00:01
To solve this problem, first time writing the equation for the acceleration.
00:05
So, a is equal to 3 .22 minus 0 .004 v squared.
00:13
Let it be equation 1.
00:16
Now, as we know that acceleration in differential form can be given by a is equal to v multiplication dv by ds.
00:28
Let it be equation number 2.
00:31
Now going forward from equation 1 and 2, so from equation 1 and 2, i can write the value as v, d .s, by, v.
00:50
Dv by ds is equal to 3 .2 -2 -2 -minus -0 -4 .4 .2.
01:00
4 v square.
01:02
Now going forward, now i will apply integration on both side with velocity limit from 0 to vb and displacement limit from 0 to 600 feet as given in the problem.
01:16
So i can write 0 to 600 d s is equal to integration of 0 to b b b by 3 .2 2 .2 to minus 0 .004 d b.
01:41
Now going forward and solving it further, i can write the value of as s from 0 to 600 is equal to on integration...