A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity of the skier at the bottom of the hill
we can use the kinematic equation [tex](v_f)^2 = (v_i)^2 + 2*a*d[/tex] where Vf is what we are looking for Vi is 0 since we start from rest a is acceleration and d is the distance
we get (Vf)^2 = (0)^2 + 2*(2)*(500) (Vf)^2 = 2000 Vf = about 44.721 or 44.7 m/s [if you are rounding this by significant figures]
