Key Concepts in Kinematics for Physics

Hello everyone, I'm your academic coach Yoo Hyun. I'm a UCLA grad, a professional tutor, and a digital creator. I post videos on YouTube and livestream on a regular basis to provide free educational content. If you find my videos helpful, please subscribe to my channel, like my videos, and leave me a review on Google. Let's get started. So today I'm going to talk about Unit 1 of AP Physics 1 Kinematics. kinematics is the most fundamental and then one of the most important part when it comes to mechanics um in kinematics we study how an object moves but we're not going to study why an object moves in such a way we're only going to focus on how it moves including in which direction it moves moving to moving upward moving down or moving to the left moving to right is it doing a linear motion or projectile motion is it speeding up or slowing down is it accelerating or decelerating is it changing our direction that's the wrong So I'm going to start with three of the most important concepts in kinematics, which is displacement, velocity, and acceleration. Remember that all of these three concepts are vector quantity rather than scalar quantity, which means they always come with a direction. Both magnitude and direction are important in the vector quantity. So displacement, how it is different from distance, is that distance takes all of the roots that you have. path that you have traveled while displacement only cares about the initial position and the final position. So displacement is literally just change in the position and a lot of times it's represented as delta x which is x final minus x initial. You can replace x with y if it's moving up and down in vertical motion and when it comes to distance you will have to take every single part of the path you took. Let's say you move between point A and B. So let's say you're moving between point A and point B and then they are 100 meters apart. Then when you go directly from A to B, the displacement is 100 meters and distance is 100 meters as well. But if you go from point A... to let's say point C and then going back to point B, then your displacement is still 100 meters while your distance is more than 100 meters, probably 150 or 200 meters. So distance is not always the same as displacement. Or let's say when you move between A and B, you went to B and then you're back to A. Then your distance is 200 meters, 100 meters. from A to B, another 100 meters from B to A, 200 meters in total. But your displacement is zero because your initial position and final position are the same. There is no change in the position. Again, displacement has nothing to do with the route you take. You can take whatever the route you take and then when you're back to A, your displacement is always zero while distance is not. And in physics, we always use displacement to calculate all the information you want, rather than the distance. Now I can try to compare velocity to speed, just like I compared the displacement to the distance. The velocity is the vector quantity while the speed is the scalar quantity. And how I calculate the velocity is it is the instantaneous rate of change in the displacement. So a lot of times when we calculate the speed, what we do is we take the total distance and then divide it using the total amount of time it takes. So we calculate the average speed. So that's not what we do in velocity in AP Physics 1. So we usually only talk about the instantaneous rate rather than the average rate. So how you're going to calculate the velocity is by taking the change in the distance. by taking the change in the displacement and divided by the amount of time it takes to travel the amount of the displacement. And later I will talk about how to calculate the velocity or how to find a velocity on the position versus time graph. And next is acceleration. So acceleration tells us how fast a velocity is changing. So velocity is not always constant. An object might speed up or slow down. Its velocity is constantly changing most of the times in real world. So acceleration tells us about how fast the velocity is changing. So for example, for a sports car versus a regular sedan. So sports car has a very high acceleration, means within one second its speed can increase by a large amount. So that's why it has a lot of advantage when you just start um in a race um on the other hand for the sedan even though it can still reach the high speed of let's say 100 meters 100 miles per hour but at the very beginning um it's accelerating slower um so for every second its speed is not going to increase that much so that's when we compared acceleration the rate of change in the velocity and then we always used on delta v, the change in velocity divided by time to represent acceleration. And again, acceleration is also a vector quantity. So in AP Physics 1, we always relate displacement, velocity, and acceleration all together. Okay, and now let's talk about how to analyze the graph of the velocity versus time and then how to analyze the graph of the displacement versus time. So when you see a graph like this, it means that it's displacement or the position is constantly changing. So it is a curved line, it does not mean that the object is moving along this curved line, it just means that its position is changing. um along this curved line um as the time changes so most likely this is a one dimensional motion or a linear motion so this object must be moving along a linear line it can be a horizontal line it can be a vertical line it can even be a scandia line and use on the linear line you usually set up the positive direction and then the negative direction and usually i set the right side as the I set the right as the positive and left as the negative. And its position is equal to zero, which means at the very beginning, it starts at the position of zero. And then after one second, it has moved to this position. After another second, it has moved to this one. After another second, it has moved to here. After another second, it has moved to there. Okay, so it represents the motion of this object along this linear line. And this is the vector diagram. The direction of the arrow represents the direction of the velocity, the direction of the motion. And then the length of the arrow represents how fast it is moving, the magnitude of the velocity. Okay, and based on the position versus time graph, how I draw the velocity versus time graph is by taking the slope of the tangent line. Okay. So the slope of the tangent line at a specific point, let's say at time t is equal to three seconds. So how I find the velocity is by drawing a tangent line to this position versus time graph at a specific point and then calculate its slope. So this slope represents the velocity and it's obvious the velocity when time t is equal to zero it's a zero because it has a horizontal tangent line and time t is equal to zero and then the tangent line gradually um is tilted up which means the velocity is gradually increasing so based on this position versus time graph i can come up with a velocity person and the velocity versus time graph on this velocity versus 10 graph Velocity must be increasing from zero. And on this velocity versus time graph, I can also find the acceleration by taking the slope of this velocity versus time graph line. So the slope of this linear line is acceleration. And since this is a linear line, its acceleration is always constant from the very beginning to the end, which means it's a constant acceleration motion. so in ap physics one you will rarely see um the scenario of a changing acceleration usually we assume that acceleration is constant to make sure that calculation is a little bit easier so on this velocity versus time graph The area of this triangle represents the change in the position or displacement. So similarly based on this velocity versus time graph, I can also draw an acceleration versus time graph. And since acceleration is constant, acceleration graph is just a horizontal line. and this horizontal lightness above x-axis because it is a positive slope and again the area under the acceleration graph represents the change in the velocity now i'm going to give you a couple of examples of the linear motion in real world um so one of the most common um linear motion would accelerate with a constant acceleration is So free flow means an object is moving in air under the influence of the gravitational force only. So it only experiences an acceleration that is pointing downward. So gravitational acceleration means it's a 9.8 meters per second squared downward acceleration. So let's say if I have if i hold an object and then release my hands it just drops and once once it drops its velocity increases downward so right at the moment when i release it its velocity is equal to zero and then because it's experiencing um gravitational acceleration so its velocity increases and then its velocity increase increases by nine point meters per second Sometimes we use negative 9.8 meters per second square to represent the gravitational acceleration. This is because I set upward, the vertical upward, as a positive direction, and then downward as negative direction. But I can always set the direction arbitrarily. So sometimes we can just set downward as positive direction to make the calculation a little bit easier. What happens if I toss an object upward? Okay, so if I toss an object upwards, then it has a non-zero upward initial velocity. So it has upward initial velocity, and then its velocity decreases because of the downward acceleration. And then at the maximum point, it's temporarily at stop. It's temporarily at rest, and then it starts to move downward. Again, due to the downward gravitational acceleration. So on its way going downward, it takes exactly the same amount of time as it goes upward. So you can notice that when it goes up and down, the velocity has only switched the direction. The magnitude of the velocity has not changed at all. So that's why the amount of time taken to go upward is the same as... the amount of time taken to go downward and remember that what acceleration is in the same direction as the velocity then this object is going faster and if the acceleration in the opposite direction of the velocity then this object is slowing down it goes slower and also remember that speeding up and slowing down um is not the same term it's not an alternative to accelerating or decelerating So speed up and slow down is referring to the speed only or the magnitude of the velocity. While accelerating and decelerating means the velocity is increasing or decreasing. Okay so for example, so an object might be speeding up while it is actually decelerating. So let's say an object is moving in the opposite direction or in the negative direction. And its speed is increasing. Okay, so obviously its speed is increasing and speeding up. But when it comes to acceleration, it's a negative acceleration. Its initial velocity is, let's say, 0 meters per second. And its final velocity is negative 10 meters per second. So its velocity, the number itself is decreasing, going from 0 to a negative value. So that's what I call by the negative acceleration. Or an object might be slowing down but accelerating. So this usually occurs when an object is moving in the negative direction. Okay so again it's moving in the negative direction and its speed is gradually decreasing. And then finally it comes to a stop. So let's say its speed has, um its velocity has changed from negative five meters per second to zero meters per second. then the velocity itself is increasing from a negative value to 0. So according to the formula of the acceleration, delta V divided by T, delta V is 0 minus negative 5, which is positive 5 meters per second. So the velocity is increasing, so acceleration is positive. So up until now, I was only dealing with a linear motion, which means the object is moving on a horizontal surface or moving up and down only. And now let's talk about the two dimensional motion. So when an object moves on a surface. So the best example for two dimensional motion in kinematics is the projectile motion. So projectile motion is actually one type of free fall, which means that object is moving in air under the influence of the gravitational acceleration only without any of the air resistance. But as for the linear motion, the initial velocity is either zero or its velocity is in the same direction or the opposite direction of the acceleration. While in the projectile motion, its initial velocity is not in the same direction as the vertical line or um the gravitational acceleration so its velocity changes so its velocity is changing the direction and it's also changing the um and it's also changing the magnitude so it's gradually pointing um toward horizontal and then it becomes horizontal at the maximum point and then it gradually points downward So this is a representation of the projectile motion and how its velocity changes at different positions. So velocity is the vector quantity and as for the vector we can always separate into the x component and then the y component. So for the initial velocity I'm going to rewrite it as the addition of the two vectors vx and then vy so two of the components and then along this projectile motion. only the gravitational acceleration is applied on this object. So from the very beginning to the end, it's always the gravitational acceleration that is pointing downward, which means only the vertical