 this possible? The answer is distance time. Distance cannot decrease. It will always accumulate. So the slope of distance time graph will always be of distance time graph. The distance always increases with time and if suppose this is your displacement time graph. Can you conclude this displacement time graph into distance time graph? How to do that? Somewhat like this. The distance will keep on increasing. Your distance will be 0 somewhere here. So it will slope was 0. So these are some final things which if I ask you in an isolated manner you will be very proactive this is not possible. But if I give you a question where this is being utilized probably one small part of the question this is where you have to pay attention and identify yes this is not possible this is impossible. Then you may not be probably able to identify it or active enough to identify it. That kind of thing will come only when you solve a lot of questions. A good strain accelerating uniformly on a straight railway track. This is straight. Approaches and this is the story. Its velocity is u and the large room passes portion is passing with velocity. If you would have a question probably. It does not matter. Uniform acceleration. I think this similar question we did. I think that was correct. I think this is where you will be right now. You have to express in terms of u. Here front part you cannot take point. Suppose velocity of the middle part is when this from this pole lets say the train is accelerating with acceleration a. So I can say that v1 square is equal to u square plus 2a into x by 2. There is no back portion. So I can also say that. So I will get first equation which is what u square which is a to x and substitute it over here. So v square is equal to v1 square plus v1 square minus u square. So v1 is equal to v square plus u square by 2. Exactly same as previous. Do you remember the first and second question? So same question can be asked in 100 different ways. That is why you could have studied in class 10 where you try to type a 2 question type 3 problem. There will be some many type of questions. What is the middle of the distance? The first thing is actually the after class. When I am giving you an option, all of you are staying back just one of left. So do not feel pressurized until you feel that since others are staying back I should also stay back. I am not pushing you out by the way. Here is the question. All of you listen. Listen carefully. How do we get into the question again and again for the question itself. Then it starts all over again. And that will not come suddenly. So listen, anyways in the exam you should not spend more than 3-4 minutes type max in a question. But we are spending 10 minutes on this single question because probably now we are going to get this type of question for first time. Right now two parties start moving. Seven parties start moving from point K with velocities 15 meter per second again respectively. Again both P and Q starts from point A meter per second. The two particles move with acceleration equal in magnitude but opposite in direction. It is not given. What is given is, accelerations are equal opposite but the velocity. Listen here. When P overtakes Q at point B its velocity is 30 meter per second. I don't know the answer. I will solve it. Okay. Two of them for the same answer. And here is one of the options. Five. Yeah, so five. And the last one you will get. Yes, same. Five in the answer. Again it is already in a positive direction. When P overtakes Q at point B. So I can use V for Q. So V, I don't know, I want to find it. Q is 20 minus A into T. And A into T is what? Okay. So, done with the chapter. Till you solve minimum 250 to 10. 100 question 4 hours becomes 7. After 7 you can spend 3-4 hours. So not 100 you can solve 50 question minimum. If you are not solving 50 question you are not serious for 10 hours. Okay. That's the thumb rule. Today if you are able to solve 50 then you have some seriousness towards the cause. Otherwise you are just saying it. Thank you.