 So, this is a video about the NCRT series, we will be doing NCRT questions. The way to use this video is you pause at the questions, try the questions yourself. If you are not able to do the question is when you do along with me, play the video and do along with me. And let me again put this disclaimer that there is no substitute to you actually practicing the questions yourself, okay? So, this video is just designed for you to help, you know, to approach the questions at least, okay? So, some people are having difficulty approaching the question itself, okay? So, take this as video as a guide to approach those NCRT questions and once you have a foundation on NCRT, you start knowing how to approach even competitive level questions, okay? So, this is just a foundation to more and more questions after this video, okay? And keep enjoying, okay? So, let us start with this NCRT problem. This is from chapter number three in NCRT which is motion in a straight line, okay? In competitive study, we call it motion in 1D, okay? So, this is NCRT level, let us start with this first question. The position of an object moving in the x-axis is given to us as a function of time, okay? So, displacement is given as a plus bt square. You should notice here immediately that this displacement is not a constant, so I cannot find out velocity by just dividing it by time, okay? Since it's a function of time, it's a variable, we have to take help of calculus and the operator that we are using from calculus is going to be differentiation. But here's few of the quantities are given to us as in the value of a is given to us as 8.5 meters, b is given to us as 2.5 meter per second square, and t is measured in seconds. So, what is the velocity? So, they are asking the velocity at certain times. They are asking us what is the velocity at t is equal to 0 and at t is equal to these two seconds, okay? So, let's first figure out the velocities at these times and then we will sort the next part which is asking about average velocity. Now, I think multiple times this has been told in class key velocity is nothing but the differentiation of displacement with respect to time. And if I try to differentiate this function of time with respect to time, so let us differentiate d by dt of, displacement in this case is a plus bt square, okay, where a and b are constants. So, here you have to understand key if I try to differentiate a constant, the first term would become 0, okay? And the second term would be b is a constant, so it will come out of the differentiation and the differentiation of t square will be 2 times t. So, the formula for differentiation of x raised to power n with respect to x was n times x n minus 1. This is the formula I used here, t square 2 comes down and 2 minus 1 is 1. So, this is how I did this. So, velocity is 2 times bt, okay? So, this is the function for velocity. Now, they had asked me velocity for 2 points, t is equal to 0 and t is equal to 2 seconds, okay? So, if I try to find out the velocity at t is equal to 0, it will be simply 0 because when I plug in the value of 0 into this equation, you will see the velocity will become 0, okay? This will be equal to 0. So, how do I want to write this? So, I will write this like this, b is 0. This means velocity at 0. v at 2 will be 2 times b into 2 because in place of t, I will plug in 2. So, this will be 4b. Now, b's value is given in the question, okay? b's value is 2.5, okay? So, this is 4 into 2.5. That means this is equal to 10 meters per second, 10 meters per second. So, the first part of the question is done. They had asked me what is the velocity at t is equal to 0 seconds and t is equal to 2 seconds. Next, they are asking what is the average velocity between 2 seconds and 4 seconds. Now, for average velocity, you don't need to differentiate because average velocity is defined as, what is the final displacement? What is the initial displacement? And divided by the time taken, divided by the time taken. So, I already have a function for displacement which is a plus bt squared. Okay? So, we are asked from t is equal to 2 seconds to t is equal to 4 seconds. So, if I try to find out what is the initial displacement, it will be simply a plus b into 2 square. Why 2 square? I just plugged in the value of 2 in this and I got what is the displacement at exactly 2 seconds I found out. Okay? Similarly, if I want to find out what is the displacement at 4 seconds, all I have to do is plug in the value of 4 in this equation, which means these are the equations. So, the first equation is a plus 4b. The second equation is a plus 16b. Now, for finding out average velocity, I need the difference of these two. So, xf minus xi would be a plus 16b minus a plus 4b. Okay? This should be in brackets. This should be in brackets. Now, this will be a and a will cancel out will have 16 minus 4 equal to 12b. Okay? Now, what is the change? What is the time taken? So, from t is equal to 2 seconds to t is equal to 4 seconds. Time taken is how much? 2 seconds. Time taken is simply 2 seconds. 