Question

A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadily, then continues on at 6.0 m/s. What is her acceleration on the rough ice?

Answer 1

The speed skater's acceleration on the rough ice, calculated using the kinematic equation, is -2.8 m/s². This value indicates deceleration.

Step-by-step explanation:

The student's question is asking to find the acceleration of a speed skater as she slows down while crossing a rough patch of ice. To solve this, we can use the kinematic equation which relates initial and final velocities, acceleration, and the distance over which the acceleration occurs:

v2 = u2 + 2as

Where:

• v is the final velocity (6.0 m/s)
• u is the initial velocity (8.0 m/s)
• a is the acceleration, which we want to find
• s is the distance over which the acceleration occurs (5.0 m)

Rearranging and solving for acceleration (a):

a = (v2 - u2) / (2s)

By plugging in the given values:

a = (6.02 - 8.02) / (2 × 5.0)

a = (36 - 64) / 10

a = -28 / 10

a = -2.8 m/s2

The negative sign indicates that the acceleration is in the opposite direction of her initial motion, meaning it is deceleration.

Answer 2

Acceleration,
`a=-2.8\ m/s^2`

Step-by-step explanation:

It is given that,

The initial speed of a speed skater, u = 8 m/s

The final speed of a speed skater, v = 6 m/s

Width of patch of rough ice, s = 5 m

We need to find the acceleration on the rough ice. Acceleration can be calculated using third equation of motion as :

`v^2-u^2=2as\\\\a=(v^2-u^2)/(2s)\\\\a=((6)^2-(8)^2)/(2* 5)\\\\a=-2.8\ m/s^2`

So, the acceleration on the rough ice is
`2.8\ m/s^2`. Negative sign shows that its speed is decreased.