component of the velocity changes. So vertical component of the velocity is gradually decreasing. And then at the maximum point, the vertical component of the velocity is equal to zero. and then gradually the vertical component of the velocity becomes negative pointing downward. On the other hand, vx, the horizontal component of the velocity is not changing along the whole the motion and this is because there is no acceleration along the horizontal axis. So in the productile motion what we need to pay attention to is only the y component, so v y, only v y changes due to the gravitational acceleration which is alone on the vertical and v x never changes. So this explains why the path of the projectile motion is parabola because v x is not changing v y is always decreasing so that's why it is when it goes up the next moment it's slightly pointing um toward a horizontal and next it's more horizontal and next it's more flat and then now it's pointing down more down and more down okay so sometimes um it's um it might it might not look like a project emotion at first but actually they are so for example when you are trying when you are when you throw a ball um in a horizontal direction on a cliff so its initial velocity is pointing on in the horizontal direction the next moment due to the vertically downward gravitation acceleration its velocity points down and then the next moment it's pointing more downward next it's pointing more downward and next it's pointing more downward okay um so again it's a vx is not changing during the whole process because there is no acceleration along the horizontal axis but the vy is always changing so at the very beginning vy is equal to zero because the velocity is just in horizontal direction next moment vy is pointing downward um but the magnitude is small and then the next moment vy is pointing down and then the magnitude is larger next moment it's pointing down and then the magnitude got even larger and larger So now it's obvious that only VY decreases while VX is always constant. So no matter which one it is, this one or this one, they both represent a projectile motion. Okay, now let's try to use the equation. So there are only three equations that you need to know, and then they are all shown on the equation sheet. So three equations that we need to know are... v final is equal to v initial plus the acceleration times t. The second one is x final, the final position is equal to initial position plus v initial t plus one half a t squared. And then the last one is v final squared is equal to v initial squared plus two times a times the change in the position order displacement. Okay, so these three are in the equation sheet. And I'm going to add one more. So all of these, remember that all of these three equations are based on the condition that acceleration is constant during the whole process of the motion. And when I say acceleration is constant, it doesn't just mean the magnitude is constant. It also means the direction of the acceleration is constant. So, for example, if the acceleration magnitude is the same, but it's changing that. direction, then this is not constant acceleration. Constant acceleration means both the magnitude and direction are constant. Okay, so given this condition, I can add one more equation. So sometimes I can use the fourth equation to make the calculation faster. The last one is the change in the position or displacement is equal to one half of the initial velocity plus final velocity. So this part actually represents the average velocity. Average velocity. Average velocity multiplied by time is equal to the change in the displacement. And I can use the velocity versus time graph to represent it. So assuming that this is a constant acceleration, and assuming that this is a positive acceleration, so velocity increasing, then velocity versus time graph is a linear line. Because on the velocity time graph, the slope represent the acceleration constant acceleration means it's a linear line and then remember that the area under the graph represents the air the the change in the position or displacement and this point represents the initial velocity so when time is equal to zero in this point represent the final velocity okay so obviously this is a trapezoid and then how i find the area of a trapezoid is one half times a plus b uh multiplied by the height h okay one half times the upper base and then the bottom base and then multiply by the h so a is represented by this part b is represented by this part so now i can just plug in v initial and v final then I get 1 half times the initial plus the final multiplied by the height, which is time t. So this is how you get the very last equation, the fourth equation. Now, let's try to use this equation and then all of this graph to apply them to actual AP style questions. So in this question, initial velocity is given even though it's not explicitly stated. It says the ball is released from rest, which means its initial velocity is equal to zero. And then one second represents the time. And then the one floor has fallen one floor. Okay, so even though this is not stated in terms of the meters or the feet or inches, but because each of the floor is evenly spaced, so one floor actually represents the displacement, the amount of distance that has traveled or the change in the position. So I'm going to use delta y to represent it. then in the given combining all of these three centers okay and one more thing one more hidden information here is um this is a free fall because it says the ball is just released so as for the free fall acceleration is always equal to the gravitational acceleration which is 9.8 meters per second squared pointing downward so now i have acceleration initial velocity time and the change in the displacement then combining all of these four again initial velocity time um acceleration and then the change in the position or displacement okay so i know i'm not going to use the equation one because v final final velocity is not given so Again, v final is not given, so I cannot use equation 3. And then I cannot use the fourth equation because final velocity is not given. So the only equation that I can use is equation 2. So I'm just going to replace x with y. Then I get this formula of delta y, change in the position or displacement, is equal to v initial t times 1 half at squared. Or 1 4 is equal to initial velocity 0. zero plus one half times g times one second squared where one floor is equal to one half g okay and it says and it's asking us to find the number of the floors so delta y it's asking us to find the the displacement uh when the ball has fall like three seconds now i'm going to use exactly the same formula but now delta y is a different um displacement delta y prime is equal to v initial t prime plus one half a t squared so the new um change in a position or the new displacement is equal to zero plus one half g times three seconds squared so that is nine times one half g so it's nine times of um the amount of the displacement that has trouble in one second which is nine floors It has traveled nine floors. So the answer is C, between seven to ten floors. Question number two, a student wants to investigate the motion of a ball by conducting two different experiments. And the student released the ball from rest and used a slow motion camera to film the ball as it falls to the ground. Okay, so in experiment one, this is just a linear line motion of free fall. It just released from rest, so it goes through vertically down. And then in experiment two, the student horizontally rolls the same ball off a table and used a video analysis to plot the balls horizontal position x and vertical position y. Okay, so in experiment two, the ball starts at a horizontal velocity, so it's gonna follow this parabola curve. It's gonna follow the parabola and then it is doing the projectile motion. So this is initial velocity and this is how its velocity changes over time. The graph from each experiment are shown, two, which of the following conclusions can the student draw from the graphs, and y. Okay, so now when I compare the experiment one and two, they have exactly the same y versus t graph. So their vertical position is changing at the same rate. The only difference is the horizontal position. So for experiment one, its horizontal position is not changing at all because it's just moving up and down along the vertical line um in figure two in experiment two it it has it started in a horizontal direction so its um horizontal position is also changing but at a constant rate this is a linear line so its slope is represents the acceleration it's a linear line so it's slope represents the velocity and then the velocity is constant um a the horizontally rolled um ball you travels a greater horizontal distance and takes a longer amount of time to reach the ground. Okay, so the first half is correct. It travels a greater horizontal distance, right? But it does not take a longer amount to reach to the ground. So no matter which one you pick, it's going to take exactly the same amount of time because the amount of time taken to reach the ground is depending, it depends on its initial vertical velocity and then the vertical displacement. and then the vertical acceleration. And experiment one and two have exactly the same vertical components. They have the, they have traveled the same vertical displacement. Their initial vertical velocity are exactly the same. Both have initial vertical velocity of zero. And then both of them are experiencing the same vertical acceleration of 9.8 meters per second squared. So the second part is not correct. It's the same amount of time. So A is not true. B. the release ball travels with a slower speed and takes a longer amount of time um to reach the ground okay so again not longer amount of time it's the same amount of time and see um since both balls have their horizontal acceleration um they reach the ground at the same time and d since the balls have the same vertical position at any given time they reach the front at the same time so out of c and d d is the right one so c is actually true um they have the zero horizontal acceleration yes it's true that both of them um have zero horizontal acceleration so in experiment two the ball is doing a constant horizontal motion while in experiment one is just not moving in horizontal direction but it does not give us any information regarding the time taken to reach the crown. So we need to take a look at the vertical components. And when we take a look at the vertical graph, the vertical position versus time graph, at every single moment, they are at the same position, the same vertical position. So this explains why they will spend the same amount of time to drop to the bottom. Three, in experiment one, what is the speed of the ball the instant it makes contact with the ground? Okay, so in experiment one, so, okay. So how we calculate or how we figure out the velocity is by drawing a tangent line on the position versus time graph and then find the slope of that tangent line. So it's going to be this linear line. This linear line is tangent to the graph. and then its slope is okay and then its slope is a negative value which means it's moving downward and then in how i'm going to calculate the slope is by picking any of the two points on this on this slope so approximately um okay so i can pick these two points i can pick this point and this point so approximately um it's slope at when it has reached the ground is going to be delta V divided by is going to be 0 minus 1.6 divided by 0.55 minus 0.25. So negative 1.6 divided by 0.3. So that's up. approximately negative five meters per second. So as for the speed, speed only takes the actual value. So if the velocity is negative five meters per second, then its speed is approximately five meters per second. Question number four, suppose that experiment one and two are conducted at the same time. One student drops the ball from rest at the same instant that a second student horizontally rolls an identical off the table. After both balls have traveled half their vertical distance to the floor, what is the acceleration of the center of mass of the two-ball system relative to the Earth? Okay, so now let's talk about what it means by the center of the mass and how we calculate the center of mass of the two-ball system. So the center of mass is usually used to represent um the where the mass is on how the mass is distributed within an object so not all the object has a regular shape like a circle or uh or a square so most of the object in real world has irregular shape so let's say for like human beings um the center of the mass is uh is more toward the upper body is closer to the upper body because the skull is pretty heavy it occupies a great percentage a large percentage in the human body weights. So compared to the lower body, the upper body is heavier. The center of mass is a little bit above your belly button. And for irregular shaped object, the center of mass is more toward the part where more of the mass is concentrated. So for example, if you have a stick like this, if one side is thicker and the other side is skinnier, and given that the density is the same then the center of the mass is more um on the left side of this stick so this is for one single object but in ap physics one you barely get tested on finding the center of the mass within an object we usually assume that that's just the center of the circle or the or the center of um