4 minus 2 is just 2. Okay? So, for this, the average velocity, the average velocity is the difference is that, you know, is the ratio of this 12b divided by 2. Okay? So, I will get 6b and b value I already know it is 2.5. So, this means this is equal to 15 meter per second. Okay? 15 meter per second. So, I found out both the values of both the average velocity and the instantaneous velocity. For this second question, they are asking us to obtain the equations of motion for a constant acceleration using the method of calculus. Okay? Again, this is one of the very favorite question for school teachers. They are always asking this. Little bit of knowledge about calculus is required for this. Basic integration and basic differentiation is what is required. Okay? So, there are three equations of motion. v is equal to u plus ad. The second equation is v square minus u square is equal to 2 as. And the last one is s is equal to ut plus of ad square. Okay? So, for deriving the first one, it is very, very simple. We are going to start with a basic definition of acceleration. Okay? Acceleration is defined as differentiation of velocity with respect to time. This in the brackets, I will write that this is from definition of acceleration from definition of from definition of acceleration. Okay? This is how you will write in duties. Okay? This is how you have to write. I have already talked about this in one of the videos where I told you that there is a set pattern to writing answers in CBCU. Okay? And this is that pattern you need to follow. Okay? After that, after you write this, what you can do is cross multiply, I can write adt is equal to dv. Okay? And both sides, I can integrate. I can integrate. So, I will not do it in the same step. I will just write here integrating both sides. Both sides. Both sides of what? Right hand side and left hand side of the equal side. Okay? Both sides. So, on the left hand side, it will be integration of adt. And on the right hand side, it will be simply integration of dv. Now, limits I have to put. Limits. Okay? So, whenever I'm talking about velocity, initial limit of velocity I represent by u. Generally, u is used to define initial velocity. Final velocity is given by v. Similarly, the observation starts from 0 and some random time t. Okay? Now, it's already given in the question key acceleration is constant. So, that means a can be taken out of the integration and this will be simply 0 to t dt. Okay? This will be simply 0 to t dt. On the right hand side, integration of dv is going to be. So, let's just write it here like this. Okay? Now, what is integration of just dx? The answer is x. Why? Because I told you if there is nothing, you can imagine that there is x raise to call 0 here. And from there, you can write 0 plus 1 divided by 0 plus 1 making integration of dx equal to x. Okay? So, if integration of dx is x, on the left hand side, you'll simply get integration of dt will be t. Okay? I'll have to plug in the values t and 0. And on the right hand side, you'll get v. Integration of dv would be v and I have to plug in the values of v and u. Okay? So, I hope you know how to put the limits into the equation. Okay, first you put in the upper limit and you subtract the lower limit. So, on the right hand side, on the left hand side, you'll have at minus a0. And on the right hand side, you will have v minus u. If you simplify this, a into 0 becomes 0 and you end up with that equation v is equal to u plus at, which we were looking for. This is the first equation of motion. Okay? Now, let us derive the second one, which is, you know, v square minus u square is equal to 2 as. v square minus u square is equal to 2 as. Okay? So, for this one, again, I'm going to start with a is equal to, a is equal to dv by dt. Okay? But I'm going to make a few changes here. I'm going to make a few changes here. Okay? I'm going to multiply dx in the numerator as well as in the denominator. So, dv by dt and then I'm going to multiply dx by dx. Okay? Nothing is going to change because I multiplied dx in the numerator and the denominator as well. Okay? So, sorry about this. Okay? So, what I'm going to do is pair this and this dt and this dx. So, I'll end up with this dx by dt and then multiplied by dv by dx. I only rearrange terms in this step. Okay? dx, this dx I took it here and this dt I took it. So, I have paired dx by dt. Now, what is dx by dt? The rate of change of displacement with respect to time is what? It's velocity. So, acceleration can also be written as v dv by dx. Okay? This is one important result that is useful in other situations also. So, I'm going to make this a box. Obviously, it's not very like this is from, this is not the equation of motion I'm looking for. I'm still boxing it because this is important for future. Okay? So, from here what I'm going to do is what I did in the previous section. Okay? I'm going to cross multiply. I'll have adx equal to v dv. Okay? Then what do I do? I will integrate both sides. Both sides. Okay? If I integrate both sides here, I'll end up with adx on the left hand side and on the right hand side I'll have v dv. I'll put the limits again. Okay? For displacement, I'm going to put from zero to some value s as the limits. Okay? And for velocity, obviously I'm going to put u as the initial velocity and v as the final. Now, I have to perform the integration. On the left hand side, acceleration is constant given the question. So, acceleration can come out. Integration of dx will be x. I just discussed above. Okay? Integration of dx will be x. So, I'll write x and what are the limits I'll put s