So we're giving this differential equation That's described his emotion of a set of particles moving along the X axis, and that's as a function of time and were told as an initial condition that the position is one at time equal to two. And we were asked to find the velocity s o what? We know that if, uh, X is a function of time is positioned than the first derivative over that is velocity. And the first derivative, uh, of excellent, respected time is exactly what we're given. All right, so this equation is philosophy. All right.
So, uh, our initial condition tells us that our position at time too. There's one all right. And so we just need to evaluate our given differential equation at those values that time. People too. Okay.
So let's like evaluating g cubed minus x of t cubed when time is too. All right. So we have two cubed minus. This would be ex of two. Huge, and we know, except two is one.
So it's two cubed minus one cube, which is seven. So when time is too, the particles that position one and its velocity is seven. That's the answer to part a next. We're asked to show that the acceleration of the particle is described by this guy and to find the acceleration that's the derivative with respect, the time of velocity. And that's exactly what we're given is our velocity equation.
So we can just take the time derivative of, um, equation the differential equation that we were given. So we'll take the first derivative of this guy. Okay. And so, taking that derivative, um, the first piece three t squared, and we have three x squared. And by the chain role since X is a function of time, we need to take, um, the derivative of that function.
Okay, Well, DX DT is exactly the function that were given, so we can just make that substitution. Okay, so we have three t squared, minus three x squared times the x t t, which is t cubed minus X cubed and just with some algebra three t squared, minus three. He cubic square. Then we get a plus three x 50. And that is exactly what we were trying to show for part B part See says our position is, too.
When time is 2.5, we're asked to say whether or not we think this particle could go back down to a position of one at any time beyond T equals 2.5. And then they give us this hint that tiki of my sex cube can be factored this way. Okay, so, um, if you're thinking about why they give you the scent, um, thinking that t squared anything swear it is always gonna be positive. So same thing for X squared. Um, based on this initial condition tee Times X is also always gonna be positive.
So this whole piece is always gonna be positive for this initial condition. And so the only way that we could ever get some negative value would be if X is ever larger than T. But we know that t cubed and X cubed thes are increasing functions. Okay, so the only way we can get a negative value again is like I said, that if the X is gonna be ever larger than T for these given initial conditions, um, excess too and t is 2.5. Okay, Also, tea is already bigger than x.
Uh, and since these are increasing functions, or because this this piece governs the sign we could never get, uh, the velocity or the whole differential equation to be a negative value. So this part, the set of particles, could never decrease in position. Ah, And so since we started out at a position of two, it could never decrease to a position of one. Okay. And so that's how you can describe that this particle will not be able to ever reached a position of wanted any future time.
So the final answer for part C is now..