uh of this like rectangular object um but you will be tested on center of the mass of the multiple object or the two objects so let's say when you have two objects um so i'm going to use two rectangles to represent it and then one of the mass is one of the rectangle has the mass of 10 kilograms the other one has the mass of 20 kilograms and then they are let's say um one meters apart then how i calculate your center of mass is i'm going to use this formula of x center of mass is equal to x 1 times minus 1 plus x 2 times minus 2 divided by the sum of the mass the total is so of one and x to represent the position of the center of mass for each of the object so i'm gonna draw i'm gonna draw this reference frame um and then let's say this represents zero and then this represents one meters then the position of the 10 kilogram object is zero the position of the 20 kilogram object is one meter and now i just plug in um these numbers in the formula Then x1 the position of the mast 1 times 10 kilogram plus x2 1 meters the position of the mast 2 multiplied by its mast and then divided by the total mass of 10 kilogram plus 20 kilogram then it is 20 divided by 30 so 2 3 approximately 0.67 meters. It means on this reference frame um 0.67 meters so here so this represent so this 0.67 meters this is the position of the the center of mass of this um two object system and when you're asked to find the velocity of the center of mass you are going to use on the same method the velocity of the center of mass is equal to velocity of of each of the mass multiplied by its mass and then you are going to add up the product and then divide it by the total mass. So going back to that question, okay so in experiment one the object, the ball is moving straight down in object two the ball is moving along doing a projectile motion moving along the parabola and then they have exactly the same vertical position the only difference the horizontal position um okay then um it's acceleration of the center of the mass relative to the earth um okay so when it comes to acceleration acceleration is a vector quantity so we are we can calculate the horizontal acceleration and then calculate the vertical acceleration and then add them up together so when it comes to the vertical acceleration since both of them are moving down at the same velocity and then the same acceleration so for both of them a y the vertical acceleration is g um just on gravitational acceleration on the other hand for the horizontal acceleration okay so one is not moving horizontally the other one is moving at a constant velocity horizontally then the velocity of the center of mass in horizontal um component is equal to v x of um of one of the ball um multiplied by v x of another ball okay so um in experiment one the ball is not moving at all in horizontal direction so its horizontal velocity is always zero um so and then i'm going to divide it by the total mass of m plus m so eventually this is one half of the horizontal um velocity of of the ball in experiment two so it means this is a constant velocity and as for the constant velocity acceleration is equal to zero because acceleration is changing velocity over time so this is zero meters per second squared okay so in horizontal direction acceleration is zero in vertical direction acceleration is g so when i combine them together so this is this is a y And this is AX. And since AX is 0, so when you add them up together, the acceleration is still equal to AY only. Next, question number five. Students provide with a battery-powered toy that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim. So when it comes to measuring the velocity or the position or acceleration of a moving object, we use either the photogate or the motion detector. So I know for sure that Which of the following measuring tools can a student use to test the validity of the claim? Select two answers. Okay so I know this is not right, a mass balance. So measuring the mass has nothing to do with its position or velocity or acceleration. So photogate and motion detector are used pretty often when it comes to measuring the velocity or the position or acceleration of a moving object. but remember that no matter which one you use either the photogate or the motion detector you always have to put it along the track where this object is moving so the object is moving in the horizontal direction that you will have to put the photogate along this horizontal track either here or here or here and it's the same for the motion detector if you place it in the vertical and in in a vertical position compared to the it's moving track then it's not going to measure the position order velocity So oriented perpendicular is not correct for sure. So the answer is A and B. The photo gate placed at the beginning and then the end and then along the various locations along the track. And B, a meter stick to measure the distance of the track that the car travels on. And the next question, do students want to determine the speed at which a ball was released? when it's thrown vertically upward into the air. So the ball is released but it's thrown vertically upward so its initial velocity is pointing upward and then its velocity gradually decreases until it comes to a stop and then it starts to move down until it hits the ground. One student throws the ball into the air while one other student measures the total time that the ball is in the air. Okay the total amount of time starting from this to this. The student used the meters to measure the release height of the ball. So the release height of the ball. Okay, so given the release height, I can calculate the change in the position or the displacement delta y, which is the final position minus the initial position. So that's going to be negative of the release height, negative h. And then given the time, okay, so it's asking us to find the speed, its initial speed. And then this is the free fall. so of course we assume that acceleration is the gravitational acceleration pointing downward so we're given one two three three information and that you're required to find the last one um okay then obviously um c and d does not apply because the final velocity is not even given so i cannot use the field or d to figure out um the initial so not c or d so it must be either a or b and then it's from the moment in time in which the ball was released to the moment in time in which the ball reaches the highest point or until it hits ground. Okay, so the amount of time it measures is from initial all the way to it going back. It hits the ground, so I will have to choose B. Seven, an object undergoes an acceleration as it travels along a straight horizontal section of a track. Which of the following graphs could represent the motion of the object? Select two answers. So A, speed versus time graph. So usually in the velocity versus time graph, the slope of the line represents the acceleration. So this means it is a constant positive acceleration. So A is the right answer. And B, position versus time graph. On the position versus time graph, the slope represents the velocity. So it just means it's a constant positive velocity. So not B. And C. um position uh and time squared okay and d is velocity versus time so i know d is not the right answer for sure because this is constant velocity or when i take a look at the slope which is zero um slope of the velocity time graph represent acceleration so the right answer is a and c so for the position versus time squared graph if i convert it into the regular position versus time graph then it's going to be a curved line. So it means that it has a different tangent light. It has the tangent light of different slopes and different time, which means the velocity is constantly changing. Question number eight, a monitor sensor is used to create the graph of a student's horizontal velocity as a function of time as the student moves toward and away from the sensor. The positive direction is the final acceleration away from the sensor, which of the following describes the student's final position x final in relation to the starting position x zero and the student's average horizontal acceleration ax between 0.0 and 3.0 seconds. Okay, so first off, this is a velocity versus time graph. So the area under the graph represents the change in position or the displacement. So this is a starting position, and this is where the final position must be. So if I add up, if I calculate the area under the graph, then obviously the positive part is greater than the negative part. So delta x. is going to be greater than zero, which means X final minus X initial is greater than zero. So X final is greater than X initial, which means the final position is more away from the monitor sensor. So either A or B. And then when we calculate the acceleration, A acts between zero and three seconds. Okay, zero and three seconds. Mm, so. Okay, so when it comes to the horizontal acceleration, so how we calculate the average acceleration is by calculating delta V divided by delta T. Okay, and then at position 0,0, its velocity is represented by this slope. Okay, so this slope is, this is a positive, this is a positive slope, which means v initial is positive value and then this represents a okay so at three a time t is equal to zero when time t is equal to three so this represents the velocity at three seconds um so v on final is equal to zero okay so delta v is v final minus v initial so it is zero minus a positive value divided by t so it must be a negative acceleration so b is the right answer a car initially at rest accelerates at 10 meters per second squared the car speed after it has traveled 25 meters is most nearly so initially at rest means its initial velocity is equal to zero and its acceleration is 10 meters per second squared so after his travel 25 meters which means its final position is 25 meters assuming that its initial position is zero meters then um okay so now i have these information so out of these three equations um i need to find the equation that has initial velocity final velocity acceleration and the and the displacement So the right one is equation three. Equation three is the only one that contains all of these four information. So v final squared is equal to v initial squared plus two times acceleration times the displacement. So zero squared plus two times ten times twenty-five. So that is five hundred. So the final is square root of 500 or approximately like 20 something. So it must be 22. So I know it's not 25 because 25 squared is 625. So that's way too big. But I know it's going to be greater than 20 because 20 squared is 400. So it's between 20 and 25. Question number 11. A student performs. An experiment in which the horizontal position of a toy car is recorded on ticket tape from a device that places dots on a tape at equal time intervals. In between the consecutive dots, it represents the same time interval. So the farther the dots are away from each other, which means the speed is greater. And then when the dots are closer to each other, it means the speed is smaller. the series of the dots in the figure represent the motion of an object moving from the negative direction to the positive direction along the horizontal direction the time interval between each recorded dot is one second which of the following experiments could the students have conducted to create the data shown on the ticket tape hey the toy car initially increased the speed okay so in between the dots the distance is getting greater and greater and greater and then and then it starts decreasing again so it has speed up and then slow down. It has speed up, travels at a constant speed, and then slow down. So the right answer is B. It increased the speed, travels at a constant speed, and then decreased the speed. Student wants to launch a toy dart toward a target that hangs from a light string. At time t is equal to zero, the dart is launched with an initial speed of v0 at an angle theta above the horizontal ground. At the instant the dart is launched, the string is cut such that the target begins to fall straight down. The positive horizontal direction is considered to be to the right, and the positive vertical direction is considered to be up. A student makes the necessary measurements to create the graph shown, which represent the vertical component of the velocity as the function of time for the dart for the target from t is equal to zero. This is for the dart. This is for the target. Until, for instance, the dart hits the target. The dart hits the target. At time t is equal to zero, the target is vertical distance h above the dart. curves for the dart and target each have the same area between them and horizontal axis. Both curves also have the same slope. Which of the following is the best method to determine the distance h from the graph? Okay so h is the distance between the target and the vertical distance between the target and the dart and as for the vertical velocity so this amount of area represents the um the displacement the vertical displacement um so y delta y when it comes to the target when it has hit the dart and then this area represents the delta y for the dart when it hits the target okay so when they meet each other so let's say the target is here and then the dart is here okay so this is delta y target and this is delta y dart okay um so of course one is positive and the other one is negative but when we take the absolute value for both of them then they add up to the total height of h the vertical height between the target and dart at the beginning so i just need to add up these two areas um and then taking positive for both of them so it will be the area between the dart line and then the target line um so uh that is um the area under the curve for the dart data. Okay, not A. So B, find twice the area under the curve for the dart's data. Okay, so it has stated that the area below these two curves are exactly the same. It's just that one is positive, the other one is negative. So by calculating the dart area and multiplying it by two, that is the area between these two lines. 