00:01
In this problem, we have h of t given as 25t minus 4 .9t square.
00:08
So, for the first part of the problem, we want to find what is h at 1.
00:13
So, this is simply 25 minus 4 .9 which is 20 .1 meters.
00:23
Next, we want to find what is ht minus h1.
00:31
So, ht minus h1, this is simply 25t minus 4 .9t square minus 20 .1.
00:44
So, this comes as simply minus 4 .9t square plus 25t minus 20 .1.
01:04
From here, we can take t minus 1 common and this is simply minus 4 .91t minus, sorry, plus 20 .1.
01:43
So, we see that t minus 1 is indeed a factor of ht minus h1.
01:50
Next, part b, we want to find the average velocity.
01:54
This is v average.
01:57
This is nothing but v at t minus v at 1...