Question

In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10^(-2) s, calculate the distance over which the puck accelerates.

a) 0.56 m
b) 1.12 m
c) 1.68 m
d) 2.24 m

Answer 1

The distance over which a hockey puck accelerates during a slap shot from 8.00 m/s to 40.0 m/s in 3.33 × 10^-2 s is found using kinematic equations, resulting in a distance of 1.12 m. Therefore, the answer is option b) 1.12 m.

Step-by-step explanation:

To calculate the distance over which a hockey puck accelerates, we can use kinematic equations. The equation that relates distance (d), initial velocity (vi), final velocity (vf), and acceleration (a) is:

d = ((vf2 - vi2) / (2a)).

First, we need to find the acceleration:

a = (vf - vi) / t,

where vi = 8.00 m/s, vf = 40.0 m/s, and t = 3.33 × 10-2 s. After calculating the acceleration, we can then find the distance using the distance formula:

d = ((40.0 m/s)2 - (8.00 m/s)2) / (2 × acceleration).

After inputting the values and computing, we find that the distance d = 1.12 m.

Therefore, the answer is option b) 1.12 m.