Question

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.

Answer 1

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