Question

in a slap shot a hockey player accelerates the puck from a velocity of 8.00 m/s to 30.0 m/s in the same direction. if this takes 3.33 ✕ 10−² s, calculate the distance over which the acceleration acts.

Answer 1

To calculate the distance over which the hockey puck accelerates during a slap shot, we used kinematic equations, determined the acceleration based on the change in velocity and the time, and finally calculated the distance to be approximately 53.632 meters.

Step-by-step explanation:

To calculate the distance over which the puck accelerates during the slap shot, we can use the kinematic equation for uniformly accelerated motion:

s = ut + 1/2 at2

Here, u is the initial velocity, t is the time duration, and a is the acceleration. First, we need to find the acceleration (a), using the equation a = (v - u) / t, where v is the final velocity. Plugging in the given values:

a = (40.0 m/s - 8.00 m/s) / (3.33 × 10−2 s) = (32 m/s) / (3.33 × 10−2 s) ≈ 961.54 m/s2

Now we can find the distance using:

s = ut + 1/2 at2

s = (8.00 m/s)(3.33 × 10−2 s) + 1/2 (961.54 m/s2)(3.33 × 10−2 s) 2 = 26.496 m/s + 1/2 (961.54 m/s2)(0.0110889 s2)

s ≈ 26.496 m + 5324.84 m·s2 × s2/2

s ≈ 53.632 m

Therefore, the distance over which the puck accelerates is approximately 53.632 meters.