Question

U. A hockey player takes a slap shot at a puck at rest on

Answer 1

Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s².  Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

(c) The time the puck takes to travel 10 ft without friction is:

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s