Answer :
For the first 5.0 seconds, the cruiser covers a distance of
[tex]\left(20.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(3.0\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2=137.5\,\mathrm m[/tex]
At this point, the cruiser will have achieved a velocity of
[tex]\left(20.0\dfrac{\rm m}{\rm s}\right)+\left(3.0\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)=35\dfrac{\rm m}{\rm s}[/tex]
The cruiser will take
[tex]\left(35\dfrac{\rm m}{\rm s}\right)-\left(5.0\dfrac{\rm m}{\mathrm s^2}\right)t=0\implies t=7.0\,\mathrm s[/tex]
to come to a stop as it decelerates. It will have covered a total distance of
[tex](137.5\,\mathrm m)+\left(35\dfrac{\rm m}{\rm s}\right)(7.0\,\mathrm s)+\dfrac12\left(-5.0\dfrac{\rm m}{\mathrm s^2}\right)(7.0\,\mathrm s)^2=\boxed{260\,\mathrm m}[/tex]