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Marta_Voda [28]
3 years ago
8

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continu

es on at 5.8 m/s . What is her acceleration on the rough ice?
Physics
1 answer:
irina [24]3 years ago
4 0

Recall that

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>

where <em>u</em> and <em>v</em> are initial and final velocities, respectively; <em>a</em> is acceleration; and ∆<em>x</em> is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 <em>a</em> (6.0 m)

<em>a</em> = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

<em>a</em> = -3.65 m/s²

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