13, which of the following graphs could represent the vertical component of the velocity as a function of time? um or the dart and target immediately after the dart is launched and then the target begins fall okay so immediately after the target um okay so the vertical component of velocity um so the target is going downward which means its velocity vertical velocity is negative while dart is going upward so its initial velocity must be positive um so you for the target um it's going from zero because it was at rest um to negative value okay and then on the other hand for the dart its vertical velocity is given um so it goes from positive and then it gradually decreased so d is the right answer and question number um question number 14 um so figure one shows a displacement versus time graph for the dart and figure two shows a displacement versus time graph for the target okay so this is for the target this is for the dart one is increasing from zero to a positive value the other one is decreasing from positive which displacement component horizontal or vertical is represented by the graph for each object okay so either for um dart or for target they're actually both doing a free fall motion which means they're moving under the influence of the gravitation or acceleration only um without any of the air resistance so their horizontal component must have a zero acceleration so for the dart its horizontal velocity should keep the same um so the dart is going this way so the dart is going along this curved line um it's doing a projectile motion but its horizontal component of the velocity should always be the same. So its horizontal displacement must be just a linear line increasing. and on the other hand its vertical component follows the acceleration graph. On the other hand its vertical component is gradually increasing but at a decreasing rate due to the acceleration pointing downward. So this explains why at the very beginning it has a very steep tangent curve but gradually its tangent line is getting flatter and flatter and then At the very end, its tangent line becomes horizontal, which means its velocity is zero. And this refers to the vertical velocity, not the horizontal velocity. So it represents the vertical displacement. So for both of them, it represents vertical. The answer is D. Next is question number 15. a student drops a rock from the top of a cliff such that the rock falls downward toward the earth's surface in an absence of the air resistance aka this is a free fall the downward direction is considered to be positive direction okay so when it goes down the velocity is positive the graph shows the rock's velocity as the function of time which of the following methods should be used to determine the total distance traveled by the rock after four seconds okay so within the first four seconds Okay, so this is a velocity versus time graph. And then we usually use the area under the curve to represent the displacement and that it has trebled. So from 0 to 4, it's going to be the area of this right triangle. 1 half times 4 times like 40, the answer should be 80 meters. So out of A, B, C, D, the answer is B. Use the area under the curve as it sees the rock trebled 80 meters. um in total next question number 16 an object travels along a straight line across a horizontal surface um and its motion is described by the velocity versus time graph shown in the figure which of the following methods will determine the total displacement of the object between zero and five seconds again a velocity versus time graph and it's moving only along the horizontal surface so it's going to be the area under this horizontal velocity versus time graph So that is, okay, A, finding a slope. That's not right. B, dividing the change in the velocity by the changing in time. No, that's acceleration. C, finding the area bound by the horizontal axis and the curve from 0 to 5 seconds. So C is the right answer. And D, using average speed is equal to total distance divided by the total time. And then multiply the average speed by 5 meters per second by a total. The average speed of 5 meters per second by a total time of 5 seconds. So D is actually applying the force equation. So it states that the change in the position or displacement is equal to 1 half of the V initial plus V final. So the average speed, assuming that the acceleration constant, multiplied by the time. So D is also right. So C and D are both correct answers. Question number 17. A student wants to study the motion of an object that has a constant acceleration. Which of the following experiments could a student conduct to provide a best situation in which an object has a constant acceleration? Select two answers. Um a constant acceleration and okay which of the following experiment could a student conduct to provide a best situation in which an object has a constant acceleration? Select two answers. A, release the ball from rest near Earth's surface. So this is actually one of the best situations that describes a constant acceleration because as an object goes through a free fall, it only experiences the gravitational acceleration, which is pointing downward. So A is correct. B, release the toy car from rest such that it travels along different looped sections of a track. Okay, so when it travels along a different looped section, It's not experiencing a constant acceleration. So that's not B. C, launch a water-propelled rocket from rest such that it travels into the air and falls back to Earth's surface. So it travels into the air and launches back. So C actually represents a projectile motion. And D, release a cart from rest such that it travels down an incline of 40 degrees with respect to ground. Okay. So both C and D actually represent the constant acceleration. So C represents the product of motion of an object. But because its velocity is not pointing in the same direction as acceleration, so I would say D is a better situation that describes an object with a constant acceleration. So in D, when a cart is moving downward, its speed is getting faster and faster because it experiences an acceleration pointing down along a slope. Question number 19. An object is released from rest near a planet's surface. A graph of the acceleration as a function of time for the object is shown for the four seconds after the object is released. The acceleration has always been a negative constant. The positive direction is considered to be upward. What is the displacement of the object after two seconds? Okay, so... Um... Since the acceleration is a constant value, negative 5 meters per second squared, and then the time is 2, time is equal to 2 seconds. Okay, then given that V final. and up next question number 19 an object is released from rest near a planet surface the graph of the acceleration as a function of time is shown where four seconds after the object is released the positive direction is considered to be outward what is the displacement of the object after two seconds so it states that object is released from rest which means its initial velocity is equal to zero And then its acceleration is a constant value of negative 5 meters per second squared. The time is 2 seconds. So given initial velocity, acceleration, and time, we need to find the displacement. Okay, so displacement, delta x is equal to v initial times p plus 1 half a t squared. So I can plug in all the values, then it's 0 plus 1 half times negative 5. meters per second squared multiplied by two seconds squared so that is negative 10 meters. The answer is B. And next question number 20. So an object is sliding to the right along a straight line on a horizontal surface. The graph shows the object's velocity as a function of time. What is the object's distance? displacement during the time depicted in the graph. Okay, so again, the displacement is represented by the area under the velocity time graph. Remember that if it's above the x-axis, then it's positive. If it's below the x-axis, then it is the negative. So for this is the linear velocity line, and then these two triangles are exactly the same. They're congruent to each other. The only difference is one is positive, the other one is negative. so they add up to zero and question number 21 a bowling pin is thrown vertically upwards such that it rotates as it moves through the air initially the center of the mass of the bowling pin is moving upward with a speed of v initial of 10 meters per second the maximum height of of the center of the mass of the bowling point is most nearly okay so um this is a typical free fall motion um And okay, so as for the free-fall motion, given that it's initial velocity, then due to the downward acceleration, its velocity gradually decreases until it reaches zero, until it temporarily comes to a stop. Then the change in the position, order displacement, is equal to V initial times T plus one-half A T squared. So acceleration... is pointing downward. So it's negative g, assuming that upward is a positive direction. Then delta y is equal to, okay, so only acceleration, initial velocity, and final velocity are given. So combining these three, When initial velocity, final velocity and acceleration is given, I need to figure out the maximum height, so the vertical component of the position or the displacement. Then I'm going to cross out equation one because it I am then I'm going to cross out equation one because time is not the subject of interest. I am interested in the delta y. And then again, I'm going to cross out equation two because, again, time is not the subject of interest. So I'm going to choose equation three, which sees that the final squared is equal to the initial squared plus two times eight times delta y. So 0 is equal to v initial squared plus 2 times negative g multiplied by the height h. Then v initial squared is equal to 2gh. Then h is equal to v initial squared divided by 2g, which is a. Question number 22. At time t is equal to zero, a moving cart on a horizontal track is at position 0.5 meters. Using a motion detector, students generate a graph of the cart's velocity as a function of time. At time t is equal to 2.5 seconds, the cart's position is most nearly... Okay, so the final position, it's asking for the final position. And the final position minus initial position is equal to the displacement. So final position is equal to initial position plus the displacement. So initial position 0.5 plus the displacement. And displacement is represented as the area below the curve, the velocity versus time graph. And this is between 0 and 2.5 seconds. So these two parts are exactly the same. These two triangles are exactly the same. So they cancel out each other. I only need to calculate the area between 0 and 0.5 seconds, this area. Then that area is equal to 1 half times the upper base plus the bottom base. So 1 half times 2 plus 3 multiplied by our height is 0.5. So that is 0.5 plus 5 divided by 4. or 1.25 so it's 1.75 meters the next two blocks are tied together with a string they're thrown onto a layer of ice such that they spin around their center of mass c as they slide horizontally across the icy surface um a graph of the two block systems velocity as a function of time is shown based on the graph which of the following claims are correct about the system. So when it says system, it's taking these two objects as one single system. So what really matters here is the center of the mass. Okay, and the direction of that. So they are spinning, and then it's moving to the right at the same time. And this shows the system's center of mass velocity. So its velocity is gradually decreasing. A, the total displacement of an individual block is zero. because it travels in a perfect circle around the center of the mass of the system so that's not zero um because while it's spinning around it's also moving toward the right so it's moving along this track so its displacement is not going to be zero b the acceleration of the center of the mass of the system is constant so b is actually correct so according to this um center of mass velocity versus time graph um the velocity decreased at a constant rate so it's a slope is a constant value. It's a negative constant. So B is correct. C, the change in the velocity per unit of time for the system center of mass changes as time increases. Okay, so again, the change in the velocity per unit of time, it represents acceleration. And acceleration is constant. It does not change. And D, the system's acceleration vector is opposite to the direction of the system's velocity vector. That is correct. So When acceleration and the velocity vector are pointing toward the opposite direction, then velocity is actually, the velocity is decreasing. So velocity is a positive value. So while it's moving toward the right, if the acceleration is a negative value, then at the next moment the velocity will gradually decrease. So B and D are the right answer.

