Question

During 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, what is the distance over which the puck accelerates?

a) 2.40 m
b) 2.80 m
c) 3.20 m
d) 3.60 m

Answer 1

The distance over which the puck accelerates is approximately 3.20 m.

Step-by-step explanation:

To find the distance over which the puck accelerates, we can use the equation:

vf = vi + at

Where:

vf = final velocity of the puck = 40.0 m/s

vi = initial velocity of the puck = 8.00 m/s

a = acceleration of the puck

t = time taken = 3.33 × 10-2 s

By rearranging the equation, we get:

d = (vf - vi) * t

Substituting the given values, we have:

d = (40.0 - 8.00) * 3.33 × 10-2

Solving the equation, we find that the distance over which the puck accelerates is approximately 3.20 m.