00:02
Hi, in the given problem, mass of the block is m is equal to 2 .00 kilogram and the force, variable force, acting on it along x -axis is fx is equal to beta minus alpha into t, where the value of beta is 4 .00 newton.
00:32
And the value of alpha is 6 .00 newton per second.
00:41
So putting these values here, the expression for the force comes out to be 4 minus 6 t newton.
00:53
In the first part of the problem, we have to find a maximum possible distance covered by the object under the influence of this force and the time when does it? it happened and the force acting on the block at that instant.
01:10
So first of all, using newton's second law of motion which says force is equal to the product of mass with its acceleration.
01:32
So here for acceleration we can say this is the differentiation of velocity with respect to time hence putting the value of force as 4 minus 6 t is equal to m into dv by d t so it becomes 4 minus 6 t into d t is equal to m dv then integrating both the sides with proper limits here in the left hand side it will be integrated under the limits of time and as initial time is zero on and finally let it be any time t.
02:37
Similarly right -hand side will be integrated with respect to velocity initial velocity is zero and finally it may have any velocity v now integration of left hand side gives us 4 t minus 6 t squared by having the limits of to t.
03:02
Similarly here, taking this m as a constant out, the integration of bv is just v from 0 to v.
03:11
So simply it becomes 4t minus 3 t square is equal to m v.
03:19
We can mark it as equation number one, which may be used later on.
03:23
Now again using calculus, this v is actually the differentiation of displacement with respect to time.
03:33
So again rearranging the terms it becomes 4 t minus 3 t square into d t is equal to m d x.
03:44
So integrating both the sides again with proper limits.
03:49
It becomes 4t minus 3 t square d t again having the limits of time from 0 to p is equal to m integration of bx from 0 to x as initially the displacement was 0.
04:09
So finally the integration of 4 t will come out to be 4 into t square by 2 having the limits from 0 to t minus 3 integration of t square will come out to be t cube by 3 again having the limits of time from 0 to t m integration of d x is x only from 0 to x.
04:35
So putting all these values here, it comes out to be 2 t squared minus t q is equal to m x.
04:49
So finally expression for the displacement or distance whatever comes out to be 2 t squared minus t cube divided by mass of the particle.
05:01
Hence for maximum displacement or distance covered by the particle differentiation of this x with respect to time should come out to be 0 from this expression...