The pilot stops the plane in 484 m using a constant acceleration of -8.0 m/s2. How fast was the plane moving before the braking began?

Answer 1

Vi = initial speed
484 = average speed during stop * t
484 = (Vi/2) * t
v = Vi + a t
0 = Vi -8 t
t = Vi/8
so
484 = (Vi/2)(Vi/8) = Vi^2 / 16
Vi^2 = 7744
Vi = 88 m/s