in a slap shot,a hockey player accelerates the puck from a velocity of 5.5m/s to 36m/s. if the shot take 3.16*10^-2, find the distance

Answer 1

a = (Vf-Vo)/t = (36-5.5)/0.0316 = 965.2 m/s^2.

d = Vo*t + 0.5a*t^2 =
5.5*0.0316 + 0.5*965.2*0.0316^2 = 0.656 m.

Answer 2

something very sophisticated