Consider a skateboarder who starts from rest at the top of a 12.5m-long ramp that is inclined at an angle of 15.0degrees to the horizontal. Assuming that the skateboarder's acceleration is g sin 15.0degrees, find his speed when he reaches the bottom of the ramp in 3.00 seconds

Answer 1

That would be V = a t, where a is the acceleration rate (which is given) and t = 3.00 s. Using this method, I get
V = a t = 7.6 m/s

But they also give you the length of the ramp X, and the acceleration rate. From that, one gets
V = sqrt (2 a X) = 8.0 m/s

They have given you more data than you need and the data are somewhat inconsistent.