00:01
So here we're going to use the definition for the average velocity.
00:07
This would be the average velocity vector would be equal to the position vector at time 2 minus the position vector at time 1 divided by time 2 minus time 1.
00:20
And so this is going to be equal to, rather this would be the equation that we're using.
00:27
We know that the position vector in unit vector form would be.
00:31
4 .0 centimeters plus 2 .5 centimeters per second squared times t squared times i hat and this would be plus 5 .0 centimeters per second times t j hat and so at time equaling zero we know that then our hat is equaling 4 .0 centimeters i hat at t equals 2 .0 seconds r hat is giving us 14 .0 centimeters i hat plus 10 .0 centimeters j hat and so we can then say that in the the average velocity in the x direction would be equal to 14 minus 4 so 10 centimeters divided by 2 .0 seconds.
01:40
This is giving us 5 .0 centimeters per second.
01:45
And then it would be for the average velocity in the y direction.
01:49
This would be delta y over delta t.
01:52
There isn't any y component at t equals zero.
01:55
So it would be 10 centimeters minus zero.
01:57
So essentially just 10 centimeters.
01:59
Again, divided by 2 .0 seconds.
02:03
And so we are getting 5 .0 centimeters per second for the average velocity in the y direction as well.
02:10
For then, this would be for part a.
02:13
So continuing on for part a, we find that the magnitude of the average velocity vector would be equal to the square root of the x component squared.
02:23
So 5 squared plus the y component squared, 5 squared.
02:28
And this is equaling to two significant figures, 7 .1 centimeters per second.
02:33
This would be the magnitude.
02:35
We know that then the angle theta would be arc tan of the y component 5 .0 divided by the x component 5 .0.
02:48
This is equaling 45 degrees.
02:50
And then if we wanted to then if we wanted to sketch it, we could say that then the sketch, this would be our.
03:08
Average velocities for part a and then essentially we could say that this would be for the magnitude and then drawing it we have the y direction the x direction and we have that this would represent the average velocity in the y direction this would represent let's do equal equal length here this would be the average velocity in the x direction and then this would be the magnitude of the average velocity and we could say that this would be our angle theta not drawn to scale but this would be at least close to 45 degrees and so now we can say for part b we can solve for the instantaneous velocity so here the velocity vector would be equal to to the derivative of the position vector with respect to time, this is going to be equal to 5 .0 centimeters per second square times t i hat plus 5 .0 centimeters per second times j hat.
04:32
So the velocity in the wide direction is constant.
04:35
And we can say that at t equals zero, we have that the velocity in the x direction is going to equal zero meters per second.
04:49
Or 0 centimeters per second.
04:53
And then we can say for the velocity, the magnitude of the velocity is simply equalling the y component of the velocity and this is equaling 5 .0 centimeters per second because the y component of the velocity is of course a constant.
05:11
And so we can easily tell that here the angle theta is 90 degrees.
05:16
And so here at t equaling one second we find that v sub x is equaling 5 .0 centimeters per second, v.
05:27
Sub y is again constant at 5 .0 centimeters per second.
05:32
And so just like the part a, we can see that the angle here is 45 degrees because the components, the x and my components are equal to one another.
05:42
And then a little bit more complicated at t equals two seconds.
05:46
We have the velocity in the x direction being here 10 .0 centimeters.
05:51
Meters per second.
05:53
Velocity in the y component stays constant at 5 .0 centimeters per second.
05:58
And so we can say that then the magnitude of the velocity would be 10 squared plus 5 squared all to the one half power...