it's moving faster in each case to the right so the acceleration since it's showing how the velocity is changing is to the right and it would have we'd have a positive acceleration let's take a second example the object is slowing down but it's moving to the left so it's initial velocity v1 is the largest velocity or at least the highest speed and then it's slowing down as it goes to v2 and v3 so which way is the velocity changing well it's changing to the right because you're taking off some of that velocity vector in the right direction so the object's moving to the left but still accelerating to the right a third case would be if the object were moving to the right but slowing down any guesses which way the acceleration would be you probably got this one right the acceleration would be to the left it's negative it's what we typically call deceleration but we see that case two is also deceleration so deceleration is a little tricky because it means an object slowing down but from a physics perspective we really care more about whether the acceleration is positive or negative based on the reference frame and we can see that it could be speeding up or slowing down with different signs of acceleration so the last case is an object moving to the left but speeding up this in this case the acceleration would be to the left because the velocity is changing more to the left as time goes on so that we we'd say that would be a negative acceleration in this reference frame again it has nothing to do with where the object is at this point the acceleration is based on how the velocity is changing again when people say accelerate they typically mean speeding up but don't confuse that with the way we use the term acceleration physics we just mean that the velocity is changing so typically that could be for an object whose velocity is speeding up or its velocity could be slowing down or even a change in direction because velocity is a vector as we've shown with our arrows we'll get more into the changing directions part in another unit there's also a couple different ways of measuring velocity and acceleration and that would be talking about average or instantaneous values an average value isn't really average like we think about in math class when we have we're looking to find our average of three test scores right an average in this case happens over an interval of time and these are important definitions that are not on the ap physics 1 equation sheet so you really want to know these definitions the average velocity is the displacement over the time interval so it's being measured over a definite time interval in the same way acceleration is the change in velocity over a time interval and those are average values you can't say that you know exactly how fast an object's moving at a certain point in time just by knowing the average you just know that in that region it was moving an average of 20 miles per hour the instantaneous quantity is how fast or what the acceleration is at a certain moment in an instant of time and that's a different quantity that we can't use these equations for so if we're looking for if we're looking at the motion over a period of time and using either the displacement or the change in velocity then we're really finding average quantities so we're going to look at how to find the instantaneous values and these are not equations that you need to know they're on the ap physics 1 equation sheet they're the first three equations on the sheet so the instantaneous velocity in the x direction is the initial x velocity in the x direction plus the acceleration in the x direction times the time it's important to note that we can change the direction we can make those all y components it really doesn't matter but you do have to stick with one coordinate system so that's how we're going to find the instantaneous velocity at any moment in time the second equation tells us how to find the position at any point in time it's the initial position where we start plus the initial velocity times the time plus one half the acceleration times the time squared be careful only the time is squared students ask me this a lot is it the acceleration n time squared and it's not only the time is being squared and the third equation is the equation that has no time in it but that's okay we can still find instantaneous quantities if we know different variables this one relates the final velocity to the initial velocity the acceleration and the displacement so v squared equals the initial velocity squared plus 2a delta x again they don't necessarily need to be used to solve for whatever's on the left side they relate three they relate four variables to each other so as long as you know three variables you can always use one of these expressions to find the fourth when can you use these equations these equations can only be used when the acceleration is constant if the acceleration is not constant you can't use these equations they just won't hold true but you're not going to be asked how to find the velocity at some point in time if the acceleration's not constant those are questions for another course these equations can be used if the acceleration is zero if you don't have any acceleration in other words when you're when you have a constant velocity but they just give one equation with all three of them which is that your your position is your velocity times your time okay so what do the graphs what graphs are important here the graphs that are important are usually graphed with time so the first one would be a position time graph when you graph a position time graph or position as a function in time you might get a straight line if you get a straight line that's great it means that you know how to find the average velocity based on the slope right because the slope is the rise which is your change in position which is displacement divided by the time interval and that's our definition of average velocity but it turns out that if your average velocity is the same everywhere because a line has only one slope to it you have a little bit more detail about the type of motion if you have a straight line on a position time graph you have a constant velocity motion