00:01
In this problem you have a fish moving in a horizontal plane, two dimensional plane.
00:05
We're given the position of the fish relative to a rock, which we take as the origin.
00:12
And we'll call this time equals zero.
00:15
We're given the velocity at that time equals zero, the initial velocity.
00:19
We're also told the velocity 20 seconds later.
00:24
And we're going to be asked some questions about acceleration and so on at this point.
00:29
And so let's get to each question.
00:33
So the first one is, what actually is the acceleration? now it tells us constant acceleration.
00:40
That means the average and instantaneous are the same.
00:45
So instantaneous a and the average are the same thing, which is delta v over delta t.
01:04
And so this would be just v1 minus v initial over delta t.
01:10
That's writing out this vector form.
01:11
Now i'm just going to leave it in vector form here just to show you that you can do it.
01:15
But in a later part i'll break it up into components.
01:19
A lot of times if you leave too much, there's too much overhead.
01:23
You got the i and j unit vectors.
01:25
You have to be comfortable with all that writing.
01:28
And it can be overwhelming sometimes, especially if you get into bigger problems.
01:36
So let's put in the two vectors.
01:39
20i minus 5j meters per second, that's v1, minus 4i plus 1j meters per second, that's the initial velocity, over 20 seconds.
01:58
And this becomes 0 .800i minus 0 .300j meters per second squared.
02:10
So this gives me that ax is 0 .800 meters per second squared, ay minus 0 .300 meters per second squared.
02:26
Could i have just broken this up into ax equals v1x minus vix over delta t? that's what it represents.
02:35
These vector equations always represent a set of the component equations.
02:43
Now let's look at this graphically though before i move on just to understand what's going on here.
02:48
Here is vi, here is v1.
02:58
Delta v is what you'd have to add to vi to get v1.
03:04
But delta v vector, 1 over delta t is a scalar.
03:10
All that would do is lengthen the vector.
03:15
Whatever delta v is, a is a certain length based on 1 over delta t.
03:21
It could be longer, it could be shorter, it could be the same.
03:26
So the acceleration vector is in the same direction as delta v, but like i said, the length could be different.
03:37
The length could be different, but it's going to be along this line, along delta v's direction.
03:42
Nothing else.
03:43
Not going to be over here, not going to be over here.
03:45
1 over delta t only will change, will only give you a different length.
03:50
That's all it will do.
03:56
So that was part a.
03:57
Part b wants to know the angle that the acceleration vector makes with the unit vector i, which is the x -axis.
04:06
So we have this as a fourth quadrant vector.
04:09
So here's a, a -y, a -x, and let's call this theta.
04:19
So tangent theta, and i just want an angle between 0 and 90, so i just take the absolute value of it.
04:27
So theta is equal to inverse tangent, 0 .300 meters per second squared, over 0 .800 meters per second squared.
04:40
And this works out to be 20 .6 degrees.
04:44
Now, how you enter this, if you're entering this electronically.
04:48
See, if you're doing this on a piece of paper, this shows you where it is.
04:54
It's in the fourth quadrant and so on.
04:56
But if you've got to enter this electronically, you have to read the instructions carefully...