Good morning! This is my review of Unit 1: Kinematics, for AP Physics 1. ♪ Flipping Physics ♪ But actually, before we even begin discussing the unit 1 topics, let’s review 2 things. First, significant figures, or sig figs. Actually, can we not talk about sig figs? Yeah. What? Sure. That’s actually my point. The AP Physics 1 exam pretty much ignores sig figs, so I suggest you ignore them as well. As long as you use roughly 3 sig figs, you should be fine. I like it. Yeah. Oh man! Second, conversions. Understanding how conversions work does actually help with understanding physics. So, let’s do one to make sure you remember how. Bo, please convert 8,765 kilograms per cubic meter to grams per cubic centimeter. Sure. Let’s start by multiplying by 1000 grams over 1 kilogram because 1000 grams equals 1 kilogram, therefore 1000 grams over 1 kilogram equals 1 and you can multiply anything by 1. Kilograms cancel out. And now we can multiply by 1 meter over 100 centimeters because 1 meter equals 100 centimeters. But we have cubic meters in the denominator, so we need to cube the conversion factor. Then we cube everything inside the parenthesis and cubic meters cancel out. And that works out to be 8.77 grams per cubic centimeter. Well done Bo, thanks. Also, if you could take a moment to like this video and subscribe to my channel, that would be lovely. And now let’s begin the unit 1 review in earnest with vectors and scalars. Class, vectors have both …? Magnitude and direction. And scalars have …? Magnitude but no direction. Billy, please give me some examples of vectors in physics. Some examples of vectors are displacement, velocity, acceleration, force, momentum, torque, angular momentum, and those are all vectors. And Bobby, please give me some examples of scalars in physics. Examples of scalars are time, distance, mass, speed, volume, density, work, energy, rotational inertia, and that’s all I can think of. Thank you both. Now, there are three basic ways you will see variables identified as vectors; an arrow over a variable, a subscript on the variable, or the variable in boldface. The boldface one you probably will not see on the AP Physics 1 exam, however, it is plausible. Also, we often illustrate vectors in diagrams and illustrations using arrows. The longer the arrow, the larger the magnitude of the vector. For example, if an initial velocity vector with a magnitude of 15 meters per second is illustrated like this, then a final velocity vector with a magnitude of 30 meters per second needs to be drawn with a length which is roughly twice as long because 30 meters per second is 2 times 15 meters per second. Alright, next let’s talk about displacement. Billy, tell me everything about displacement. Absolutely! The displacement of an object is the straight-line distance between the initial location of the object and the final location of the object. In other words, displacement is the change in position of an object. A symbol for displacement is delta x, where x means position and delta means change in. Change in position equals position final minus position initial. If the initial and final positions are in the same location, then the displacement of the object is zero, no matter what path the object took to come back to where it started. That shows that the distance an object travels will always be greater than or equal to the magnitude of the displacement of the object. And magnitude just means the amount of something. For example, if someone travels 5 meters North, the magnitude of their displacement is 5 meters. And displacement is a vector. It has both magnitude and direction. Wonderful Billy, thanks. Bobby, tell me about speed and velocity. Okay. Average speed is the distance traveled over the time duration of that travel. That means the equation is speed equals distance over time. But that time in the denominator is actually the time duration over which the distance was traveled, not a specific time. And speed is a scalar. Average velocity equals displacement over change in time. Typical units for velocity and speed are meters per second, kilometers per hour, or miles per hour. And velocity is a vector. Thank you Bobby. I will also point out that, if the time interval over which the average velocity is taken is very small, then the velocity is considered to be an instantaneous velocity. Where an instantaneous velocity is the velocity at a specific time, rather than the average velocity which is the velocity over a time period. Okay, Bo, please tell us about acceleration. Sure. Average acceleration equals change in velocity over change in time. Acceleration is a vector. Typical units for acceleration are meters per second squared. Actually, that’s all we really ever see for the units for acceleration. And I guess I’ll point out that, if the time interval over which the average acceleration is taken is very small, then the acceleration is considered to be an instantaneous acceleration. Where an instantaneous acceleration is the acceleration at a specific time, rather than the average acceleration which is the acceleration over a time period. Nice. Thanks. Yeah. Next, when the acceleration of an object does not change, when the acceleration remains constant or uniform, we can use these uniformly accelerated motion equations or UAM equations. These three UAM equations are the ones provided