00:01
Hi, here in this given problem for the particle given here initially means when the time is 0 second its position is such that it is 4 meter along x axis and it is 3 meter along y axis.
00:31
So, initial position vector that will be given as 4i cap plus 3j cap meter.
00:42
Initial velocity, we represent it as vi bar.
00:48
That is 2i cap minus 9j cap meter per second.
00:53
And constant acceleration that is 4 i cap plus 3 j cap meter per second square.
01:04
So in the first part of the problem at a time t is equal to 2 second we have to find final velocity achieved by this particle for which we will be using first equation of motion which which says vf, final velocity, that is equal to vi, initial velocity, plus acceleration a multiplied by time.
01:32
So, plugging in the known values for vi, this is 2i cap minus 9j cap plus acceleration 4i cap plus 3j cap multiplied by time, which is 2 seconds.
01:49
Second.
01:49
So expanding the bracket we get 2 i cap minus 9 j cap plus 8 i cap plus 6 j cap.
02:02
So the velocity final velocity in vector form that is obtained to be equal to 10 i cap minus 3 j cap meter per second and that is the answer for the first part of the given problem.
02:18
Then in the second part of the problem we have to find final position vector.
02:25
We need not to find displacement vector.
02:28
So initial position vector will not be required here...