Answer:
[tex]\boxed{\text{\sf \Large 3.0 s}}[/tex]
Explanation:
Use distance formula
[tex]\displaystyle d=ut+\frac{1}{2} at^2[/tex]
[tex]u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}[/tex]
[tex]\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2[/tex]
[tex]t=3[/tex]
