In a slap shot a hockey player accelerates the puck from a velocity of 9.00 m/s to 60.0 m/s in the same direction. if this takes 3.33 ✕ 10−2 s, calculate the distance over which the acceleration acts.

First, you have to calculate the value of the constant acceleration according to the equation below:

a = (v,final - v,initial)/t
Substituting the values,
a = (60 m/s - 9 m/s) / 3.33×10⁻² s
a = 1,531.53 m/s²

We use the value of a to determine the distance using the equation below:
2ad = v,final² - v,initial²
2(1,531.53 m/s²)(d) = (60 m/s)² - (9 m/s)²
Solving for d,
d = 1.15 meters

Domain eukarya; kingdom plantae should be the answer.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward