A speed skater moving across frictionless ice at 9.0 m/s hits a 5.0 m -wide patch of rough ice. She slows steadily, then continues on at 6.5 m/s . What is her acceleration on the rough ice?
The Correct Answer and Explanation is:
To determine the skater’s acceleration on the rough ice, we can use the following kinematic equation:vf2=vi2+2adv_f^2 = v_i^2 + 2a dvf2=vi2+2ad
Where:
- vfv_fvf is the final velocity (6.5 m/s)
- viv_ivi is the initial velocity (9.0 m/s)
- aaa is the acceleration (what we want to find)
- ddd is the distance traveled over the rough ice (5.0 m)
We can rearrange the equation to solve for acceleration:a=vf2−vi22da = \frac{v_f^2 – v_i^2}{2d}a=2dvf2−vi2
Substituting the known values:a=(6.5)2−(9.0)22(5.0)a = \frac{(6.5)^2 – (9.0)^2}{2(5.0)}a=2(5.0)(6.5)2−(9.0)2a=42.25−81.010a = \frac{42.25 – 81.0}{10}a=1042.25−81.0a=−38.7510a = \frac{-38.75}{10}a=10−38.75a=−3.875 m/s2a = -3.875 \, \text{m/s}^2a=−3.875m/s2
So, the skater’s acceleration on the rough ice is −3.875 m/s2-3.875 \, \text{m/s}^2−3.875m/s2. The negative sign indicates that the skater is slowing down (decelerating).
Explanation:
The equation used is a fundamental equation of motion that describes how velocity changes with constant acceleration over a certain distance. In this case, the skater starts with an initial velocity of 9.0 m/s, moves across the rough patch of ice (which slows her down), and finishes with a final velocity of 6.5 m/s. Since she is slowing down, the acceleration is negative, indicating a decrease in speed. The magnitude of the acceleration, 3.875 m/s², tells us how much her speed decreases for each second she is on the rough ice.
