Calculus for real time 't' and velocity 'v', (space travel).?

I'm 90,000 miles from a virtual planet at some distant point in space. From v=0, I begin accelerating to that point along a straight line. Passing through this point, I record my time and speed.

I do this again, but from 110,000 miles. I record my time and speed after traveling (accelerating) exactly 100,000 miles (10,000 miles short of this point).

While it is obvious t1 is 'less than' t2, I need to Prove V1(first trip)=v2 (second trip). 'average acceleration' will not work.

Yes, I tried astronomy/space
hi doc, the acceleration is less for the second trip, because of the increased distance from the target point, but the incresed distance will compensate for this.
'inverse square law'
Rick b. I like your approach to the problem, I was thinking the same thing a few moments ago. You missed the 'virtual' aspect however. We need a virtual point because we continue accelerating to that point. There is no surface, there is no 'real' planet. This is also nessesary because the 'inverse square law' applies to 'centers' of mass.

Is there a maximum velocity that can be reached?

Are you accelerating forever? At what rate?

I don't see that you can prove v1 = v2. If anything, it appears that v2 > v1 since you will be travelling further, and have more time to accelerate more.

Edit:

Based on your problem:

t1 < t2
d1 < d2

I suppose, theoretically, it may be possible that:

v1 = v2 if you accelerate at different rates (i.e. a1 != a2)

Since velocity = distance / time

IF the change in distance (between d1 and d2) is in proportion to the change in time (between t1 and t2), you WILL have the same velocity.

There seem to be details missing from your problem statement. I'll assume that what you mean is this:
Trip 1: you start out at 90,000 miles from the _surface_ of the planet. You measure your speed (v1) at the moment you touch the planet's surface.
Trip 2: You start out at 110,000 miles from the surface of the planet, and measure your speed (v2) when you reach a point 10,000 miles above the planet's surface.
Both trips: You're accelerating due to the planet's gravity.

I would consider the conservation of energy. The work required to move outward from a distance r (from planet's center) to a farther distance r′ is

W = ∫Fdr = GMm∫(1/r²) = GMm(1/r − 1/r′)

By the conservation of energy, this amount is also equal to the object's change in KE as it drops. So:

mv²/2 = GMm(1/r − 1/r′)
or (canceling "m"):
v²/2 = GM(1/r − 1/r′)

In Case 1, r = R (planet's radius) and r′ = R+90000
In Case 2, r = R+10000 and r′ = R+110000

That should be enough to get you started.