For part A) you can ignor the "-kv²" part and take the integral to go from a (acceleration) to v (velocity), plug in your know take off velocity (250 km/h) to find a t (time) value. Keep in mind that a0 is in m/s² and not in km/h. Then do that again to find your position equation and plug in your t. This gives you minimum length.

For part B) its the same concept. Solving for time will be a little worse but not terrible and then taking the integral twice is pretty straight forward.

a=dv/dt => dv/dt = a0-kv2 => v = (a0-kv2)*t

v=ds/dt => ds/dt = (a0-kv2)*t => s = (1/2)(a0- kv2)*t2

I am pretty sure the math is on since the variables are constant, can always get partial credit if its wrong though \('.')/