on the equation sheet, however, there is one more UAM equation I think it is helpful to remember. That equation is displacement equals one half the quantity velocity final in the x-direction plus velocity initial in the x-direction all times time. I will point out that these UAM equations are also sometimes called the kinematics equations. Billy, what are the 5 UAM variables? The 5 UAM variables are acceleration, velocity final, velocity initial, displacement, and change in time. I do not see velocity final, displacement, or change in time in any of those UAM equations. Mr.p usually uses change in time instead of just time in the UAM equations. These equations assume time initial is zero. Because change in time equals time final minus time initial, with time initial equal to zero, change in time equals just time final and the College Board drops the final subscript from the variable. So, change in time equals t for time. Oh, yeah, that means velocity with a subscript of x means velocity final in the x-direction. That’s where velocity final is. And position final, or x, minus position initial, or x with a subscript of zero, equals change in position or displacement. Right. And the variables are positive or negative depending on direction. Like a final velocity which is down is negative. Unless you’ve specifically identified down as positive. Correct. So, class, there are how many UAM variables? Five. How many UAM equations? Four. And, if you know how many of the UAM variables? Three. You can figure out the other … Two. Which leaves you with one … Happy physics student! I love that! Yeah. Sure. It is kinda fun. It sure is! Yeah! Please realize both of the velocities in the UAM equations are instantaneous. Velocity final and velocity initial are both at a specific point in time, hence instantaneous velocities. Next, let’s talk about what happens when an object, like this ball, is close to the surface of planet Earth and air resistance can be ignored. Bobby, what do we know about this object? Uh we know the only force acting on the object is the force of gravity and that means the acceleration of the object is 9.81 meters per second squared in the downward direction. This motion is often called Free Fall. Sure. And, we can actually just use 10 meters per second squared for the Earth’s free fall acceleration on the AP Physics exams. An object in Free Fall has a constant acceleration, so we can use the uniformly accelerated motion equations. Also, the velocity of the object at the top of the path in the y-direction is zero. And remember, when taking the square root, to be smarter than your calculator and consider if the answer your calculator gives you is positive or negative. That often comes up when we use the UAM equation with the squares of the final and initial velocities in it. Right. Perfect. Thanks everybody. Next, let’s talk about motion graphs. Class, what is the slope of a position versus time graph? Velocity. Uh how did delta x go from the first denominator to the second numerator? Uh … Oh! The delta x in the first denominator refers to the change in whatever is on the x axis and the delta x in the second numerator refers to the change in position of the object. Yeah. We replace the y in the first numerator with position or “x”, because that is what is on the y-axis. And we replace the x in the first denominator with time or “t”, because that is what is on the x-axis. Thanks, that makes sense. You are welcome. Class, what is the slope of a velocity versus time graph. Acceleration. To clarify, slope equals change in what’s on the vertical axis over change in what’s on the horizontal axis. For position versus time, the slope is change in position over change in time, which means the slope is velocity. For velocity versus time, the slope is change in velocity over change in time, which means the slope is acceleration. Class, what is the area between the curve and the time axis on a velocity versus time graph? Change in position. And the area between the curve and the time axis on an acceleration versus time graph is …? Change in velocity. To clarify, the area between the curve and the horizontal axis of a graph equals what is on the vertical axis times what is on the horizontal axis. Because the equation for velocity can be rearranged to be change in position equals velocity times change in time, the area between the curve and the time axis on a velocity versus time graph is change in position. And because the equation for acceleration can be rearranged to be change in velocity equals acceleration times change in time, the area between the curve and the time axis on an acceleration versus time graph is change in velocity. And remember the area above the horizontal axis is positive and the area below the horizontal axis is negative, right mr.p? Absolutely Billy. And I just want to make sure everyone realizes these three graphs cannot all describe the motion of one object. To illustrate that point, let’s remove what is on the position and acceleration versus time graphs and use the velocity versus time graph to build the accompanying position and acceleration versus time graphs for the motion of an object. Bo, tell me what you see. Sure. Well, we know the slope of a velocity versus time graph is acceleration. The slope of this velocity versus time graph is a constant, negative value. That