and zero. Similarly, on the right hand side, I have to integrate v with respect to dv. Okay? So, this is v raised to power one. So, the integration will be v raised to power one plus one which is two divided by two and the limits will be v and u. Now, if I plug in the limits, I'll get as is equal to v square minus u square by two. If I rearrange the term, I get the second equation of motion that says v square is equal to u square plus two a s. Okay? This is the second equation of, this is the second equation of motion. Okay? Let's do this next question. In this question, it's given to us a ball is thrown upwards with a velocity of 20 meter per second. That is given in the first statement. A ball is thrown vertically upwards with 20 meter per second. And it is not just thrown from the ground. It's not thrown from the ground. This ball is already at a height of 25 meters from the ground. Okay? So, this ball is already at a height of 25 meters from the ground. Okay? Again, let me tell you, let me again tell you, he has physics students making diagrams is paramount. Okay? It's highly expected of you. He has good physics students. You are able to make good diagrams. Okay? Because half the battle is won if you understand the question. And what, how, how better to understand the question by making a diagram. Okay? So, how high will the questions are? How high will the ball rise? Okay? Second is how long will it take before the ball to hit the ground? Okay? So, very easy questions. I'm going to use the three equations of motion to solve this. I know that the initial velocity of travel is 20 meters per second. Okay? And whenever some ball reaches the maximum height, at exactly the maximum height, the velocity of the body is going to change in the opposite direction. But before that happens, it is at exactly the maximum height, the velocity has to become zero. Okay? So, what I can say is final velocity at the maximum point is going to be zero. Zero meter per second. Okay? And for the first part, I have to find out how high will this rise. So, essentially, what I can do is acceleration I can take as minus 10 meter per second square. It's given in the question. Why minus? Because the gravity is always acting in the downward direction. And what I always do is I'll always take upwards as positive and downwards as negative. Okay? So, that is why I have taken the initial velocity as plus 20 meter per second. Because I threw the ball upwards. Okay? And acceleration is minus 10 meter per second square. Now, if I use the second equation of motion which says v square minus u square is equal to 2 As. In here, if I plug in the value v square is zero square, this is 20 square. Two times acceleration is minus 10 into s. Here, you will see e minus minus will cancel. And s would come out to be 20 meters. You can just simplify this. On the left hand side, you have 20 square. On the right hand side, you have 20 s. So, one of the 20s will cancel out and displacement would come out to be 20 meters. So, the answer to the first part is how high the ball will rise. It will be equal to 20 plus 25 is equal to 45 meters. Okay? Why did I add 25 to this? That's because the ball was already 25 meters from the ground. Okay? So, if they are asking how high the ball will rise, the answer is 45 meters from the ground. So, the first part's answer is the ball will be 45 meters high from the ground. Okay? From the ground. So, you have to mention this. Okay? Mention this. Where can you take this 45 meters from? From the building, from the multi-story building it is thrown from. It will rise 20 meters above that, above the building. Okay? But from the ground level, it will be total of 45 meters. Now, the next question they're asking is how long will it take before the ball to hit the ground? Okay? So, what you can do is, you can do this in parts. You can first calculate how much time it took for the ball to rise to that maximum height. And from there, you can find out how long it will take to fall, come back 45 meters. Okay? You can do that like that. But I will do it in one single step. How? Because, you know, I want, I like shortcuts. We'll take a shortcut. If the ball initially started with, let's say, 20 meter per second in upwards direction, it will go high, but it will again come back and then it will reach the ground. So, if I ask you what is the displacement? Not the distance. What is the displacement? The displacement is going to be, how much? It will be minus 25 meters. It went 20 meters up, then it again came back 20 meters and then it fell down by 25 meters. So, if I say the final displacement is equal to minus 25 meters, there is nothing wrong with this. Okay? There is nothing wrong with this. What is the acceleration? The acceleration is minus 10 meter per second square. The initial velocity is plus 20 meter per second because I threw the ball upwards. Now, if I use the equation s is equal to ut plus half at square, half at square and in place of s, I write minus 25. In place of u, I write 20t plus half into minus 10 into t square. There is nothing wrong with this equation. So, let's try to solve this. Okay? So, in terms here, so this will be minus 5t square. Plus 20t, right? And then plus 25 is going to be equal to 0. Now, if I cancel this with, what should I do here? Yes, I divide this equation by 