NO!!!!
You assume that the acceleration, a, is CONSTANT.
v = 0 + a t
at t = 1.18 s
v = 7.96 m/s
so
7.96 = a (1.18)
a = 7.96/1.18
NOW
use x = 0 + 0 + (1/2) a t^2
A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he reaches the line of scrimmage. Assume that the fullback accelerates uniformly (even during the handoff), reaching the line with a velocity of 7.96 m/s. If he takes 1.18 s to reach the line, how far behind it did he start?
Can I assume that acceleration is 0? If yes then I can use the x=v=Vi*t + 1/2at^2 equation. If not, I have no idea where to start.
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