we also look at velocity time graphs if you have a straight line on a velocity time graph a diagonal line the slope is showing how the velocity is changing so it's your change in velocity that's your rise in this case divided by your time interval which we've already defined to be our average velocity but again if you have a straight line on a velocity time graph it tells you a little bit more about the motion it tells you that there's a constant acceleration you're not it's not possible to have a diagonal line on a position time graph at the same time that you have a diagonal line on a velocity time graph to describe the same motion you can't have a constant velocity and a constant acceleration unless that acceleration was zero what else can we tell from these graphs well we can tell area but the only graph that we're going to use for the area is a velocity time graph so let's take a look at this example what do we mean by area well i shaded three different regions between the graphs and the x-axis which is the time axis and shaded those three different colors the area in each region represents the displacement during those time intervals so i would say delta x1 delta x2 delta x3 for those three different displacements notice that they're all positive displacements because they're all above the time axis it is possible that this graph goes that the velocity graph drops below the time axis in which case you would have a negative displacement and as you find the area to the time axis one thing that i find really helpful before i start to solve a problem is to make sure i can translate this graph into a clear sentence so let's see if we can do that for this graph this velocity is starting at rest because we know the velocity is zero at time equals zero and then the velocity is increasing in the positive direction which means it's got a positive acceleration at the end of this interval the velocity stays constant so now i've got a constant velocity region for region two the third region shows that the velocity is decreasing towards zero so i have a negative acceleration in this case i could say a deceleration until the object comes to stop in all points this this object is moving to the right it's moving forwards if my positive x direction is to the right okay so let's practice a couple different stitch a couple different problems so the first question that we're going to look at is is translating representations on the ap physics 1 exam you're going to be asked to make different types of representations of the same type of motion so maybe you'll have to make a motion diagram and translate that to a graph or translate that to a short description or even some numerical quantities so one thing that you won't get on the ap physics exam is a video but i've got a video of me releasing a tennis uh sorry a ping-pong ball onto a desk and watching it bounce so let's watch that video maybe i need to slow down that video a little bit for you so you can see the details so i'm releasing it and it's falling towards the table and i want to graph the motion until it hits the table or just before it hits the table the second time so there's the information that i want you to focus on so what do we need to do to be able to start to translate this information well if you said create a reference frame you're right i need to describe where the object is based on a reference frame so i made the reference frame at the point that i released the ping pong ball i called that my origin and i called the positive y direction upwards so now i'm going to try to make a motion diagram of that what hap and i think the easiest thing in a motion diagram is to try to look at the velocity first so as the object as ping-pong ball is dropping towards the table what's happening to its velocity well first the velocity starts at rest right starts at zero then it's gaining speed as soon as i release it its velocity is increasing in the downwards direction just before it hits the table its velocity is as large as it's going to be in the downwards direction now it hits the table now that happens very quickly so i can't really describe what happens as it hits the table but it does happen very fast even though it is a a time that we could measure maybe with a high speed camera that slows the motion down even more than we can see now in any case we'll just jump to right after it bounces from the table so that next velocity is upwards but is it a high or a low velocity well i would say it's moving pretty fast as soon as it returns from the table because it's going to lose speed on the way up so its fourth velocity would be upwards pretty large and then that velocity is getting smaller and smaller until it gets to v6 which is the height that it returns just before it starts to fall back down towards the table so i think the easiest graph to translate that into is a velocity graph because i have all this velocity information right i'm going to start with the velocity at t equals 0 and i know that the velocity is zero and it's increasing but it's increasing in the negative direction now it turns out that that's a diagonal line you may recognize that that diagonal line is due to the acceleration of gravity we'll get to that in a moment then this is where it gets tricky i know that v4 is a positive large value so i've got to go all the way up into this positive zone it's not as large a positive number as v3 was a negative number but it starts in the positive and then it goes to zero so this point here is v6 everything after that point the velocity kind of repeats itself and this graph would continue going downwards and then we would just see if we were to continue bounces that it would just keep being these diagonal lines going downwards but that's enough of the graph for now that gives us plenty of information so now which graph is the next easiest for us to get i would say it's the acceleration graph because we're going to be able to get this information for the acceleration from the velocity graph going back to what happens between time three and time four when ping pong balls in contact with the table again i don't have a lot of information about that so i'm just going to put a filler line