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
The average acceleration or the total time will be:
According to the question,
Velocity,
Acceleration,
(a)
As we know,
→ [tex]v^2=v_0^2+2a(x-x_0)[/tex]
By substituting the values, we get
→ [tex](40)^2=v_0^2+2(-20)(100-10)[/tex]
→ [tex]v_0^2=5200 \ ft^2/s^2[/tex]
→ [tex]= 20\sqrt{13} \ ft/s[/tex]
Now,
The average acceleration will be:
→ [tex]a_{avg} = \frac{\Delta v}{\Delta t}[/tex]
→ [tex]= \frac{20\sqrt{13}-0 }{0.01}[/tex]
→ [tex]= 2000\sqrt{13} \ ft/s^2[/tex]
→ [tex]= 7200 \ ft/s^2[/tex]
(b)
The distance will be:
→ [tex]v^2=v_0^2+2a(x-x_0)[/tex]
→ [tex](0)^2= 5200+2(-20)(x-10)[/tex]
→ [tex]0= 5200-40(x-10)[/tex]
→ [tex]x = 140 \ ft[/tex]
(c)
The time taken to travel 10 ft will be:
→ [tex]t = \frac{10}{20\sqrt{3} }[/tex]
[tex]= \frac{\sqrt{13} }{26} \ s[/tex]
The time taken over rough ice will be:
→ [tex]v = at+v_0[/tex]
[tex]0=-20t+20\sqrt{13}[/tex]
[tex]20t = 20\sqrt{13}[/tex]
[tex]t = \sqrt{13} \ s[/tex]
hence,
The total time taken will be:
→ [tex]t = \frac{\sqrt{13} }{26} +\sqrt{13}[/tex]
[tex]= \frac{27\sqrt{13} }{26}[/tex]
[tex]= 3.7 \ s[/tex]
Thus the above answers are correct.
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