ANSWER
i think i got the answer.
well it says that for the first 15sec at each sec the acceleration increases by 2m/s^2. so that means that time=0 a=0
then time=1 a=2. time=2 a=4, time=3 a=6, and so on till time =15sec so a=30m/s^2. what needs to do with this information is that they need to find out how far the ship traveled each of those seconds. but befor u can do that u need the speed at each second. so ween to calculate speed. we will use the following formula of velocity as a funtion of time

so for the time =1 we plug in a=2, t=1, and Vo=0 and we get v at 1second to be 2m/s we do this again but this time a=4 Vo=2 t=1(we are working with indivisual seconds) and we get v at 2seconds to be 5m/s. so the total velocity at 2 seconds is 5m/s not 5+1=6. we continue to do this for the first 15sec. i suggest u write down each indivisual velocity seeing that u will need them later on in the calculations. so the velocitry if u did this correctly should 239m/s at the 15 second mark. don't forget the objective of the problem we need the distance not spped. the way to get the distance traveled is by using the formula position as a function time and velocty. X=Xo+.5(Vo+V)
where x is the distance Xo is the intial distance, Vo intial velocity, and v is the velocity.
for time=0 to time=1 Vo=0 V=1 Xo=0 so x=.5
for time 1 to time=2 Vo=1 V=5 Xo=.5 so x=3.5
note the total distance is not 3.5+.5=4m( once again i say this we are ddealing with each indivisual second. so the totaldistance traveld the first 2 seconds is 3.5m
u continue this 13 more times till u reach time =15. the total distance should be 1345.5m.
this is for the first 15 seconds not the whole minute. we still have to calulate the other 45seconds now.
the problem now states that the ship goes at a constant acceleration for the next 30seconds. this means the ship with an acceleration on 30m/s^2 for 30sec. the way i got 30m/s^2 is from when we calculated the acceleration in the begining. this part is pretty easy to calculate. because the acceleration is constant we use the formula position as a funtion of time and acceleration

what we do is plug in Xo=1345.5 Vo= 239 t=30 and a=30
when we do this we get a value of 22015.5m this is the total distance traveled for the first 45sec. people may be wndering why i used 30 for the time instead of one, well the reason is that the acceleration is constant here and if the other part the acceleration was not constant it was increasing by 2m/s^2 everysecond. now for the third part of the problem we know that there is zero accleration, which means that the velocity is constant. to calculate the the distance traveld in the last 15seconds we need to first know the speed at which the ship is moving at. so we have to calculate the velocity of the ship at the 45second mark. we do that with the following formula position as a function time and velocty. X=Xo+.5(Vo+V) this formula can only be used when there is a constant acceleration which there is for the 30secs.
we plug in x=22015.5 Xo=1345.5 Vo= 239 and t=30
from this we get v= 1139m/s
we then take that velocity and plug it into the formula position as a funtion of time and accleration

we plug Xo=22015.5 Vo=1139 t=15 and a=0 and solve for x.
as a result u should get the total distance traveld by the ship to be 39100.5m in the first minute(60sec). that is the number u put into the lenny conundrum