 In this question, it's given that a car starts from rest and moves along the x axis with constant acceleration. So, third information is constant acceleration and the value is 5 meter per second square for 8 seconds. So, time is also given. If it then continues with constant velocity, what distance will the car cover in 12 seconds total since it started from the rest? Now, the first step in all such problems is to draw a diagram, a well representative diagram. So, let's say this is x axis x axis. So, car starts from rest and let's say it starts at origin. This is origin and it starts and moves along the x axis with constant acceleration. So, when it starts from rest, so hence initial velocity u is equal to 0 and it's given that it is constant axis. It moves with constant acceleration and the value is 5 meter per second square. And then time for which constant acceleration acts is t1 is equal to 8 second. So, let's say this was here is let's say t is equal to 8 second and let's say here is t is equal to 12 seconds. So, in let me, so in this phase, a is equal to 5 meter per second square in this point here from here a is equal to 0. So, this phase is uniform uniform velocity is it? Why? Because acceleration is 0. So, but uniform velocity but what will be the velocity of the value of the velocity which will be nothing but the after acceleration whatever is the value this body achieves in terms of its velocity. So, whatever is the velocity here the same velocity will be the same velocity will be throughout. So, let's say it moves in this phase with the same velocity v. Okay, now let's analyze the so hence we can divide this whole problem into two parts. Okay, let's analyze the first part what's happening here in this case clearly the same information which is mentioned here is for phase one this is for phase one. Okay, now so let's find out the velocity v. Okay, so basically we have to find out the distance. So, distance from origin to the final point at t equals to 12 seconds. So, distance total distance travelled will be s1 and s2. So, we have to find to find s1 and s2. Right? So, in case of s1 we know u t plus half a t square can we apply it? Okay, so we have let's write the three equations which we know. So, now we know v is equal to u plus a t s is equal to u t plus half a t square and v square is equal to u square plus 2 a s. So, out of all these all these we'll use that equation in which most number of variables are known. For example, u is known, a is known and time is known. So, the best v is unknown. So, best equation to be used is the second one. So, let's use that and then directly we have been asked for s1 also. So, let's find out u was 0, t was 8 plus half into a was 5 and again t was 8 square and obviously this will be in meters. Everything is in meters or meter per seconds or meter per seconds square. So, this will be in meters, isn't it? So, hence it will be this is 0. So, this is 5 into 64 upon 2, isn't it? So, this is nothing but 5 into 32 meters that is nothing but 160 meters. So, in phase 1 the body travels 160 meters. So, now what will be phase 2 distance s2? S2 is clearly since there is no acceleration over here. So, we can write the same equation number 2. Let's try that. So, hence s2 is let's say in this case the initial velocity is in phase 2, in phase 2 initial, initial velocity is v which v this v here this one right now and then we have to find out s2 which is unknown. Time is nothing but how much time did the body take to go from this point to this point? Obviously t is equal to 12 minus 8 that is 4 second 4 second. Now s2 is unknown a is given as 0 why because it's a case of uniform velocity. So, now let's deploy all these values in s2 is equal to vt plus half into. So, don't get confused between this v. This v is here just a value. So, it's like in this case the u was v. So, half into 0 into 4 square. So, v was whatever v we do not know, but it's clearly v into 4 meters. So, if we know v we can find out st. Let's find out v what will be the value of v? So, in the first phase v is the final velocity. So, v is nothing but v is clearly using equation 1 u plus at is acceleration in the first phase. So, this is nothing but 0 plus 5 into 8 why because the time taken in the first phase is t1 or let's be let me write t1 8. So, it is nothing but 40 meters per second. Now, deploy this value here. So, what do you get? s2 is equal to 40 into 4 is equal to again 160 meters. That means total distance, total distance travels or in this case distance is equal to displacement is nothing but s1 plus s2 is equal to 160 plus 160 is equal to 320 meters. Now, did you notice that s1 is equal to s2? In this case what are the inferences we can draw? s1 is equal to s2. But this took only, this took 8 seconds, but this took only 4 seconds to cover. Why? Because in this case the body was accelerating. So, hence it started from rest, but in this case it had already had some velocity. Already it has already achieved some velocity and hence because of that velocity it takes lesser amount of time to cover same amount of distance. Is it it? So, these are additional inferences from this problem.