5. This becomes minus t square plus 4t plus 5. Okay? Now, I can do a middle term break here, minus t square minus, okay? What I'm going to do here is that 5t minus t plus 5. This is the middle term break I'm doing, okay? 5 into minus 1 is minus 5. So, this is matching, okay? Quadratic equation, simple quadratic equation I'm solving. So, from the first time I can take minus t as common and I'll end up with minus t minus 5, okay? And from the second term, if I take minus 1 common, I'll get t minus 5 again, okay? So, this will be t plus 1, t minus 5. This is the factors for this. So, that means there are two values of time, 5 and minus 1. So, obviously, we will neglect this minus 1 term, okay? So, t is going to be 5. So, that means from this equation, I can simply solve for the b part, it will take 5 seconds to reach back, to reach back to the ground, reach back to the ground, okay? So, that is how you answer a question. Again, let me tell you, if you have done the question in such a way I found out from the top of the building to the highest point, this is the time it took. And from, again, from that position back to the ground, this is the time it took. And you added both of them, that way is also correct, okay? That way is also correct, okay? Next question. In this question, it says a woman starts from home at 9 a.m. walks at a speed of 5 kilometers per hour on a straight road up to her office, 2.5 kilometer away, okay? So, let's first write down. Initially, the time is 9 a.m., okay? She walks at what speed? Speed is given as 5 kilometers per hour. And the distance up to her office is 2.5 kilometers, okay? If I represent this in a diagram, this is her starting position. She goes to her office, okay? She goes to her office. She goes to her office. And then this distance is how much? This distance is 2.5 kilometers. This distance is 2.5 kilometers. Stays up to office till 5 p.m., okay? She works up to 5 p.m., good for her. Good for her. She's working only up to 5 p.m. And she returns home by auto with a speed of 25 kilometers per hour, okay? So, her returning speed is 25 kilometers per hour. And obviously, she comes back to her home. That means she again travels 2.5 kilometers. Choose suitable scales and plot the xt graph. So, that means I have to plot with all this data, I have to plot a displacement time graph. Displacement time graph. I know the initial displacement will be 0. The final displacement should also be 0, okay? But I don't have, I don't exactly know what is the time intervals. I know she starts at 9 a.m., but when does she reach? Okay. When does she reach? But I can find it out. There is no acceleration here. So, if she travels at 5 kilometers per hour and a speed and the office is located 2.5 kilometers away, then time taken for her to reach the office would be how much? It will be 2.5 kilometers divided by 5 kilometers per hour. Okay. So, that would mean it will be half an hour. Okay. So, 30 minutes. 30 minutes it will take for her to reach her office. And why am I finding out the time? Because on the x axis, I have to plot the time. I have to plot it like this. This is 9 a.m., okay? This is 9.30 a.m., okay? And this is, let's say this is 5 p.m. And I need to understand you when she reaches back home. Okay. That time will be given by 2.5 kilometers divided by 25 kilometers per hour. 25 kilometers per hour. Because when she is returning, she is returning in auto and she is traveling at 25 kilometers per hour. So, this 2.5 kilometer distance will be covered at 25 kilometers per hour. So, that means this is 1 by 10 of an hour, which would mean 60 minutes by 10, which is 6 minutes. Okay. So, she will be back home at exactly 5.06, exactly at 5.06. So, I have plotted important points on this time axis, okay? Now, you know the slope of a displacement time graph represents velocity, okay? So, initially she is traveling at 5 kilometers per hour, okay? So, the line will be something like this, okay? She reaches the office at 9.30. From 9.30 to 5.00 p.m., her displacement is going to be same. It's not going to change because she is just sitting there and working, okay? But when she is coming back, the displacement will again start decreasing and it will be back to zero. Excuse my straight lines. It's not looking straight, but this is a straight line, okay? This is a straight line. So, this is how the graph will be represented, okay? One more thing I need to add here is the slope of the line from 9.00 a.m. to 9.30 will be less compared to the one which is so from because she is traveling in auto here, here the slope will be more steeper, okay? The slope will be higher. So, here the velocity will be higher, here the velocity will be low, okay? So, this is how you can represent her entire motion, okay? This is how you can represent her entire motion, okay? Let's do this question, okay? A drunkard working in a narrow lane takes five steps forward and three steps backwards, followed by five steps forward, again three steps backwards and so on. Each step is one meter long and requires one second. Plot the xt graph for the motion and determine graphically and otherwise, how long the drunkard will take to fall in a pit of 13 meters, okay? So, if I just try to do this analytically, this is very, very simple, okay? So, if there is a guy, his initial displacement is let's say