A speed skater moving across frictionless ice at 9.0 m/s hits a 5.9 m -wide patch of rough ice. She slows steadily, then continues on at 5.9 m/s .What is her acceleration on the rough ice?

To find the acceleration of the speed skater on the rough ice, we can use the kinematic equation:

\[
v_f^2 = v_i^2 + 2a \Delta x
\]

where:
- \( v_f \) = final velocity = 5.9 m/s
- \( v_i \) = initial velocity = 9.0 m/s
- \( a \) = acceleration (which we need to find)
- \( \Delta x \) = distance traveled on the rough patch = 5.9 m

### Rearrange the equation to solve for acceleration (\( a \)):

\[
a = \frac{v_f^2 - v_i^2}{2 \Delta x}
\]

Now, substitute the known values:

\[
a = \frac{(5.9^2 - 9.0^2)}{2 \times 5.9}
\]

Let’s calculate that.The skater's acceleration on the rough ice is approximately \(-3.91 \, \text{m/s}^2\), meaning she is decelerating at that rate.

Do you want more details or have any questions?

Here are 5 related questions to expand on this concept:
1. What is the total time the skater spends on the rough ice patch?
2. How far would the skater have traveled if the rough patch had been twice as long?
3. What is the force acting on the skater if her mass is 60 kg?
4. How long would it take for the skater to come to a complete stop if the ice patch was longer?
5. How much kinetic energy does the skater lose while on the rough patch?

**Tip:** When an object slows down, its acceleration is negative, indicating deceleration in physics.

A speed skater moving across frictionless ice at 9.0 m/s hits a 5.9 m-wide patch of rough ice. She slows steadily, then continues on at 5.9 m/s. What is her acceleration on the rough ice?

Learn how to calculate the acceleration of a speed skater moving across a 5.9 m-wide rough ice patch using kinematic equations. This problem demonstrates key concepts in kinematics, deceleration, and physics, suitable for Grades 10-12 students.

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