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
Hi there!
We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).
Remember to break the velocity into its vertical and horizontal components.
Thus:
0 = vi - at
0 = 16sin(33°) - 9.8(t)
9.8t = 16sin(33°)
t = .889 sec
Find the max height by plugging this time into the equation:
Δd = vit + 1/2at²
Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²
Solve:
Δd = 7.747 - 3.873 = 3.8744 m
<span>25,000 miles per hour
hope that this helps</span>
