00:01
All right, so we are finding the velocity and acceleration of a particle at t equals six seconds if its position is given by s equal square root of 1 plus 4t.
00:12
Now, the thing to remember is that given a position function, position's derivative will be velocity, and velocity's derivative will be acceleration.
00:22
So that's what we need to find in order to answer this.
00:25
So let's go ahead and find, let's say for part a, the acceleration.
00:30
If i want to find ds over dt, looks like we're going to need a chain rule.
00:36
So let's write s equals u to the one -half, or just to make it more upfront, square root of u, and u equals 1 plus 4t.
00:49
Now, as i already kind of hinted at, we're going to do one u to the one -half right here.
00:56
And now, hey, we can derive both of these guys.
01:00
D .s.
01:00
Over d .u is one -half u to the negative 1 .1 .2.
01:04
One half, which can be written as one half, one plus four t to the negative one half.
01:19
I apologize, i'm gonna have to come over here with it.
01:23
Whereas d u over dt is just four.
01:31
Therefore, d s over dt, which is our velocity.
01:35
So i'll write v of t, which is just d s over dt.
01:42
That's just d s over d.
01:46
Times du over dt, which is just one half, one plus four t to the negative one half, all times four, which is just two over the square root of one plus four t.
02:15
Now having it in a form with a negative one half power will be better for when we want to derive it, but as far as plugging in stuff, like since we want to find the velocity at t equals six seconds i kind of like this form um right here in our answer because then what we can say is what is v of six well if we just plug that in what we'll get is two over and i plug in six square root of 25 six times four is 24 plus 1 25 so we'll end up just getting two -fifths as an answer or 0 .2 meters per second.
03:07
And in fact, i'm sure that's probably how the book will word it 0.
03:18
Okay, bam, there's your velocity.
03:22
Now let's find our acceleration.
03:32
Ah, oh well.
03:33
So if we are going to find acceleration, we need to derive velocity.
03:38
Now, as i mentioned, i'm actually going to be using this form right here, writing this...