means the acceleration graph should have a constant, negative value. That is a horizontal line which is below the horizontal time axis. (That is correct Bo. What about the position versus time graph?) Well, we know the slope of a position versus time graph is velocity. The initial velocity of the object is positive, that means the initial slope of the position versus time graph should be positive. But, where do we start the position versus time graph? I mean, is that information even on the velocity versus time graph? That is a great question Bo. And, no, a velocity versus time graph on its own does not give you any information about the initial position of the object. For our purposes today, let’s just assume the initial position of the object is zero. But, again, realize we did not get that from the velocity versus time graph, we just decided the initial position is zero. Bo, please keep going. Okay. We decided the initial position is zero, so let’s start the position versus time graph there. And we know the initial slope is positive because the initial value for velocity on the velocity versus time graph is positive. As time increases on the velocity versus time graph the velocity of the object decreases. That means the slope on the position versus time graph decreases. Halfway through the motion the velocity on the velocity versus time graph is zero, that means the slope of the position versus time curve is zero at that point. After that the velocity on the velocity versus time graph is negative and getting more and more negative. That means the slope of the position versus time curve decreases in value. Hold up, we know the area under a velocity versus time graph is change in position, right? Yeah. And? And the area above the horizontal axis is positive and the area below the horizontal axis is negative, right? Yeah. And? The positive and negative triangular areas between the velocity curve and the time axis are equal in magnitude. That means they add up to zero. That means the displacement of the object for the whole graph is zero. And that is why the final position on the position versus time graph is zero. The total displacement is zero and the initial position is zero. It goes back to where it started. Nice. And the area between the acceleration curve and the horizontal time axis is change in velocity. That whole area is below the horizontal time axis. That means the change in velocity for the object is negative. Which you can see on the velocity versus time graph. Cool. Nice. Well done everybody! Now, I just want to take a moment to identify what type of motion this is. Any thoughts? Uh … Yeah. Well, the acceleration is a constant, negative number. … Right. It is free fall motion. Oh, as long as that constant acceleration is about negative 10 meters per second squared, right? Actually, if it’s a different number, this could just be near the surface of a different planet. Sure. The initial and final points are the same, and the object goes up and then comes down again. And the velocity for the first half is positive and for the second half is negative. So, it’s probably an object thrown upward near the surface of a planet with negligible air resistance as long as that object was caught at the same height it was thrown. And it’s uniformly accelerated motion. Yep. Yeah. But it could just be something with a constant acceleration. Yeah. Well done. Yes, that is correct. {Phone rings.} I’m sorry. Give me a sec. I gotta take this. Good morning. What do you both think of the Ultimate Review Packet? I like it. Seems pretty useful. What’s the Ultimate Review Packet? Billy and What’s the Ultimate Review Packet? Yeah. What’s the Ultimate Review Packet? The Ultimate Review Packet is only the single greatest resource for studying AP Physics 1. It’s got review videos, study guides, more review videos, multiple-choice problems, helpful videos which go over common difficult topics in AP Physics 1, a practice AP Physics 1 exam, solutions to free response questions, actually, it’s got detailed solutions to everything in the Ultimate Review Packet, and more. It’s awesome! Back me up here Bobby! Sure. I’m just trying to figure out how Bo does not know what it is, considering he’s in it. I am? We all are. That does not surprise me. Is it free? No, it is not free. Mr.p’s got to make money somehow. Giving away so much of his content for free does not really pay the bills. Speaking of mr.p. Welcome back. Thanks. And sorry about that. Next up, two-dimensional motion. Often, in 2D motion, you will have to break, or resolve, vectors into their component vectors. Bobby, please break this vector A, which is at an angle theta with the vertical, into its x and y-direction components. Okay. Sine theta equals opposite over hypotenuse. The side opposite theta is vector A in the x-direction, and on the hypotenuse is vector A. That means vector A in the x-direction equals A sine theta. Cosine theta equals adjacent over hypotenuse. The side adjacent theta is vector A in the y-direction, and the hypotenuse is still vector A. That means vector A in the y-direction equals A cosine theta. I thought the x-direction used cosine and the y-direction used sine? That’s only true when the angle is with the horizontal. Yeah, you really should walk through the equation each time to make sure you don’t mess it up. I’ve messed