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 forVi is 0 since we start from resta is accelerationand d is the distancewe get(Vf)^2 = (0)^2 + 2*(2)*(500)(Vf)^2 = 2000Vf = about 44.721or 44.7 m/s   [if you are rounding this by significant figures]