00:01
This question, the position of a training helicopter and its weight is given and a test is given by this value and we have to find a net force on the helicopters.
00:07
So for that we need to find the acceleration and we know that the acceleration is nothing but the second derivative of the position.
00:14
So first we need to find the first derivative and the first derivative is nothing but the velocity.
00:20
So let's find the first derivative first derivative is going to be 0 .02 and that will be 3 t squared, the differentiation of t cube.
00:30
But then we have 2 .2.
00:32
That's it because the differentiation of t is 1.
00:34
And then we have minus 0 .06 and this is nothing but 2t over here.
00:41
Then we differentiate again.
00:42
So let's first simplify it.
00:44
This is 0 .06 t squared plus 2 .2 minus 0 .122t.
00:52
So this is 0 .06 2t plus 0 minus 0 .12.
00:57
So this is 0 .12t minus 0 .12.
01:02
This is the value of acceleration.
01:04
We have to find the value of the force at 5.
01:07
So we need to find the acceleration as well at 5.
01:12
So the acceleration is equal to 0 .12 times 5 minus 0 .12 .12.
01:19
So this becomes 12 times 5 is 60.
01:25
So we have 0 .6 minus 0 .12 .12.
01:28
So this is coming as 0 .6 minus 0 .12 is equal to 0 .48 meter per second square...