Try this solution:
1. rule for this type of movement:
[tex]S=v_0t+\frac{at^2}{2}, \ where \ a-acceleration,S-distance,v-startSpeed,t-time.[/tex]
2. according to the rule described above:
[tex]S=0+\frac{7.4*2^2}{2}=7.4*2=14.8(m)[/tex]
answer: 14.8 meters.