Let's get started. So today I'm going to talk about Unit 1 of AP Physics 1 Kinematics. kinematics is the most fundamental and then one of the most important part when it comes to mechanics um in kinematics we study how an object moves but we're not going to study why an object moves in such a way we're only going to focus on how it moves including in which direction it moves moving to moving upward moving down or moving to the left moving to right is it doing a linear motion or projectile motion is it speeding up or slowing down is it accelerating or decelerating is it changing our direction that's the wrong So I'm going to start with three of the most important concepts in kinematics, which is displacement, velocity, and acceleration.

Remember that all of these three concepts are vector quantity rather than scalar quantity, which means they always come with a direction. Both magnitude and direction are important in the vector quantity. So displacement, how it is different from distance, is that distance takes all of the roots that you have. path that you have traveled while displacement only cares about the initial position and the final position. So displacement is literally just change in the position and a lot of times it's represented as delta x which is x final minus x initial.

You can replace x with y if it's moving up and down in vertical motion and when it comes to distance you will have to take every single part of the path you took. Let's say you move between point A and B. So let's say you're moving between point A and point B and then they are 100 meters apart. Then when you go directly from A to B, the displacement is 100 meters and distance is 100 meters as well.

But if you go from point A... to let's say point C and then going back to point B, then your displacement is still 100 meters while your distance is more than 100 meters, probably 150 or 200 meters. So distance is not always the same as displacement.

Or let's say when you move between A and B, you went to B and then you're back to A. Then your distance is 200 meters, 100 meters. from A to B, another 100 meters from B to A, 200 meters in total. But your displacement is zero because your initial position and final position are the same.

There is no change in the position. Again, displacement has nothing to do with the route you take. You can take whatever the route you take and then when you're back to A, your displacement is always zero while distance is not.

And in physics, we always use displacement to calculate all the information you want, rather than the distance. Now I can try to compare velocity to speed, just like I compared the displacement to the distance. The velocity is the vector quantity while the speed is the scalar quantity.

And how I calculate the velocity is it is the instantaneous rate of change in the displacement. So a lot of times when we calculate the speed, what we do is we take the total distance and then divide it using the total amount of time it takes. So we calculate the average speed. So that's not what we do in velocity in AP Physics 1. So we usually only talk about the instantaneous rate rather than the average rate. So how you're going to calculate the velocity is by taking the change in the distance.

by taking the change in the displacement and divided by the amount of time it takes to travel the amount of the displacement. And later I will talk about how to calculate the velocity or how to find a velocity on the position versus time graph. And next is acceleration.

So acceleration tells us how fast a velocity is changing. So velocity is not always constant. An object might speed up or slow down.

Its velocity is constantly changing most of the times in real world. So acceleration tells us about how fast the velocity is changing. So for example, for a sports car versus a regular sedan. So sports car has a very high acceleration, means within one second its speed can increase by a large amount.

So that's why it has a lot of advantage when you just start um in a race um on the other hand for the sedan even though it can still reach the high speed of let's say 100 meters 100 miles per hour but at the very beginning um it's accelerating slower um so for every second its speed is not going to increase that much so that's when we compared acceleration the rate of change in the velocity and then we always used on delta v, the change in velocity divided by time to represent acceleration. And again, acceleration is also a vector quantity. So in AP Physics 1, we always relate displacement, velocity, and acceleration all together. Okay, and now let's talk about how to analyze the graph of the velocity versus time and then how to analyze the graph of the displacement versus time.

So when you see a graph like this, it means that it's displacement or the position is constantly changing. So it is a curved line, it does not mean that the object is moving along this curved line, it just means that its position is changing. um along this curved line um as the time changes so most likely this is a one dimensional motion or a linear motion so this object must be moving along a linear line it can be a horizontal line it can be a vertical line it can even be a scandia line and use on the linear line you usually set up the positive direction and then the negative direction and usually i set the right side as the I set the right as the positive and left as the negative. And its position is equal to zero, which means at the very beginning, it starts at the position of zero.

And then after one second, it has moved to this position. After another second, it has moved to this one. After another second, it has moved to here.

After another second, it has moved to there. Okay, so it represents the motion of this object along this linear line. And this is the vector diagram.

The direction of the arrow represents the direction of the velocity, the direction of the motion. And then the length of the arrow represents how fast it is moving, the magnitude of the velocity. Okay, and based on the position versus time graph, how I draw the velocity versus time graph is by taking the slope of the tangent line.

Okay. So the slope of the tangent line at a specific point, let's say at time t is equal to three seconds. So how I find the velocity is by drawing a tangent line to this position versus time graph at a specific point and then calculate its slope. So this slope represents the velocity and it's obvious the velocity when time t is equal to zero it's a zero because it has a horizontal tangent line and time t is equal to zero and then the tangent line gradually um is tilted up which means the velocity is gradually increasing so based on this position versus time graph i can come up with a velocity person and the velocity versus time graph on this velocity versus 10 graph Velocity must be increasing from zero.

And on this velocity versus time graph, I can also find the acceleration by taking the slope of this velocity versus time graph line. So the slope of this linear line is acceleration. And since this is a linear line, its acceleration is always constant from the very beginning to the end, which means it's a constant acceleration motion. so in ap physics one you will rarely see um the scenario of a changing acceleration usually we assume that acceleration is constant to make sure that calculation is a little bit easier so on this velocity versus time graph The area of this triangle represents the change in the position or displacement.

So similarly based on this velocity versus time graph, I can also draw an acceleration versus time graph. And since acceleration is constant, acceleration graph is just a horizontal line. and this horizontal lightness above x-axis because it is a positive slope and again the area under the acceleration graph represents the change in the velocity now i'm going to give you a couple of examples of the linear motion in real world um so one of the most common um linear motion would accelerate with a constant acceleration is So free flow means an object is moving in air under the influence of the gravitational force only.

So it only experiences an acceleration that is pointing downward. So gravitational acceleration means it's a 9.8 meters per second squared downward acceleration. So let's say if I have if i hold an object and then release my hands it just drops and once once it drops its velocity increases downward so right at the moment when i release it its velocity is equal to zero and then because it's experiencing um gravitational acceleration so its velocity increases and then its velocity increase increases by nine point meters per second Sometimes we use negative 9.8 meters per second square to represent the gravitational acceleration. This is because I set upward, the vertical upward, as a positive direction, and then downward as negative direction. But I can always set the direction arbitrarily.

So sometimes we can just set downward as positive direction to make the calculation a little bit easier. What happens if I toss an object upward? Okay, so if I toss an object upwards, then it has a non-zero upward initial velocity. So it has upward initial velocity, and then its velocity decreases because of the downward acceleration. And then at the maximum point, it's temporarily at stop.

It's temporarily at rest, and then it starts to move downward. Again, due to the downward gravitational acceleration. So on its way going downward, it takes exactly the same amount of time as it goes upward. So you can notice that when it goes up and down, the velocity has only switched the direction.

The magnitude of the velocity has not changed at all. So that's why the amount of time taken to go upward is the same as... the amount of time taken to go downward and remember that what acceleration is in the same direction as the velocity then this object is going faster and if the acceleration in the opposite direction of the velocity then this object is slowing down it goes slower and also remember that speeding up and slowing down um is not the same term it's not an alternative to accelerating or decelerating So speed up and slow down is referring to the speed only or the magnitude of the velocity. While accelerating and decelerating means the velocity is increasing or decreasing. Okay so for example, so an object might be speeding up while it is actually decelerating.

So let's say an object is moving in the opposite direction or in the negative direction. And its speed is increasing. Okay, so obviously its speed is increasing and speeding up.

But when it comes to acceleration, it's a negative acceleration. Its initial velocity is, let's say, 0 meters per second. And its final velocity is negative 10 meters per second. So its velocity, the number itself is decreasing, going from 0 to a negative value.

So that's what I call by the negative acceleration. Or an object might be slowing down but accelerating. So this usually occurs when an object is moving in the negative direction. Okay so again it's moving in the negative direction and its speed is gradually decreasing. And then finally it comes to a stop.

So let's say its speed has, um its velocity has changed from negative five meters per second to zero meters per second. then the velocity itself is increasing from a negative value to 0. So according to the formula of the acceleration, delta V divided by T, delta V is 0 minus negative 5, which is positive 5 meters per second. So the velocity is increasing, so acceleration is positive.

So up until now, I was only dealing with a linear motion, which means the object is moving on a horizontal surface or moving up and down only. And now let's talk about the two dimensional motion. So when an object moves on a surface. So the best example for two dimensional motion in kinematics is the projectile motion.

So projectile motion is actually one type of free fall, which means that object is moving in air under the influence of the gravitational acceleration only without any of the air resistance. But as for the linear motion, the initial velocity is either zero or its velocity is in the same direction or the opposite direction of the acceleration. While in the projectile motion, its initial velocity is not in the same direction as the vertical line or um the gravitational acceleration so its velocity changes so its velocity is changing the direction and it's also changing the um and it's also changing the magnitude so it's gradually pointing um toward horizontal and then it becomes horizontal at the maximum point and then it gradually points downward So this is a representation of the projectile motion and how its velocity changes at different positions.

So velocity is the vector quantity and as for the vector we can always separate into the x component and then the y component. So for the initial velocity I'm going to rewrite it as the addition of the two vectors vx and then vy so two of the components and then along this projectile motion. only the gravitational acceleration is applied on this object.

