00:01
In this problem, we have a particle that moves with an acceleration, some constant capital a here, times a square of time.
00:09
This capital a is defined as two meters per second to the five halves, which is a bit of an odd unit.
00:18
But it makes sense when you consider that when you multiply this constant by a square of seconds, it needs to give you meters per second squared.
00:27
So that's why we get this unit here.
00:30
We also know the initial velocity and position of this particle.
00:36
And what we're going to do is find the velocity and position functions of time for this particle.
00:44
So we'll start out with finding the velocity.
00:49
So velocity, we know is the integral of acceleration.
00:53
So what this will be is capital a times the integral of the square of t, dt.
01:03
To make this integral a little bit easier, we can write this as t to the one -half d -t.
01:09
And now we can solve this like we solve integrals of t to the a, right? we add one to the power and then divide by the new power.
01:19
So this will be a -t to the three -halfs divided by.
01:25
Three halves or multiplying by two -thirds equivalently.
01:30
And then of course this is an definite integral so we have to add a constant which we know is the initial velocity.
01:36
So we're adding in 7 .5 meters per second to this.
01:42
So that is the velocity with respect to time...