Post a New Question

Physics

posted by .

A speed skater moving across frictionless ice at 9.0 hits a 5.1 -wide patch of rough ice. She slows steadily, then continues on at 6.4 .
What is her acceleration on the rough ice?

  • Physics -

    a = (V^2-Vo^2)/2d.
    a = (40.96-81)/10.2 = -3.93 m/s^2.

    "a" is negative because she was slowing down.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question