in orange here i know that it's got some really large upward acceleration right the table is pushing it upwards with a pretty significant force at that point so the acceleration is going to be very very large and even if we can't determine what it is exactly because i don't know the time frame i know that some time passes and i'm just going to connect those if you just want to put a dashed line to connect those two diagonal lines that would work as as well so now going to the acceleration graph going back to this diagonal line we know that the slope because it's the acceleration of gravity turns out to be negative 9.8 meters per second squared right that's the acceleration of gravity in this case downwards so we know we have a constant negative i call it 10 meters per second squared worth of acceleration in that first region okay how about on the way up let's skip the orange between t3 and t4 region for a moment and look at from v4 v5 v6 we know we have the same slope on in that section so again it's not a positive because it's moving up it's a negative acceleration because the slope of a velocity time graph is the acceleration so it's a negative slope of 10 meters per second squared now i can kind of fill in what happens in between and t3 and t4 as we mentioned that's a huge positive acceleration probably can't even represent it too accurately here because the acceleration would be a much higher positive value than i'm showing but again it's it's okay i'm just putting a filler region in between those two constant accelerations the main point is it doesn't have a constant acceleration from v0 to v6 it has a constant acceleration from v0 to v3 and then the ping pong ball is a constant acceleration of from v4 to v6 in between is when some interesting things happen but not things that we can necessarily graph so now the last graph that we have is our position time graph i started this graph and made it very long in the negative direction because i know that the position never becomes positive right again it's easy to tell where we start we define the starting point to be our zero position i didn't have to but lots of times it's easiest to do that okay but now i have an acceleration so i know that this is not going to be a straight line going down on my position graph right it's going to be a curve and that curve shows that i'm picking up speed in the negative direction i'm hitting the table so here i can't even clearly represent the orange region very well but i'm just drawing a a parabola going down to its the point where it hits the table and then obviously it hits the table starts to return towards the zero position and then it starts to fall you'll notice that at the very top here my direction is changing and that's where the velocity is zero at that at that instant you'll also notice this is important that the acceleration doesn't change at that moment right when i'm at that high point at v6 gravity the force of gravity is still acting on the ball and so it still has this acceleration of negative 10 meters per second squared lots of information from that graph okay let's look at a multiple choice type question so this question talks about a cart being released from a ramp sometimes images are drawn for you on the exam sometimes they're not and when they're not i think it's a great idea to make a quick sketch you don't need to be an artist to do this you just need to identify what's going on in the situation and see if you can kind of figure out a way of picturing it so i've got this ramp i know i start at the the object starts at the top of the ramp at zero and it ends at the bottom of the ramp with a final velocity we know that a object going down a ramp has a constant acceleration well it tells us that in this problem and so the question is i don't have to find the velocity exactly at the midpoint but i do need to figure out if it's less than half the velocity more than half the velocity or equal to the final velocity at the end of the ramp i think we can throw away option d right it doesn't all of a sudden go at its highest velocity and then continue at that velocity down the ramp we know it's a constant acceleration which means its velocity is constantly changing so option d is not a good answer but a b and c seem pretty reasonable so i think i want to make a different sketch and that other sketch is a quick velocity time graph because i know this is a constant acceleration and so the question is what is the velocity at this half at this midpoint and where is the object at the midpoint i know that the shaded regions represent the displacements so you know what how the position has changed in this half period of time so in the first shaded interval i can see that that's much smaller than the second shaded interval which means i definitely have not traveled halfway down the ramp at this point right i've traveled much less than halfway down the ramp in that tf over two point but i do know that that's where the velocity is vf over 2. so i think i've got my answer now i know that i have to be moving faster at the midpoint than vf over 2 because i hit vf over 2 way back here so that would be one way of answering this question without doing a lot of math and without really getting into any of the equations too much because i don't have enough numbers to plug in there and really work through i'm just trying to solve this one conceptually the other interesting thing to point out here is that if i look at the displacements in a little bit more detail i know that not only do i travel farther in this second half but if i draw that rectangle and then cut that rectangle in half i know that each of those three triangles these rough approximations of triangles each of those three triangles is the same area as the red shaded triangle which means in the second half of the time i've traveled three times farther than in the first half of the time right so i guess that i've only gone down one quarter of the ramp in tf over two there's a lot of information you can get with some quick sketches okay the other type of kinematics problem that we deal with we've so far looked at one-dimensional problems are two-dimensional problems and a lot of people think that