five, okay? The next displacement will be five minus three, which is two meters, okay? The next displacement will be this value plus five, which means seven and then the fourth displacement will be seven minus three will be four, okay? So, let's keep going, let's see where it ends, okay? X five would be four plus five, which is nine, right? And X six, the sixth step, the sixth iteration of this step will be nine minus three, which is six. So, X seven would be six plus five, 11 and X eight is eight iteration here will be 11 minus three equal to eight, okay? So, the next here, X nine would be eight plus five, which is eight plus five is how much? Nine, ten, thirteen and this is where he falls into the pit, okay? Because if you read the question completely, it will say the drunkard is going to fall in a pit, which is 13 meter away, okay? Now, what I have to do is put this in a graph, put this in a graph, okay? Plot the xt graph for his motion. You have to understand here, the velocity is given to us, the velocity is given to us. Each step is one meter and it takes time of one second, okay? So, the velocity is how much? It is one meter per second, okay? So, the slope of the lines will be 45 degree because I told you in a displacement time graph, slope represents what velocity? So, slope is one in this case, okay? Slope is one in this case. So, slope will be equal to one and slope is also represented by tan theta, okay? So, each lines, the angle would be how much? 45 degrees, okay? It will be 45 degree lines we will be making, okay? So, let's make a very nice graph here, okay? And then you will plug in the values, what are the important values here? 2, 4, 6, you have to plot, okay? So, I will keep on plotting till 14. Well, let's make it a little bit big, okay? Also, let's make it a little bit pretty also, okay? So, let's use this line here. Now, it's looking nice, okay? So, let's plug in the values here. Let's take it 2, 4, 6, 8, 10, 12, okay? And this is where the pit is, 13, okay? Let's put with the orange marker the in-between values. Let's call 4 between, this is 5, this is 5, 6, 7. This is going to be 9, this is going to be 9, this is 11. And yeah, that's 13 should be in orange actually, okay? 13 is in orange, okay? The reason I use different colors is just because I don't want, you know, you to get confused, okay? So, initially the block goes up to 5 at 45 degrees, okay? Remember, at 45 degrees. I have to make the line as 45 degrees. So, I go up to 5 and then I come back to 2. It's not, I'm not the drunkard, okay? So, the drunkard is going up to 5 and it is coming back to 2. Again, it will go from 2, how much it will go? 2 to, it will go to 7. So, from 2 it will go up to, I'm trying my best here to make straight lines. So, he'll go up to 7, okay? So, let me plot this here. And then he is going to come back to 4, okay? And so on, okay? So, next he will go to 9, okay? Then he is going to come back to 6. Then he is going to go to 11. And then he is going to come back to 8. And then he goes to 13 and he falls in the pit. So, the displacement time graph, let me write down the values here. This is time and this is going to be the displacement, okay? So, remember, this line, every time this will be 5, 5 seconds because each step takes 1 second. And this will be 8, 8 seconds because 5 plus 3 is 8. Time will not get subtracted. You do understand when I'm going in this direction, like 5, I went in this direction and I came back 3 in this direction. Doesn't mean that total time taken will be 5 minus 3, okay? It was, time will keep on increasing whether you're coming back or going, okay? So, this will be 8 plus 5, which is 13 seconds. Then again, 8, 13 plus 3, 16 seconds and so on, okay? So, if you, this is how the graph will look like. This is how the graph will look like, okay? So, you can, you know, give all designations to this and this is how the graph should look like, okay? Any doubts regarding this? Put it in the comments. Let's do this next question. A player throws a ball upwards with initial speed 29 meter, 29.4 meter per second, okay? So, there is a ball, there is a ball which is thrown upwards with 29.4 meter per second velocity. The first question is, what is the direction of acceleration during the upward motion of the ball? This, this words upward motion is kind of redundant here, okay? Because the direction of acceleration does not depend on where you throw the ball, okay? If you have a ball in air, it will always be attracted towards the earth. No matter whether it is going upwards or downwards, right? The ball will always get accelerated downwards, okay? So, the direction of acceleration for the first option, the answer is gravitational acceleration equal to 9.8 meter per second square will always be directed downwards no matter where the ball is going, okay? So, this is the first, the answer to the first question is that the acceleration of the ball will be 9.8 meter per second in the downward direction. What is the next question? The next question is asking, what are the velocity and acceleration of the ball at the highest point of its motion, okay? Again, there is a redundancy in asking about the acceleration. I don't care where the ball is. I don't care where the ball is. The ball, if it's not static, okay? If it has not fallen on the ground, if it is