it up a lot of times. I have too. Sure. Thank you. Now, projectile motion is where an object is moving in two-dimensions near the surface of a planet with the only force acting on the object being the force of gravity. Billy, what is a typical approach for solving projectile motion problems? When solving projectile motion problems, we typically separate our known values into the x and y-directions. In the x-direction the acceleration of the projectile is zero because there are no forces acting on the projectile in the x-direction. That means the velocity of the projectile in the x-direction is constant. And the equation for that is velocity in the x-direction equals displacement in the x-direction over change in time. There are 3 variables in that equation which means if we know 2 of the variables we can find the other 1. In the y-direction the projectile is in free fall because the only force acting on the projectile in the y-direction is the force of gravity. That means the acceleration of the projectile in the y-direction equals negative little g, or the negative of free fall acceleration, so negative 9.81 meters per second squared, however, on the AP Physics exams we can just use 10 meters per second squared to make all the calculations easier. In other words, in the y-direction, the acceleration of the projectile is constant and we can use the Uniformly Accelerated Motion equations. Those have 5 variables and 4 equations, which means, if we know 3 of the variables, we can figure out the other 2, which leaves me! Me too. And me. I’m full of mirth. The only variable which is the same in both directions is change in time because change in time is a scalar. So, often we are solving for change in time in one direction and using it in the other. Also, we often need to resolve the initial velocity into its components like you just had Bobby do. Unless the initial velocity is completely horizontal, then the initial velocity of the projectile in the y-direction is zero. Oh, and the velocity of a projectile at the very top of its path in the y-direction is zero! And if the projectile starts and ends at the same height, in other words the displacement in the y-direction equals zero, then the change in time while going up is the same and the change in time while going down. And the velocity in initial in the y-direction equals the negative of the velocity final in the y-direction, again where the initial and final points are at the same height. Well done y’all. Thanks. Last up we have relative motion. The basic idea here is that the description of the motion of an object changes depending on the frame of reference of the person observing the motion. What? Ehhhh? Uhhh? Yeah, I know that sounds complicated. It really is not. It’s just vector addition and, for the purposes of AP Physics 1, they have stated that relative motion is “restricted to motion along one dimension.” Likely, they will ask you something along the lines of “A stationary observer measures car A to be moving at 60 km/hr East and car B to be moving at 35 km/hr East. What would a person in car B measure the velocity of car A to be?” Bo, please solve this one. Sure. We know the velocity of car A relative to the Earth is 60 km/hr East and the velocity of car B relative to the Earth is 35 km/hr East. And we are solving for the velocity of car A relative to car B, because the frame of reference in the question is the observer who is in car B. Looking at it in terms of vector addition, the velocity of car A with respect to car B plus the velocity of car B with respect to the Earth equals the velocity of car A with respect to the Earth. You can see that in the vector addition diagram and we can remember that the common variable, car B, drops out and we are left with car A and the Earth or the velocity of car A with respect to the Earth. .. Solving for the velocity of car A with respect to car B, we get that that equals the velocity of car A with respect to the Earth minus the velocity of car B with respect to the Earth or 60 km/hr East minus 35 km/hr East or 25 km/hr East. If a passenger in car B looks out the window as car A passes it, car A will look like it is moving at 25 km/hr East or forward at 25 km/hr. What if the question asks, “What would a person in car A measure the velocity of car B to be?” That really is too many A’s and B’s. Yeah it is. The velocity of car B with respect to car A is just the negative of the velocity of car A with respect to car B. So, the velocity of car B with respect to car A equals negative 25 km/hr East, and negative East is West. So, the velocity of car B with respect to car A equals 25 km/hr West. A passenger looking out the window of car A as it passes car B will see car B moving at 25 km/hr West or backwards at 25 km/hr. Should it be 20 km/hr and not 25 km/hr because 60 only has 1 sig fig? Nope. Remember the AP Physics exams do not really concern themselves with sig figs. Oh yeah, right. Thanks. Absolutely. Well, that concludes my Unit 1 AP Physics 1 review. I do want to take a moment to let you know about my AP Physics 1 Ultimate Review Packet. We already did that. (We, We did?) While you were on the phone. Oh. Right. Well, thank you for that then. You are welcome! No problem. Wait a second, was that a real phone call? Thank you very much for learning with me today, I enjoy learning with you.