these problems are significantly harder than one-dimensional problems well they could be but you have to remember you just need to separate these problems into what kind of motion the object has in each dimension so let's take a look at this free response question i've got two different objects and they're launched with different speeds from different heights but they both get to the same place when they hit the ground okay so the question's asking me to relate the second the higher initial velocity or the the velocity of the higher object to the velocity of the lower object initially again the important ideas are separate out the x and y directions because those types of motion are independent from each other what kind of motion does it have in the y direction well we know that there is the force of gravity acting on it and only the force of gravity and in the x direction there are no forces we're ignoring air resistance and we can do that in projectile motion problems unless the problem says consider resistive forces so again we know x veloci x direction has a constant velocity and y direction has a constant acceleration we also need to say what's the same and what's different about these two situations right and as we mentioned the value of d how far it goes is the same for both objects the initial y velocity is zero for both objects these are launched horizontally the y displacement's different therefore the time it spends in the air is different in these two situations so in order for us to really get to the heart of the matter here we need to figure out how much time each of these objects takes to fall so we're going to look at the vertical motion to try to figure out how much time each of these objects falls for i can change my reference frame a little bit between the two motions because i only care about the time so if i'm looking at my the first object which falls from a height h i'm going to call that initial position zero and just see where is this object at when does this object hit the ground as it falls a distance h so i called the downwards direction positive in this case so i know that h equals really just this last term because its initial y velocity is zero it's not v one because that's only in the horizontal direction so h just equals one half a t squared or in this case g t squared so t one is the square root of two h over g i'm going to apply that same equation to the second object which has a height of 4h it falls 4h with the same kind of acceleration again no initial velocity so when i solve for that it's that time it's 2 times the square root of 2h over g in other words it's twice as much as t 1. now i now now that i know that the second object falls for twice the time i can look at the horizontal motion and decide what the velocity is again it's a constant velocity problem the acceleration is zero so i can use this equation in the x direction but it simplifies out to x equals v times t so the range is the same in both times d is the same for both of these expressions so i can ex i can equate v2 t2 to v1 t1 since i know that t2 is twice v1 i can just substitute that in and solve for v2 as v1 over 2. right the t1s cancel out so i know i think this makes sense that the second velocity the velocity of the higher object can be less than the velocity of the first object but i know how much less in this case it's v1 over two so a lot of the problems that you will face have to do with this kind of functional dependence where you're just substituting in values because you may not have numbers to solve for specific quantities but you can always compare one situation to the next by just substituting the variables into the equations and then relating those equations to each other okay another kind of representation problem this one is a little bit more complicated talks about uh it gives you the the numerical situation first so when i read this uh situation i see that okay this rocket accelerates upwards at five meters per second squared for 10 seconds and then the engine shuts off and falls freely when it says falls freely that's a big hint that the acceleration is going to be the acceleration of gravity it's free fall towards the earth's surface so be careful when it asks you for graphs you really need to know what the graphs are of right this first graph is going to be an acceleration time graph which given this information i think this is the easiest one to start with as well right initially it has an acceleration of 5 meters per second squared for 10 seconds but that's a constant acceleration so think about what your scale is right you need to think all the way through this one we already talked about the 5 meters per second squared upwards but i know that this rocket is going to accelerate at negative 10 meters squared as it free falls so i've got to think about my scale and in this case in this case each box is worth 1 meter per second squared on the vertical axis so i know i have a constant acceleration of 5 for the first 10 seconds in between or right at 10 seconds again it's one of those situations i don't really have a lot of information for it but if we assume that the engine shuts off nearly instantaneously very short period of time and then the acceleration just becomes negative 10. it doesn't gradually change its acceleration from 5 to negative 10. so that would be the acceleration graph now it's asking to graph the velocity as a function of time for all 20 seconds again we could go one of two ways we could look at this graph and say well the area of an acceleration time graph is the change in velocity that makes sense because the area of velocity time graph is the position or displacement so we're just taking the same kind of step to go backwards from acceleration to velocity but i think in this case i know that the object starts at rest so i know it starts at 0 and i'm going to use the fact that it just accelerates at a constant 5 to know that i'm going to end at 10 seconds going at 50 meters per second and since it's a constant acceleration it's going to be a diagonal line again i think i need to figure out that number first so i can set the scale right so don't start graphing until you figure out really what