00:01
So, let us start with the concept which we are going to use here for this question that is second equation of motion third equation of motion v square is equals to u square plus 2 a s where we find the speed u initial speed x s in its displacement.
00:16
So, we if we consider v is a maximum speed and let the acceleration by travelling of x of meters can be calculated as v square is equals to starting from rest that is why u will be 0 2 times of acceleration we have been given 1 .5 times of the displacement we just let to be as x.
00:37
So, x would become equals to v square by t right.
00:41
So, distance travelled in deceleration this is what for acceleration time now in deceleration what will be the distance travel.
00:53
So, it should be total distance is of 1 kilometer right already it is covering the x kilometers.
00:59
So, obviously the remaining is going to be deceleration and that is why it will be 100 1000 minus of x.
01:05
So, it is starting from rest that is why final speed is going to be 0 because it is going to stop and the initial one we have just found.
01:17
So, this will be v square minus 2 into 2 which is the deceleration times of the distance 1000 minus of x.
01:25
So, you will get v square is equals to 4000 minus 4 into observes the values that is v square by 3 you will get a required value of v is equals to 41 .4 meter per second.
01:41
So, basically we if we try to find the time of acceleration t1.
01:46
So, for that we have to use first equation of motion v is equals to u plus 80.
01:49
So, let us say this as 80 of 1 right whereas, as we have calculating the time for acceleration...