anywhere in its motion from upwards to downwards, the acceleration will be equal to 9.8 meter per second square, okay? Some people will be saying, but sir, at the maximum height, the velocity will be zero, no? Yes. So, why is the acceleration still 9.8? The answer is, does acceleration depend on the velocity? No. Acceleration, we cannot have a body whose acceleration is downwards and the body. So, imagine if a body is at rest and we give an acceleration to it, at that particular instant, the initial velocity is zero and it has acceleration, right? So, this is the same case. It may be at the highest point, but its acceleration has not become zero. Acceleration is independent of velocity and it is dependent on the force. Acceleration comes from the force, okay? And what is this force here? Gravitational force. The earth is attracting the ball. So, when the ball is going and stopping mid-air, does the earth stop pulling it? No, it does not. It's still pulling it. So, that means the acceleration at even the highest point will still be 9.8 meter per second square in the downward direction, okay? Let's do the third question. The third question says, choose the x equal to zero and t is equal to zero to be the location and the time of ball at its highest point. Vertically, downward direction to be positive direction of the x-axis and give signs of position, velocity and acceleration of the ball during its upward and downward motion, okay? So, here they are asking us to do a certain assumption or certain convention we have to follow here, okay? So, let me draw the diagram here, okay? So, here I think this is where we need to definitely make a diagram, okay? And I'm going to make a diagram with a different color, okay? So, this is the ground. This is the ground. So, there was this one ball which was here, okay? And it was sent upwards. It was sent upwards and it reached a maximum height. It reached a maximum height and then it started coming down. It started coming down. It started coming down and it fell somewhere here, okay? Actually, it will fall back at the exact same spot, but just because I want to, you know, designate the different directions, that's why I made it like this. Actually, this is one demotion. That means the ball will follow the same path going up and same path going down, but they will overlap on top of each other. That is why I have made it a little bit on the far side to just, that is just for my convenience to write down the things. Actually, the motion is just on top of each other. So, technically, this ball is on top of this ball, okay? So, I hope you understand that, okay? Now, the question says, let's assume this is x equal to 0 and this is t equal to 0. This is given in the question. If you look at this, take x is equal to 0 and t is equal to 0 to be the location and time of the ball at its highest point. So, at this highest point, this is now considered the origin and this is where the experiment is going to start, okay? And they have also mentioned one more thing. Take downwards as positive and upwards as negative. This is what they have given here. Vertically downward direction to be the positive direction of the x axis. So, any measurement done in the upward direction would be negative. Any calculation or any vector which is pointing downwards will be a positive vector, okay? So, now they're asking, give the science of position, velocity and acceleration of the ball during its upward and downward motion, during its upward and downward motion. So, let's first analyze what is the upward motion look like. So, during the upward motion, the velocity is pointing upwards. Okay, let me use a different color here. Let's switch back to white, okay? So, during this ball, when the ball is going up, the velocity is upwards. This is 29.4 meter per second, okay? And when going up, acceleration downward is 9.8 meter per second square. The displacement here is in the upward direction. Let's say plus s, let's call this plus s, okay? Why plus? Because, oh no, actually this should be minus s. Why? Because upward displacement will be negative, okay? So, this is going to be minus s. When coming down, this displacement will be plus s, why? Because downward we have considered plus, okay? And acceleration will be what? Plus 9.8 meter per second square, okay? Going up also, the acceleration will be plus 9.8, but the velocity will be minus 2.9, 29.8 meter per second, okay? So, all the conventions I have written, okay? So, let me write down this in an orderly fashion. So, upward motion, upward motion. During upward motion, the initial velocity is minus 29.8 meter per second. Why? Because upward is considered negative and downward is considered positive. And the ball was going upwards, okay? What is the acceleration? Acceleration is plus 9.8 meter per second square. And how much is the displacement during the upward motion? It is minus s, okay? I don't know how much this value is. Actually, I can find it out, okay? So, let's find it out. If a ball is sent upwards with a velocity of 29.8 meter per second, what is the displacement? It will happen. I can use b square minus u square is equal to 2 as, 2 as, okay? So, from here, final velocity is obviously 0. At the highest point, the velocity is 0. Minus u square 29.8 square is equal to 2 into 9.8 into s, okay? So, from here, this