you want your scale to be or else you're like you're going to have to make a lot of erasers and redo the graph anyway so at 10 seconds it ends up going 50 meters per second and now it starts to fall down do i jump down to a negative velocity no after the engine shut off it's already going 50 meters per second it's going to continue at that moment that's what its velocity is and the acceleration of gravity will just act to reduce that velocity and eventually right that's our diagonal line the slope is going to be negative 10 meters per second squared eventually the velocity hits zero but then it goes into negative and i just need to draw this until 20 seconds because that's all the problem asks for from t to 20 seconds so the next part of the question will get into a quantitative analysis and this one is actually an argument type question because you need to justify whether something is true or not true based on what they're describing in this case they're saying that a student has come up with an equation for the position at any point in time from 0 to 20 seconds or actually until it lands back on the earth so why is this equation incorrect well that equation assumes that the initial position is zero that's fine the initial velocity is zero that's fine and the acceleration is a constant five meters per second squared for the entire interval well that's not fine that's only true for the first 10 seconds that ignores the fact that its engine shuts off and the motion continues to happen until the object hits the ground so this one expression can't be used because remember these equations that we have are only true for a constant acceleration problem and here we have multiple regions of a constant acceleration so i can't lump that into one equation another type of argumentation question is to describe what's going on in in the graph and how we would find that so let's let's take a look at this question this one's asking to turn to determine the time that it reaches its maximum height so where in this graph are you reaching your maximum height a lot of people first guess that the peak is where you reach the maximum height because that's the highest point but this is not a position graph this is a velocity time graph and you're continuing to move in the positive direction even after the engine shuts off so the rocket's continuing to move upwards you have to keep going until the velocity is zero when the velocity is zero and it went from positive velocity just before that to negative velocity that means that's where the object changed directions so it's very high point is when that velocity is zero so that happens at fifth at the 15 second mark well okay i've got the first part of the answer i know when it's at its maximum height but now the second part is is asking how can you figure out what that maximum height is again i wouldn't you can't use a single equation to get that because the acceleration is not constant from 0 to 15. what can we use what physical feature of the graph supports our claim well that feature will actually be the easy way for us to figure out what this height is the feature of a velocity time graph that tells us what the displacement is is the area so all i need to do is figure out this area and this is much easier than using a kinematics equation the area is a triangle and the area of a triangle is a half base times height so a half of 15 times 50 is 375 meters therefore that's the maximum height that the rocket achieves so those are a couple of problems that hopefully help you help you validate that you know what's going on and and review here's some of the key points that i think you need to take away from this first is sketch a motion diagram sketching that is the easiest thing to do to help visualize what's happening to the velocity and then you can decide what the acceleration vectors look like at different times again you don't need to be an artist just draw dots and arrows and as long as your arrows represent the chain the length the relative lengths between the vectors you've got enough information there the second thing i think helps is sketching a graph some of these problems that i solved i solved by using information from the graph so the velocity time graph i find to be the easiest one to draw the nice thing about the velocity time graph is that will be able that will allow you to figure out both the position and the acceleration from the information on that with either the slope or the area another key thing to try to determine early on in a problem is whether the situation is a constant velocity right if it's constant velocity that means that the acceleration is zero so again there's really only one expression to use for constant velocity that's our displacement equals velocity times time if the velocity is changing it's an acceleration if it's changing out a steady rate then the object has a constant acceleration be careful using the word constant by itself you can't just say the motion is constant that's not a clear enough description because again either you can have a constant velocity type problem or perhaps you have a constant acceleration type problem and if the acceleration isn't zero those are two very different types of motion so be careful using the word constant it's a great term but always use another word after it that helps describe the motion constant velocity or constant acceleration separate the motion into horizontal and vertical directions right they're independent of each other this is uh important for projectile motion problems so i think these are all the the main things to take away the other thing to take away is know your definitions some of the equations are on the ap physics 1 equation sheet and some of them aren't and you need to know the definitions of terms because sometimes the hardest problems take you all the way back to knowing those definitions if you practice these keep watching videos there's lots of great videos out there and i know you've got lots of practice problems i'm sure you're going to do great i'm looking forward to seeing you in a couple follow-up videos and preparing helping you prepare for the exam and you're almost there you're going to do great thank you