 So this is the ninth question and it says derive the second equation of motion, okay? So we have to derive the second equation of motion and the second equation of motion is given like this s is equal to ut plus half 80 square graphically. So before that let's understand what this meaning of second equation is all. So as we had seen in the previous problem, let's draw a path, straight path, okay? So let's say this is the journey, okay? And this is point A and the body is moving from point A to point B. So at point, when it was point A, time was let's say T1 and at B, time was let's say T2, okay? And in this case, let's say T1 is equal to 0 or let it be like this. General case we'll describe and then we'll come to this equation, okay? So T equals to T1, particle is at A, T is equal to T2, particle is at B. And at the same time if you see the initial velocity U at this point is U, at A is U and the velocity at B is V, right? And we are assuming that there is a case of uniform acceleration and acceleration is in this direction, right? So the particle was here at T equals to P1, particle is at here at T equals to T2 in a straight line motion. Now, second equation says S is equal to UT plus half AT square. Now what is the meaning of S? So in this case, you already know, this distance traveled in this T2 minus T1 time is S, right? So what are the terminologies used here? So let me use, yeah, so S is what? S is distance traveled, since in this case it's a straight line motion. So whether it is distance or displacement hardly matters. We are assuming straight line motion in one direction, so distance is equal to displacement. So I'm saying S is equal to distance. What is U? U is initial velocity, initial velocity, isn't it? And what is V? V is final velocity, final velocity. Though in the second equation of motion, V doesn't figure out or it doesn't appear in the second equation of motion. But these are anyways the five players in all equations of motion. So this is acceleration, correct? And T is time. So these are the usual rotations. Now, graphically it says, you have to find out graphically. So let's understand what all graphs have we, anyways, learned. One is position time graph or displacement time graph. And another one is velocity time graph. So in this case, we will be using velocity time graph because we have an idea that velocity time graph area under the velocity time graph gives you displacement or distance traveled in case of straight line motion. So let's say this is V on the y-axis, T on the x-axis. And let me draw a graph. This is the graph of the given motion, okay? So let's say we are starting from T equals to T1. So here, instead of any other T1, let us say T1 is equal to 0, right? And T2 is equal to T, okay? So this is T is equal to 0. And let's say this is T is equal to T, okay? Time is T, general time T. So what is the velocity of the particle at T equals to T? So velocity here is U, initial velocity. And velocity here will be V. Let's say this is my point, this is V. And you have to expand this curve going up, okay? Sorry, not this color. Let me use the same color which I used for this axis and here it is, right? So this is V2, okay? Let me use this color, okay? So this is V2, V1, not V2, sorry, my bad, it is V anyways. So U and V, right? And this is the time T. So at time T, velocity is V. At time 0, velocity is U, correct? And we have to simply find out area of this triangle. So let's name this triangle. So what is this? Let us say this is O, A, B, C. Or not to confuse, so that you don't get confused between these A and B. I'll use other terms, so other names. Let's say P, Q, RS, P, Q, N, R. So basically what do I know? Distance traveled. So let me use color, so distance traveled or displacement. Displacement or distance, in this case same thing, so distance or hence I'm writing distance or displacement because straight line motion is equal to A area under VT curve or VT graph, correct? This is our previous knowledge. So let's find out this area. How do we find out this area? So basically if you see O, P or O, P, Q, R is a trapezium, O, P, Q, R is a trapezium, so either you can find out using the area of the trapezium, so hence what is area of trapezium or you could have done this. So let's say you drop this perpendicular here and name this point as P, capital T. So hence area is nothing but area of triangle. Let's say you don't know the area of trapezium, you can do it from that also. Let's say you don't know what is the area of trapezium formula. So area of triangle RTQ plus area of, area of rectangle O, P, P, R, is it? This is the two thing. So what is the area of triangle RTQ? So we need base and height, you know that. So this gap here is how much? Clearly V minus U, check. Right? Total is V and this one is U, isn't it? So let me draw here. So this is V and this one is U, you can check from the figure. So hence this is V minus U, this part is V minus U, no doubt about it. What about this distance? This one? This gap, this gap is simply T, in the T axis if you see, this is simply T. So let me now write here. Okay? So area of triangle RTQ is half into base which looks like T over there and V minus U is the height, value wise, isn't it? Plus area of this red triangle. So hence this height is simply U, you can see, then U into T, right? And now this is nothing but the displacement S. So S is I will write, rearrange it, you will write UT plus half T times V minus U and we already know that V minus U from first equation, from, from, from first equation of motion we know V is equal to U plus AAT, right? Or V minus U is equal to, or V minus U is equal to simply AAT, isn't it? Right? Or from here also you can see the slope, A is given by, if you don't want to take the first equation use, let's say you don't want to use that first equation. So A is given by slope of VT curve. So A is nothing but V minus U upon T, slope is V minus U divided by this T, correct? So using this two, you can write again V minus U is equal to AAT. So call it equation number one and call it equation number two and here let's complete the proof. So from, from one and two, from one and two, what do you see? So you simply substitute V minus U, okay? So half T and instead of this V minus U here, I can write this AAT, so write AAT, okay? And hence, oh sorry, there will be a T here, yeah. So hence finally what will you get? You will get UB plus half A T square. This is the second equation, right? This is how you have to find out from graph. So I, what did we do? We first of all, you know, we know that the displacement is nothing but the area under the curve VT graph and then you divide it into two parts, triangle and a rectangle. You found out the area with respect to the values given in the graph and acceleration is also known as the slope of the VT graph. So hence, acceleration you calculated V minus U and then finally, club the two equations and you got the final result. So this is what was the objective of this question. Understood? So this is how you should be solving this question.