Hi Nadine,
Since we are given:
a: acceleration= 3.22 m/s2
t: time= 32.0 m/s2
vi: initial velocity= 0 m/s
And we are trying to find the change in distance Δx
We can solve:
Δx=(vi)(t) + (½)(a)(t2)
Δx= (0 m/s)(32.8 s) + (½)(3.22 m/s2)(32.8 s)2
Δx=1764.90 m
Therefore, the distance traveled before take off was 1764.90 meters.
I hope this helps! If you have further questions, please contact me.
Best,
Star