What was the acceleration of the bubble?

Question

A small air bubble separated from the the bottom of the pool
filled with water, and reached the surface in T = 5s later.
The depth of the pool is d = 1m. Estimate how much time τ
did it take for the bubble to acclerate to the speed v= 1cm/s.

I am confused. I tried to apply the formula from the textbook
a = d/2T² , and then τ = v/a = 2vT²/d and get = 50 ms, but
it's nowhere near the answer, not even close. Please help.

Thank you guys, now I see where I screwed up. Unfortunately τ = v/a = vT²/2d =125 ms is even further from the correct answer..

Thank you guys, now I see where I screwed up. Unfortunately τ = vT²/2d = 125 ms is even further away from the correct answer.

Answer 1

d = 1/2 at²
2d = at²
a = 2d/t²

Your first equation was wrong. Try it again.

Answer 2

I think it took 0.125 seconds (1/8 of a second) to reach v = 1cm/s