 So, to say higher level questions, ok. So, here is a question to find out is acceleration or standard and just introducing question to you. This is not a test where I will wait till, right. I will wait for couple of minutes. If you are not getting it, then I will solve it. You will understand what is the concept, ok. So, here square both side we get v square equal to 2x plus 3 square. So, compare it with v square equal to u square plus 2 ax. It is a constant acceleration. So, v square is u square and a is 1. Acceleration which is derivative under root 2x plus 9 into 2, into the ends of this, that will get cancelled and then it has 1. So, v square equal to u, compare it and get the answer. Understand in 4 second, distance of the space band, right, the space band, space band also. In fact, both are same. You see this question. Ok. So, how you have done this? T is T is 4. T is? 4. Put T equal to 4, you get. Yes, you. If it is constant, then velocity into time is distance. Acceleration is 1 and then. Like that, ok, fine. And using graph, the best way is to use graph. The area of v t curve is, so this is your time and this is velocity. So, this is let us say 2. So, at plus 2 to 4, it will be T minus 2 dt, very unique. So, a simple one as well. We just create equation, we will not solve it completely. Direction of motion is not changing. So, 5 seconds. 5 seconds. 5 seconds. 5 seconds. 5 seconds. Ok. 5 seconds. So, 5 seconds. 5 seconds. 3 seconds. 3 seconds. 3 seconds. 5 seconds. 5 seconds. 5 seconds. No. B is also moving. B is fixed. But it is moving. But it is moving. You are guessing the answer, you have not solved it. You are guessing the answer. There is a difference. Even though that may be a correct answer, it will not help Now, I will help you to solve the problem. You solved it? How? Line joining between any two points in the straight line only. Always in the straight line. Yeah, exactly. Papandikul, there is no point in having Papandikul. The only one line is there. Line joining A. This is a standard type of question. Let me give you a hint. Find out the distance between them after time t. It is equal to x2 minus x1 whole square, y2 minus. Why do we have to apply physics? What is the coordinate geometry? It is x2 understand physics only. Minus to t. The distance between them is 10 minus 2t whole square. So, this is the distance minimum.