So from the very beginning to the end, it's always the gravitational acceleration that is pointing downward, which means only the vertical component of the velocity changes. So vertical component of the velocity is gradually decreasing. And then at the maximum point, the vertical component of the velocity is equal to zero.

and then gradually the vertical component of the velocity becomes negative pointing downward. On the other hand, vx, the horizontal component of the velocity is not changing along the whole the motion and this is because there is no acceleration along the horizontal axis. So in the productile motion what we need to pay attention to is only the y component, so v y, only v y changes due to the gravitational acceleration which is alone on the vertical and v x never changes. So this explains why the path of the projectile motion is parabola because v x is not changing v y is always decreasing so that's why it is when it goes up the next moment it's slightly pointing um toward a horizontal and next it's more horizontal and next it's more flat and then now it's pointing down more down and more down okay so sometimes um it's um it might it might not look like a project emotion at first but actually they are so for example when you are trying when you are when you throw a ball um in a horizontal direction on a cliff so its initial velocity is pointing on in the horizontal direction the next moment due to the vertically downward gravitation acceleration its velocity points down and then the next moment it's pointing more downward next it's pointing more downward and next it's pointing more downward okay um so again it's a vx is not changing during the whole process because there is no acceleration along the horizontal axis but the vy is always changing so at the very beginning vy is equal to zero because the velocity is just in horizontal direction next moment vy is pointing downward um but the magnitude is small and then the next moment vy is pointing down and then the magnitude is larger next moment it's pointing down and then the magnitude got even larger and larger So now it's obvious that only VY decreases while VX is always constant. So no matter which one it is, this one or this one, they both represent a projectile motion.

Okay, now let's try to use the equation. So there are only three equations that you need to know, and then they are all shown on the equation sheet. So three equations that we need to know are... v final is equal to v initial plus the acceleration times t. The second one is x final, the final position is equal to initial position plus v initial t plus one half a t squared.

And then the last one is v final squared is equal to v initial squared plus two times a times the change in the position order displacement. Okay, so these three are in the equation sheet. And I'm going to add one more.

So all of these, remember that all of these three equations are based on the condition that acceleration is constant during the whole process of the motion. And when I say acceleration is constant, it doesn't just mean the magnitude is constant. It also means the direction of the acceleration is constant. So, for example, if the acceleration magnitude is the same, but it's changing that. direction, then this is not constant acceleration.

Constant acceleration means both the magnitude and direction are constant. Okay, so given this condition, I can add one more equation. So sometimes I can use the fourth equation to make the calculation faster.

The last one is the change in the position or displacement is equal to one half of the initial velocity plus final velocity. So this part actually represents the average velocity. Average velocity. Average velocity multiplied by time is equal to the change in the displacement.

And I can use the velocity versus time graph to represent it. So assuming that this is a constant acceleration, and assuming that this is a positive acceleration, so velocity increasing, then velocity versus time graph is a linear line. Because on the velocity time graph, the slope represent the acceleration constant acceleration means it's a linear line and then remember that the area under the graph represents the air the the change in the position or displacement and this point represents the initial velocity so when time is equal to zero in this point represent the final velocity okay so obviously this is a trapezoid and then how i find the area of a trapezoid is one half times a plus b uh multiplied by the height h okay one half times the upper base and then the bottom base and then multiply by the h so a is represented by this part b is represented by this part so now i can just plug in v initial and v final then I get 1 half times the initial plus the final multiplied by the height, which is time t.

So this is how you get the very last equation, the fourth equation. Now, let's try to use this equation and then all of this graph to apply them to actual AP style questions. So in this question, initial velocity is given even though it's not explicitly stated. It says the ball is released from rest, which means its initial velocity is equal to zero.

And then one second represents the time. And then the one floor has fallen one floor. Okay, so even though this is not stated in terms of the meters or the feet or inches, but because each of the floor is evenly spaced, so one floor actually represents the displacement, the amount of distance that has traveled or the change in the position.

So I'm going to use delta y to represent it. then in the given combining all of these three centers okay and one more thing one more hidden information here is um this is a free fall because it says the ball is just released so as for the free fall acceleration is always equal to the gravitational acceleration which is 9.8 meters per second squared pointing downward so now i have acceleration initial velocity time and the change in the displacement then combining all of these four again initial velocity time um acceleration and then the change in the position or displacement okay so i know i'm not going to use the equation one because v final final velocity is not given so Again, v final is not given, so I cannot use equation 3. And then I cannot use the fourth equation because final velocity is not given. So the only equation that I can use is equation 2. So I'm just going to replace x with y. Then I get this formula of delta y, change in the position or displacement, is equal to v initial t times 1 half at squared. Or 1 4 is equal to initial velocity 0. zero plus one half times g times one second squared where one floor is equal to one half g okay and it says and it's asking us to find the number of the floors so delta y it's asking us to find the the displacement uh when the ball has fall like three seconds now i'm going to use exactly the same formula but now delta y is a different um displacement delta y prime is equal to v initial t prime plus one half a t squared so the new um change in a position or the new displacement is equal to zero plus one half g times three seconds squared so that is nine times one half g so it's nine times of um the amount of the displacement that has trouble in one second which is nine floors It has traveled nine floors.

So the answer is C, between seven to ten floors. Question number two, a student wants to investigate the motion of a ball by conducting two different experiments. And the student released the ball from rest and used a slow motion camera to film the ball as it falls to the ground.

Okay, so in experiment one, this is just a linear line motion of free fall. It just released from rest, so it goes through vertically down. And then in experiment two, the student horizontally rolls the same ball off a table and used a video analysis to plot the balls horizontal position x and vertical position y. Okay, so in experiment two, the ball starts at a horizontal velocity, so it's gonna follow this parabola curve.

It's gonna follow the parabola and then it is doing the projectile motion. So this is initial velocity and this is how its velocity changes over time. The graph from each experiment are shown, two, which of the following conclusions can the student draw from the graphs, and y. Okay, so now when I compare the experiment one and two, they have exactly the same y versus t graph.

So their vertical position is changing at the same rate. The only difference is the horizontal position. So for experiment one, its horizontal position is not changing at all because it's just moving up and down along the vertical line um in figure two in experiment two it it has it started in a horizontal direction so its um horizontal position is also changing but at a constant rate this is a linear line so its slope is represents the acceleration it's a linear line so it's slope represents the velocity and then the velocity is constant um a the horizontally rolled um ball you travels a greater horizontal distance and takes a longer amount of time to reach the ground.

Okay, so the first half is correct. It travels a greater horizontal distance, right? But it does not take a longer amount to reach to the ground. So no matter which one you pick, it's going to take exactly the same amount of time because the amount of time taken to reach the ground is depending, it depends on its initial vertical velocity and then the vertical displacement.

and then the vertical acceleration. And experiment one and two have exactly the same vertical components. They have the, they have traveled the same vertical displacement. Their initial vertical velocity are exactly the same.

Both have initial vertical velocity of zero. And then both of them are experiencing the same vertical acceleration of 9.8 meters per second squared. So the second part is not correct. It's the same amount of time. So A is not true.

B. the release ball travels with a slower speed and takes a longer amount of time um to reach the ground okay so again not longer amount of time it's the same amount of time and see um since both balls have their horizontal acceleration um they reach the ground at the same time and d since the balls have the same vertical position at any given time they reach the front at the same time so out of c and d d is the right one so c is actually true um they have the zero horizontal acceleration yes it's true that both of them um have zero horizontal acceleration so in experiment two the ball is doing a constant horizontal motion while in experiment one is just not moving in horizontal direction but it does not give us any information regarding the time taken to reach the crown. So we need to take a look at the vertical components. And when we take a look at the vertical graph, the vertical position versus time graph, at every single moment, they are at the same position, the same vertical position.

So this explains why they will spend the same amount of time to drop to the bottom. Three, in experiment one, what is the speed of the ball the instant it makes contact with the ground? Okay, so in experiment one, so, okay. So how we calculate or how we figure out the velocity is by drawing a tangent line on the position versus time graph and then find the slope of that tangent line.

So it's going to be this linear line. This linear line is tangent to the graph. and then its slope is okay and then its slope is a negative value which means it's moving downward and then in how i'm going to calculate the slope is by picking any of the two points on this on this slope so approximately um okay so i can pick these two points i can pick this point and this point so approximately um it's slope at when it has reached the ground is going to be delta V divided by is going to be 0 minus 1.6 divided by 0.55 minus 0.25. So negative 1.6 divided by 0.3.

So that's up. approximately negative five meters per second. So as for the speed, speed only takes the actual value.

So if the velocity is negative five meters per second, then its speed is approximately five meters per second. Question number four, suppose that experiment one and two are conducted at the same time. One student drops the ball from rest at the same instant that a second student horizontally rolls an identical off the table. After both balls have traveled half their vertical distance to the floor, what is the acceleration of the center of mass of the two-ball system relative to the Earth?

Okay, so now let's talk about what it means by the center of the mass and how we calculate the center of mass of the two-ball system. So the center of mass is usually used to represent um the where the mass is on how the mass is distributed within an object so not all the object has a regular shape like a circle or uh or a square so most of the object in real world has irregular shape so let's say for like human beings um the center of the mass is uh is more toward the upper body is closer to the upper body because the skull is pretty heavy it occupies a great percentage a large percentage in the human body weights. So compared to the lower body, the upper body is heavier. The center of mass is a little bit above your belly button.

