 All right friends welcome again. This is Dheeraj from Centrum Academy So we are creating video solutions of all the past year J questions. Okay, so I take care of physics So the we are taking the first chapter of physics, which is kinematics Okay, so these are the questions which have been asked in previous year J exams and as you know that Earlier J used to be a subjective paper. Fine. So I am assuming all of these subjective questions are useful in the current scenario of the J advance paper also So we'll discuss all of these one by one. So stay tuned and You will get to learn many concepts from these videos. Okay, so let us see what is this question is all about, right? So car accelerates from rest at a constant rate alpha for some time, right? So it has an initial acceleration of alpha okay, and It decelerates at a rate of beta to come to rest If the total time elapsed p second you need to find out maximum velocity and total distance traveled. Okay now Where do you think the maximum velocity will be attained? The velocity will keep on increasing till The object is accelerating. Isn't it? So till the time there is an acceleration Alpha the velocity will keep on increasing. Isn't it? So the maximum velocity will be the point where acceleration Goes down and it starts to decelerate Okay, so if we visualize it in a graph let's say I plot Velocity versus time graph because it talks about time and velocity, right and If we draw the graph, let's say it accelerates at an angle Sorry, it accelerates at constant acceleration alpha. So this line Represents the constant acceleration line. Okay, let's say it makes an angle theta Okay, and then after this point onwards, let's say this is time t1 Fine after that point onward it decelerates. So what will happen? It's velocity will slowly decrease and it will go down to zero fine So suppose it goes down to zero at this time and this time is given as t Getting it. So this is deceleration beta. Let's say this is a B and this point is C So a to B is constant acceleration and B to C is constant deceleration right If I say that this time is t1 time of acceleration if I say is t1 then time of deceleration What will I say? It will be equal to t minus t1 Isn't it t is a total time t minus t1 will be the total time of deceleration Okay, now what is given here is that the This line a to B line The slope is what the slope should be the acceleration only alpha Isn't it? So if I say that at point B It has velocity of VM, which is the maximum velocity then the slope of AB is VM divided by t1 Okay, so this is alpha that is given and B to C if you take mod of the slope This should give you beta Okay, so the mod of the slope is what? VM divided by t minus t1 Although the slope is negative All right, but then we are just taking a magnitude of the slope because magnitude of the acceleration and decelerations are given So this is equation one and that is equation two, right? You can also solve this question by using v is equal to u plus a t and you'll get the same equation Okay, now if I divide one and two what will I get I'll get t minus t1 divided by t1 is equal to alpha by beta Okay, so t minus t1 will come out to be alpha by beta times t1 So t1 will come out to be equal to beta divided by alpha plus beta times t Okay, so this is t1 or this is a time till it will accelerate Let's say this is equation number three So if I use this t1 in equation number one, I will get maximum velocity, right? The maximum velocity is alpha into t1. So that will be equal to alpha beta divided by alpha plus beta times t Okay, so like this you can solve the first part of the question Okay, now, how will you answer the second part of the question? All you have to do is to divide this entire journey into two parts You just find the distance traveled from A to B and Then distance traveled from B to C. Why I'm dividing it into two parts because in entire journey if you take The accession is not constant But if you split it up like if you split A to B accession is constant and even B to C it is constant deceleration fine, so like that if I See distance traveled when it goes from A to B It will be what? ut1 Plus half alpha t1 Square right initial velocity is zero isn't it since initial velocity is zero it will be equal to half Alpha into t1 which you have already got from equation One sorry equation three, but this will be equal to beta square divided by alpha plus beta Whole square into t square fine. So this is distance traveled a to B You can also find out distance traveled from B to C you know the velocity is this at point B, right? So it's the same equation when you apply as From B to C is going to ut u is what? Alpha beta divided by alpha plus beta This into t is you Okay into t is what t will be equal to t minus t1 now Okay, so if you use the value of t1 from equation three and put it there this and now there will be a Negative sign here because it is decelerating so minus of half Beta into t1 square guys Okay, so just add up as AB and as BC you'll get the total distance traveled by the car Okay, we are assuming one thing here is that car is moving in a straight line. All right So that's it with respect to this particular question. It will soon come up with the next question