Asked by Anonymous

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

Answers

Answered by Henry
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.

Answered by haydon
something very sophisticated
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