And for irregular shaped object, the center of mass is more toward the part where more of the mass is concentrated. So for example, if you have a stick like this, if one side is thicker and the other side is skinnier, and given that the density is the same then the center of the mass is more um on the left side of this stick so this is for one single object but in ap physics one you barely get tested on finding the center of the mass within an object we usually assume that that's just the center of the circle or the or the center of um uh of this like rectangular object um but you will be tested on center of the mass of the multiple object or the two objects so let's say when you have two objects um so i'm going to use two rectangles to represent it and then one of the mass is one of the rectangle has the mass of 10 kilograms the other one has the mass of 20 kilograms and then they are let's say um one meters apart then how i calculate your center of mass is i'm going to use this formula of x center of mass is equal to x 1 times minus 1 plus x 2 times minus 2 divided by the sum of the mass the total is so of one and x to represent the position of the center of mass for each of the object so i'm gonna draw i'm gonna draw this reference frame um and then let's say this represents zero and then this represents one meters then the position of the 10 kilogram object is zero the position of the 20 kilogram object is one meter and now i just plug in um these numbers in the formula Then x1 the position of the mast 1 times 10 kilogram plus x2 1 meters the position of the mast 2 multiplied by its mast and then divided by the total mass of 10 kilogram plus 20 kilogram then it is 20 divided by 30 so 2 3 approximately 0.67 meters. It means on this reference frame um 0.67 meters so here so this represent so this 0.67 meters this is the position of the the center of mass of this um two object system and when you're asked to find the velocity of the center of mass you are going to use on the same method the velocity of the center of mass is equal to velocity of of each of the mass multiplied by its mass and then you are going to add up the product and then divide it by the total mass.

So going back to that question, okay so in experiment one the object, the ball is moving straight down in object two the ball is moving along doing a projectile motion moving along the parabola and then they have exactly the same vertical position the only difference the horizontal position um okay then um it's acceleration of the center of the mass relative to the earth um okay so when it comes to acceleration acceleration is a vector quantity so we are we can calculate the horizontal acceleration and then calculate the vertical acceleration and then add them up together so when it comes to the vertical acceleration since both of them are moving down at the same velocity and then the same acceleration so for both of them a y the vertical acceleration is g um just on gravitational acceleration on the other hand for the horizontal acceleration okay so one is not moving horizontally the other one is moving at a constant velocity horizontally then the velocity of the center of mass in horizontal um component is equal to v x of um of one of the ball um multiplied by v x of another ball okay so um in experiment one the ball is not moving at all in horizontal direction so its horizontal velocity is always zero um so and then i'm going to divide it by the total mass of m plus m so eventually this is one half of the horizontal um velocity of of the ball in experiment two so it means this is a constant velocity and as for the constant velocity acceleration is equal to zero because acceleration is changing velocity over time so this is zero meters per second squared okay so in horizontal direction acceleration is zero in vertical direction acceleration is g so when i combine them together so this is this is a y And this is AX. And since AX is 0, so when you add them up together, the acceleration is still equal to AY only. Next, question number five. Students provide with a battery-powered toy that the manufacturer claims will always operate at a constant speed.

The student must design an experiment in order to test the validity of the claim. So when it comes to measuring the velocity or the position or acceleration of a moving object, we use either the photogate or the motion detector. So I know for sure that Which of the following measuring tools can a student use to test the validity of the claim?

Select two answers. Okay so I know this is not right, a mass balance. So measuring the mass has nothing to do with its position or velocity or acceleration.

So photogate and motion detector are used pretty often when it comes to measuring the velocity or the position or acceleration of a moving object. but remember that no matter which one you use either the photogate or the motion detector you always have to put it along the track where this object is moving so the object is moving in the horizontal direction that you will have to put the photogate along this horizontal track either here or here or here and it's the same for the motion detector if you place it in the vertical and in in a vertical position compared to the it's moving track then it's not going to measure the position order velocity So oriented perpendicular is not correct for sure. So the answer is A and B. The photo gate placed at the beginning and then the end and then along the various locations along the track. And B, a meter stick to measure the distance of the track that the car travels on.

And the next question, do students want to determine the speed at which a ball was released? when it's thrown vertically upward into the air. So the ball is released but it's thrown vertically upward so its initial velocity is pointing upward and then its velocity gradually decreases until it comes to a stop and then it starts to move down until it hits the ground.

One student throws the ball into the air while one other student measures the total time that the ball is in the air. Okay the total amount of time starting from this to this. The student used the meters to measure the release height of the ball.

So the release height of the ball. Okay, so given the release height, I can calculate the change in the position or the displacement delta y, which is the final position minus the initial position. So that's going to be negative of the release height, negative h. And then given the time, okay, so it's asking us to find the speed, its initial speed. And then this is the free fall.

so of course we assume that acceleration is the gravitational acceleration pointing downward so we're given one two three three information and that you're required to find the last one um okay then obviously um c and d does not apply because the final velocity is not even given so i cannot use the field or d to figure out um the initial so not c or d so it must be either a or b and then it's from the moment in time in which the ball was released to the moment in time in which the ball reaches the highest point or until it hits ground. Okay, so the amount of time it measures is from initial all the way to it going back. It hits the ground, so I will have to choose B.

Seven, an object undergoes an acceleration as it travels along a straight horizontal section of a track. Which of the following graphs could represent the motion of the object? Select two answers. So A, speed versus time graph. So usually in the velocity versus time graph, the slope of the line represents the acceleration.

So this means it is a constant positive acceleration. So A is the right answer. And B, position versus time graph. On the position versus time graph, the slope represents the velocity. So it just means it's a constant positive velocity.

So not B. And C. um position uh and time squared okay and d is velocity versus time so i know d is not the right answer for sure because this is constant velocity or when i take a look at the slope which is zero um slope of the velocity time graph represent acceleration so the right answer is a and c so for the position versus time squared graph if i convert it into the regular position versus time graph then it's going to be a curved line.

So it means that it has a different tangent light. It has the tangent light of different slopes and different time, which means the velocity is constantly changing. Question number eight, a monitor sensor is used to create the graph of a student's horizontal velocity as a function of time as the student moves toward and away from the sensor.

The positive direction is the final acceleration away from the sensor, which of the following describes the student's final position x final in relation to the starting position x zero and the student's average horizontal acceleration ax between 0.0 and 3.0 seconds. Okay, so first off, this is a velocity versus time graph. So the area under the graph represents the change in position or the displacement.

So this is a starting position, and this is where the final position must be. So if I add up, if I calculate the area under the graph, then obviously the positive part is greater than the negative part. So delta x. is going to be greater than zero, which means X final minus X initial is greater than zero.

So X final is greater than X initial, which means the final position is more away from the monitor sensor. So either A or B. And then when we calculate the acceleration, A acts between zero and three seconds.

Okay, zero and three seconds. Mm, so. Okay, so when it comes to the horizontal acceleration, so how we calculate the average acceleration is by calculating delta V divided by delta T. Okay, and then at position 0,0, its velocity is represented by this slope.

Okay, so this slope is, this is a positive, this is a positive slope, which means v initial is positive value and then this represents a okay so at three a time t is equal to zero when time t is equal to three so this represents the velocity at three seconds um so v on final is equal to zero okay so delta v is v final minus v initial so it is zero minus a positive value divided by t so it must be a negative acceleration so b is the right answer a car initially at rest accelerates at 10 meters per second squared the car speed after it has traveled 25 meters is most nearly so initially at rest means its initial velocity is equal to zero and its acceleration is 10 meters per second squared so after his travel 25 meters which means its final position is 25 meters assuming that its initial position is zero meters then um okay so now i have these information so out of these three equations um i need to find the equation that has initial velocity final velocity acceleration and the and the displacement So the right one is equation three. Equation three is the only one that contains all of these four information. So v final squared is equal to v initial squared plus two times acceleration times the displacement. So zero squared plus two times ten times twenty-five. So that is five hundred.

So the final is square root of 500 or approximately like 20 something. So it must be 22. So I know it's not 25 because 25 squared is 625. So that's way too big. But I know it's going to be greater than 20 because 20 squared is 400. So it's between 20 and 25. Question number 11. A student performs. An experiment in which the horizontal position of a toy car is recorded on ticket tape from a device that places dots on a tape at equal time intervals. In between the consecutive dots, it represents the same time interval.

So the farther the dots are away from each other, which means the speed is greater. And then when the dots are closer to each other, it means the speed is smaller. the series of the dots in the figure represent the motion of an object moving from the negative direction to the positive direction along the horizontal direction the time interval between each recorded dot is one second which of the following experiments could the students have conducted to create the data shown on the ticket tape hey the toy car initially increased the speed okay so in between the dots the distance is getting greater and greater and greater and then and then it starts decreasing again so it has speed up and then slow down. It has speed up, travels at a constant speed, and then slow down.

So the right answer is B. It increased the speed, travels at a constant speed, and then decreased the speed. Student wants to launch a toy dart toward a target that hangs from a light string. At time t is equal to zero, the dart is launched with an initial speed of v0 at an angle theta above the horizontal ground. At the instant the dart is launched, the string is cut such that the target begins to fall straight down.

The positive horizontal direction is considered to be to the right, and the positive vertical direction is considered to be up. A student makes the necessary measurements to create the graph shown, which represent the vertical component of the velocity as the function of time for the dart for the target from t is equal to zero. This is for the dart. This is for the target. Until, for instance, the dart hits the target.

The dart hits the target. At time t is equal to zero, the target is vertical distance h above the dart. curves for the dart and target each have the same area between them and horizontal axis. Both curves also have the same slope. Which of the following is the best method to determine the distance h from the graph?

Okay so h is the distance between the target and the vertical distance between the target and the dart and as for the vertical velocity so this amount of area represents the um the displacement the vertical displacement um so y delta y when it comes to the target when it has hit the dart and then this area represents the delta y for the dart when it hits the target okay so when they meet each other so let's say the target is here and then the dart is here okay so this is delta y target and this is delta y dart okay um so of course one is positive and the other one is negative but when we take the absolute value for both of them then they add up to the total height of h the vertical height between the target and dart at the beginning so i just need to add up these two areas um and then taking positive for both of them so it will be the area between the dart line and then the target line um so uh that is um the area under the curve for the dart data. Okay, not A. So B, find twice the area under the curve for the dart's data. Okay, so it has stated that the area below these two curves are exactly the same. It's just that one is positive, the other one is negative.

