s(t)=2.5t³-13.5t²+22.5t, s(6)=189 ft, making the final position (6,189).
s(0)=0, so the distance travelled in a straight line is the same as the position=189 ft. However, the actual distance travelled along the curved path has to be calculated from the arc length, x. This will create a line integral.
We can write, using Pythagoras:
dx²=dt²+ds², so (dx/dt)²=1+(ds/dt)².
ds/dt=7.5t²-27t+22.5=3(2.5t²-9t+7.5).
dx/dt=√[1+9(2.5t²-9t+7.5)²].
x=∫₀⁶√[1+9(2.5t²-9t+7.5)²]dt.
This is not easy to integrate but approximations for the definite integral are available by dividing the interval [0,6] into small segments. I think this method might be beyond the scope of this problem.