Physics, 12.09.2019 19:30 amastar

U. a hockey player takes a slap shot at a puck at rest on
the ice. the puck glides over the ice for 10 ft without
friction, at which point it runs over rough ice. the
puck then accelerates opposite its motion at a uniform
rate of - 20 ft/s2. if the velocity of the puck is 40 ft/s
after traveling 100 ft from the point of impact,
(a) what is the average acceleration imparted to the
puck as it is struck by the hockey stick? (assume that
the time of contact is 0.01 s.) (b) how far in all does
the puck travel before coming to rest? (c) what is the total time the puck is in motion, neglecting contact time?

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Answer from: Destiny8372

The average acceleration or the total time will be:

(a) 7200 ft/s²(b) 140 ft(c) 3.7 s

According to the question,

Velocity,

$v^2 = 40 \ ft/s$

Acceleration,

$a = -20 \ ft/s^2$

(a)

As we know,

→ v^2=v_0^2+2a(x-x_0)

By substituting the values, we get

→ (40)^2=v_0^2+2(-20)(100-10)

→ $v_0^2=5200 \ ft^2/s^2$

→ $= 20\sqrt{13} \ ft/s$

Now,

The average acceleration will be:

→ $a_{avg} = \frac{\Delta v}{\Delta t}$

→ $= \frac{20\sqrt{13}-0 }{0.01}$

→ $= 2000\sqrt{13} \ ft/s^2$

→ $= 7200 \ ft/s^2$

(b)

The distance will be:

→ v^2=v_0^2+2a(x-x_0)

→ (0)^2= 5200+2(-20)(x-10)

→ 0= 5200-40(x-10)

→ $x = 140 \ ft$

(c)

The time taken to travel 10 ft will be:

→ $t = \frac{10}{20\sqrt{3} }$

$= \frac{\sqrt{13} }{26} \ s$

The time taken over rough ice will be:

→ v = at+v_0

$0=-20t+20\sqrt{13}$

$20t = 20\sqrt{13}$

$t = \sqrt{13} \ s$

hence,

The total time taken will be:

→ $t = \frac{\sqrt{13} }{26} +\sqrt{13}$

$= \frac{27\sqrt{13} }{26}$

$= 3.7 \ s$

Thus the above answers are correct.

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Answer from: chrisraptorofficial

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

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

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