Question

A speed skater moving across frictionless ice at 8.9 m/s hits a 5.3 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?
a= _____ m/s²

Answer 1

The acceleration on the rough ice is approximately -6.74 m/s².

Step-by-step explanation:

To calculate the acceleration on the rough ice, we can use the formula:

`\rm a = (v_f - v_i) / t`

Where:
a is the acceleration

`\rm v_f` is the final velocity (6.5 m/s)

`\rm v_i` is the initial velocity (8.9 m/s)
t is the time taken to slow down

The skater's steady slowing is assumed to be constant, allowing us to rearrange the formula to solve for acceleration.

`\rm a = (v_f - v_i) / t`

a = (6.5 - 8.9) / t

The skater's velocity decreases from 8.9 m/s to 6.5 m/s over 5.3 m due to frictionless rough ice, indicating no external forces, as per the equation.

`\rm a = (v_f^2 - v_i^2) / (2 * d)`

Where:

a is the acceleration


`\rm v_f` is the final velocity (6.5 m/s)


`\rm v_i` is the initial velocity (8.9 m/s)

d is the distance (5.3 m)

`\rm a = (6.5^2 - 8.9^2) / (2 * 5.3)`

We determine the acceleration on the rough ice to be roughly -6.74 m/s² by evaluating this equation.