is automatically coming out. See, the displacement is going to come out negative. This is going to come out negative minus 29.8 into 29.8 divided by 9.8, okay? So, let's do this on a piece of calculator, okay? Calculations like these will not be in your UTS or competitive exams. They will always make the calculations easier because we are after the concept. We are after the concept, we are not after the calculations, okay? As engineer, you have to have some decent calculations, okay? But doesn't mean you will be able to do this, okay? Doesn't mean you need to be able to do this. In decimals, at least, we don't need to worry about. So, 29.8 into 29.8, this divided by 9.8, okay? So, this will be 9, close to 91, 90.5, okay? 90.5 meters. This is close to 90 point minus 90.5 meters and minus indicates what? In our reference, minus indicates upwards. So, this perfectly makes sense, okay? The body is going to go upwards by 90.5 meters, okay? So, if it went upwards by 90.5 meters, obviously, it's going to go down by 90.5 meters, okay? So, during the downward motion, during the downward motion, okay? Initial velocity is going to be zero, okay? Acceleration, again, it will be plus 9.8 meter per second square only. But this time, the displacement would be plus 90.5 meters, okay? In the upward motion, the displacement will be minus 90.5 meters, okay? So, this is how you have to solve this question. This is how you have to solve this question. Let's do the fourth one. Let's do the fourth one. To what height does the ball rise and after how long does the ball return to the player's hand? So, the first part we have already solved. What height the answer is 90.5 meters? The ball will rise up to 90.5 meters. And in this same equation, if you apply v is equal to u plus 80, you can also find out the time. P is equal to u plus 80. So, all these data I'm going to put. Final velocity is going to be zero, okay? U is going to be minus 29.8 meter per second. Acceleration is going to be plus 9.8. T will be T, okay? So, again, if you see, there is no mistake in this equation. How do I know this? Because if I take 29.8 on that side and divide it by 9.8, there is no mistake in this. If you take anything wrong in the convention, not the convention itself, for example, if you took upwards and positive and downwards is negative, and then somewhere you mess up the direction, that is when your time will come out negative, then immediately you have to understand, okay, I made some mistake in taking one of the values. If you take any convention, upward, positive, downward, negative, or downward, positive, upward, negative, the answer will come out correct no matter what if you followed that convention, okay? So, any convention you want to use, you can use, but always be careful, he don't mix the conventions. For, you know, if you take upwards as positive, all the values should be positive, which are pointing upwards, okay? That is the only thing you have to be careful about. So, I am doing 29.8 divided by 9.8, which means this is the time taken going upwards is close to three seconds, okay? So, going up and going down, the time taken will be three seconds. So, this is how you solve all the parts of these questions. If you have any doubts, put it in the comments, okay? Okay, so, let us do this one, this question, okay? In this question, there are four graphs given to us, and we have to analyze each graph here, and we have to say if the graph is possible or not, okay? And if it's not possible, we'll also have to give a proper reasoning for this, okay? So, the first graph here is a displacement time graph, it's a displacement time graph, which looks something like this, looks something like this, okay? So, in this graph, what we have to see here is that, if I make a line parallel to, you know, x-axis, that would mean, for example, this time is let's say five seconds. So, exactly at five seconds, if you see, this line intersects this graph at two points. That means, what this graph is telling me is that, at exactly five seconds, there are two positions for a single body, okay? Because what is written here, what is written here? It's, this is supposed to represent one-dimensional motion of a single particle. Now, tell me, why is this not possible? Because if you draw it like this and you say, at a given time, this graph shows that the position that there are, that there are two positions, two positions for a single body, okay? Single body. So, tell me if this is possible or not? Not possible, okay? So, this graph is not possible, it's not possible. Let's look at the next one, okay? And this one can be exactly the same exact reason I can give for this also. If I take a certain time, if I take a certain time, let me zoom in a little bit, if I take a certain time, this graph is indicating key for the same body, it has two different velocities. Two different velocities, which is not possible, okay? So, the second graph is also not possible. This is not possible because for a given time, there are two velocities for the same body. The third graph is a graph between speed and time. It's a graph between speed and time and we see at some locations, the speed are negative, the value of speed is negative. So, again, this graph is not possible, the reason is different. This is not possible because this is a graph between