So by calculating the dart area and multiplying it by two, that is the area between these two lines. 13, which of the following graphs could represent the vertical component of the velocity as a function of time? um or the dart and target immediately after the dart is launched and then the target begins fall okay so immediately after the target um okay so the vertical component of velocity um so the target is going downward which means its velocity vertical velocity is negative while dart is going upward so its initial velocity must be positive um so you for the target um it's going from zero because it was at rest um to negative value okay and then on the other hand for the dart its vertical velocity is given um so it goes from positive and then it gradually decreased so d is the right answer and question number um question number 14 um so figure one shows a displacement versus time graph for the dart and figure two shows a displacement versus time graph for the target okay so this is for the target this is for the dart one is increasing from zero to a positive value the other one is decreasing from positive which displacement component horizontal or vertical is represented by the graph for each object okay so either for um dart or for target they're actually both doing a free fall motion which means they're moving under the influence of the gravitation or acceleration only um without any of the air resistance so their horizontal component must have a zero acceleration so for the dart its horizontal velocity should keep the same um so the dart is going this way so the dart is going along this curved line um it's doing a projectile motion but its horizontal component of the velocity should always be the same. So its horizontal displacement must be just a linear line increasing.

and on the other hand its vertical component follows the acceleration graph. On the other hand its vertical component is gradually increasing but at a decreasing rate due to the acceleration pointing downward. So this explains why at the very beginning it has a very steep tangent curve but gradually its tangent line is getting flatter and flatter and then At the very end, its tangent line becomes horizontal, which means its velocity is zero. And this refers to the vertical velocity, not the horizontal velocity.

So it represents the vertical displacement. So for both of them, it represents vertical. The answer is D. Next is question number 15. a student drops a rock from the top of a cliff such that the rock falls downward toward the earth's surface in an absence of the air resistance aka this is a free fall the downward direction is considered to be positive direction okay so when it goes down the velocity is positive the graph shows the rock's velocity as the function of time which of the following methods should be used to determine the total distance traveled by the rock after four seconds okay so within the first four seconds Okay, so this is a velocity versus time graph. And then we usually use the area under the curve to represent the displacement and that it has trebled.

So from 0 to 4, it's going to be the area of this right triangle. 1 half times 4 times like 40, the answer should be 80 meters. So out of A, B, C, D, the answer is B. Use the area under the curve as it sees the rock trebled 80 meters.

um in total next question number 16 an object travels along a straight line across a horizontal surface um and its motion is described by the velocity versus time graph shown in the figure which of the following methods will determine the total displacement of the object between zero and five seconds again a velocity versus time graph and it's moving only along the horizontal surface so it's going to be the area under this horizontal velocity versus time graph So that is, okay, A, finding a slope. That's not right. B, dividing the change in the velocity by the changing in time.

No, that's acceleration. C, finding the area bound by the horizontal axis and the curve from 0 to 5 seconds. So C is the right answer. And D, using average speed is equal to total distance divided by the total time.

And then multiply the average speed by 5 meters per second by a total. The average speed of 5 meters per second by a total time of 5 seconds. So D is actually applying the force equation.

So it states that the change in the position or displacement is equal to 1 half of the V initial plus V final. So the average speed, assuming that the acceleration constant, multiplied by the time. So D is also right.

So C and D are both correct answers. Question number 17. A student wants to study the motion of an object that has a constant acceleration. Which of the following experiments could a student conduct to provide a best situation in which an object has a constant acceleration? Select two answers. Um a constant acceleration and okay which of the following experiment could a student conduct to provide a best situation in which an object has a constant acceleration?

Select two answers. A, release the ball from rest near Earth's surface. So this is actually one of the best situations that describes a constant acceleration because as an object goes through a free fall, it only experiences the gravitational acceleration, which is pointing downward. So A is correct.

B, release the toy car from rest such that it travels along different looped sections of a track. Okay, so when it travels along a different looped section, It's not experiencing a constant acceleration. So that's not B. C, launch a water-propelled rocket from rest such that it travels into the air and falls back to Earth's surface. So it travels into the air and launches back.

So C actually represents a projectile motion. And D, release a cart from rest such that it travels down an incline of 40 degrees with respect to ground. Okay.

So both C and D actually represent the constant acceleration. So C represents the product of motion of an object. But because its velocity is not pointing in the same direction as acceleration, so I would say D is a better situation that describes an object with a constant acceleration.

So in D, when a cart is moving downward, its speed is getting faster and faster because it experiences an acceleration pointing down along a slope. Question number 19. An object is released from rest near a planet's surface. A graph of the acceleration as a function of time for the object is shown for the four seconds after the object is released. The acceleration has always been a negative constant. The positive direction is considered to be upward.

What is the displacement of the object after two seconds? Okay, so... Um...

Since the acceleration is a constant value, negative 5 meters per second squared, and then the time is 2, time is equal to 2 seconds. Okay, then given that V final. and up next question number 19 an object is released from rest near a planet surface the graph of the acceleration as a function of time is shown where four seconds after the object is released the positive direction is considered to be outward what is the displacement of the object after two seconds so it states that object is released from rest which means its initial velocity is equal to zero And then its acceleration is a constant value of negative 5 meters per second squared. The time is 2 seconds. So given initial velocity, acceleration, and time, we need to find the displacement.

Okay, so displacement, delta x is equal to v initial times p plus 1 half a t squared. So I can plug in all the values, then it's 0 plus 1 half times negative 5. meters per second squared multiplied by two seconds squared so that is negative 10 meters. The answer is B. And next question number 20. So an object is sliding to the right along a straight line on a horizontal surface.

The graph shows the object's velocity as a function of time. What is the object's distance? displacement during the time depicted in the graph. Okay, so again, the displacement is represented by the area under the velocity time graph. Remember that if it's above the x-axis, then it's positive.

If it's below the x-axis, then it is the negative. So for this is the linear velocity line, and then these two triangles are exactly the same. They're congruent to each other.

The only difference is one is positive, the other one is negative. so they add up to zero and question number 21 a bowling pin is thrown vertically upwards such that it rotates as it moves through the air initially the center of the mass of the bowling pin is moving upward with a speed of v initial of 10 meters per second the maximum height of of the center of the mass of the bowling point is most nearly okay so um this is a typical free fall motion um And okay, so as for the free-fall motion, given that it's initial velocity, then due to the downward acceleration, its velocity gradually decreases until it reaches zero, until it temporarily comes to a stop. Then the change in the position, order displacement, is equal to V initial times T plus one-half A T squared.

So acceleration... is pointing downward. So it's negative g, assuming that upward is a positive direction.

Then delta y is equal to, okay, so only acceleration, initial velocity, and final velocity are given. So combining these three, When initial velocity, final velocity and acceleration is given, I need to figure out the maximum height, so the vertical component of the position or the displacement. Then I'm going to cross out equation one because it I am then I'm going to cross out equation one because time is not the subject of interest. I am interested in the delta y.

And then again, I'm going to cross out equation two because, again, time is not the subject of interest. So I'm going to choose equation three, which sees that the final squared is equal to the initial squared plus two times eight times delta y. So 0 is equal to v initial squared plus 2 times negative g multiplied by the height h. Then v initial squared is equal to 2gh. Then h is equal to v initial squared divided by 2g, which is a.

Question number 22. At time t is equal to zero, a moving cart on a horizontal track is at position 0.5 meters. Using a motion detector, students generate a graph of the cart's velocity as a function of time. At time t is equal to 2.5 seconds, the cart's position is most nearly...

Okay, so the final position, it's asking for the final position. And the final position minus initial position is equal to the displacement. So final position is equal to initial position plus the displacement. So initial position 0.5 plus the displacement. And displacement is represented as the area below the curve, the velocity versus time graph.

And this is between 0 and 2.5 seconds. So these two parts are exactly the same. These two triangles are exactly the same.

So they cancel out each other. I only need to calculate the area between 0 and 0.5 seconds, this area. Then that area is equal to 1 half times the upper base plus the bottom base.

So 1 half times 2 plus 3 multiplied by our height is 0.5. So that is 0.5 plus 5 divided by 4. or 1.25 so it's 1.75 meters the next two blocks are tied together with a string they're thrown onto a layer of ice such that they spin around their center of mass c as they slide horizontally across the icy surface um a graph of the two block systems velocity as a function of time is shown based on the graph which of the following claims are correct about the system. So when it says system, it's taking these two objects as one single system. So what really matters here is the center of the mass.

Okay, and the direction of that. So they are spinning, and then it's moving to the right at the same time. And this shows the system's center of mass velocity. So its velocity is gradually decreasing. A, the total displacement of an individual block is zero.

because it travels in a perfect circle around the center of the mass of the system so that's not zero um because while it's spinning around it's also moving toward the right so it's moving along this track so its displacement is not going to be zero b the acceleration of the center of the mass of the system is constant so b is actually correct so according to this um center of mass velocity versus time graph um the velocity decreased at a constant rate so it's a slope is a constant value. It's a negative constant. So B is correct. C, the change in the velocity per unit of time for the system center of mass changes as time increases.

Okay, so again, the change in the velocity per unit of time, it represents acceleration. And acceleration is constant. It does not change. And D, the system's acceleration vector is opposite to the direction of the system's velocity vector. That is correct.

So When acceleration and the velocity vector are pointing toward the opposite direction, then velocity is actually, the velocity is decreasing. So velocity is a positive value. So while it's moving toward the right, if the acceleration is a negative value, then at the next moment the velocity will gradually decrease.

So B and D are the right answer.

Transcript for:Key Concepts in Kinematics for Physics

Transcript for:
Key Concepts in Kinematics for Physics