 Alright friends, so here is another question, this is question number 2 from kinematics chapter which has been asked in one of the J paper, it was 1979, so let us see what it is about. Displacement of a particle moving in one dimension under an action of constant force is related by the equation t equals to root x plus 3, where x is in meters and t is in seconds, you need to find displacement of the particle when the velocity is 0. Now it is asking for the displacement when velocity is 0 but the equation which you have does not have velocity in it, so what will be the first thing you should do, you should try to find the velocity and velocity is what? Velocity is equal to dx by dt. So what I will try to do here is that I will try to differentiate this equation. So what I will get here is let me differentiate with respect to time only, so I will get 1 is equal to 1 by 2 root x into dx by dt, fine. So from here I will get velocity which is dx by dt to be equal to 2 root x, fine, so this is equation number 1. Now dx by dt is velocity, so when velocity is 0 you can see that even x should be 0, right. So the answer for the first question is that the displacement should be 0 when the velocity is 0, okay. Now let us move to the next part of the question, the work done by the force in first 6 seconds, okay. We need to find work done in 6 seconds. Now it says that work done by the force, fine. So what we will assume that it means all forces acting on it, what is the net work done by those forces, fine. Now this equation that is work energy theorem should come in your mind, work done is equal to change in the kinetic energy, fine. So let us try to find out how much kinetic energy is changed in first 6 seconds, okay. So that will be equal to what half m into v2 square minus v1 square. Now let us try to find out what is the velocity as a function of time, okay. In order to get velocity as a function of time, first I need to write this down to x to the power half, right. So what I will do, I will substitute the value of x to the power half as a function of time from this equation. Let us put it in brackets. So from this equation what I will get is root x is equal to t minus 3, fine. So if I put it here, I will get 2 times of t minus 3, this is velocity, okay. So velocity when t is equal to 0 is what? 2 into minus 3 that is minus 6 meter per second, fine. And velocity when t is equal to 6 seconds will be what? 2 into 3 which is 6 meter per second, fine. So the velocity magnitude remains the same just that the direction is reversed. So if you put it here, you will get half m into v2 which is 6 square minus of minus 6 square, fine. So you will get this as 0, right. So answer for the second part of this question is 0, right. So like this you can do this particular question. So next video we will come up with another question.