Answer:
              S =34.71 m
Explanation:
M=1500 kg
F=-7000 N
For short and safe distance stop the friction acting should be maximum
               a(max)=[tex]\frac{F}{M}[/tex]
                    =[tex]\frac{-7000}{1500}[/tex]
                     =[tex]\frac{-14}{3}[/tex] m/s2
                 u=18 m/s
Applying equation of motion
                  v²=u²+2aS
                  0= [tex]18^{2} +2*\frac{-14}{3}[/tex]
                  S =34.71 m