a scalar, it's a scalar quantity versus time, okay? So, this scalar quantity speed is never negative, it's never negative, okay? Speed can only be positive. Colossity can be negative because the negative sign will represent the direction, but speed is a scalar quantity. There's no way speed is going to be negative. So, this graph is not possible because this graph goes below the x-axis, which means it is giving me some negative values of speed. The last graph is a graph between length and time, okay? Length and time. This graph is possible because I don't see any reason. So, initially the length is increasing and then it is decreasing and again it is increasing, okay? So, length is not changing, it's never negative or any given time, there is no two lengths and length is a scalar quantity, it is always positive and it is evident for this graph, okay? So, this D graph is the only possible graph, the other graphs are not possible, okay? So, yeah, that's it, okay? Now, let's do this question. In this question, they're saying that there is a displacement time graph, xt graph of one-dimensional motion of a particle, okay? So, initially it is given that this is 1D motion, this is 1D motion, okay? This is a displacement time graph. Is it correct to say that from the graph that the particle moves in a straight line for t less than 0, okay? So, before the observation, before the observation, if you see the blue graph, the displacement is essentially not changing, okay? The displacement is not changing. So, is it true or can I say that it is moving in a straight line so the answer is no? For the first part, t less than 0 from this region to this region, I think the displacement is not changing at all, okay? So, I cannot say whether it's moving in a straight line or not, okay? And the next part they're saying is that from 0 to sometime t, the body has taken a parabolic path. Actually, this is a conceptual question. The concept behind this is that path means the actual path of the body, which has nothing to do with this graph. This is a graph between a displacement and time, okay? If I made a graph between, let's say, if I plotted the Cartesian plane of that area between, let's say, let's call this x length, this is y length, and then I gave you the path like this, then I say, oh, this path is parabolic. But here the graph is between displacement, a quantity versus another quantity, okay? This is not a coordinate plane where we are representing the movement of the body. This line here does not mean that the body is traveling like this, okay? This is not traveling like this. A body could travel in a straight line, a body could travel in a straight line, and then its displacement from the origin is calculated and plotted on this graph. So I can say that the displacement, the rate of change of displacement is parabolic in nature, okay? That I can say, but the body itself is not traveling in a parabolic path. That will only happen, that I can say only if I have a graph between x and y, okay? The actual area where the body is kept, if that coordinate plane is given to us and there it is drawn, oh, the body is moving like this, like this. Then I can say, oh, it is a circle or it is a parabola. But this is a graph between the displacement and time. It has nothing to do with the actual motion of the body, okay? If not, suggest a physical context for this graph, okay? So imagine if you have a car, if you have a car, which you start accelerating, if you start accelerating, okay? If you start accelerating, what will happen is initially the displacement will increase slowly, slowly, okay? So for a given amount of time, let's say for one second, this is the displacement at the beginning, but now as the body is accelerating, the next one second it will take a bigger chunk of displacement, then it will make even bigger chunks of displacement, then it will make even bigger chunks of displacement. This is what this graph is representing, okay? So initially this increase in displacement is very, very less. For the same amount of time, for the same amount of time, this displacement is bigger than the, the displacement in second is, so let me zoom in so that you are able to see better, okay? Let me change my eraser also, sorry about this, okay? Now it is better, okay? So what you need to understand from here is very, very simple. If I take equal amounts of intervals of time, let's say this, this, this, okay? These are equal intervals of time, okay? From here to here, the time is same. From here to here, the time is same. From here to here, time is same, approximately. Now if you check the displacement regarding this, if you check the displacement regarding this, you'll exactly understand what I mean by this. This displacement is small, this displacement is bigger than this and this displacement is even bigger than this. So what this means is that the rate of change of displacement is getting bigger and bigger, which is the physical significance of which is an accelerating car, okay? For the same interval of time, the body will cover more distance in the same amount of time interval, okay? So that is how you can answer this question, okay? Hope you understood any doubts, again put it in the comments, put it in the comments, okay? Let's move on to the next question, okay? There is no next question.