Question

A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10-m-wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s ?

Answer 1

Incomplete question.The complete question is here

A speed skater moving across frictionless ice at 8.30 m/s hits a 5.10m wide patch of rough ice. She slows steadily, then continues on at 5.20 m/s.What is her acceleration on the rough ice?

Answer:

acceleration = -4.103 m/s²

Step-by-step explanation:

Given data

Initial velocity Vi=8.30 m/s

Final velocity Vf=5.20 m/s

Initial distance xi=0 m.......(We choose xi=0 the start point of acceleration motion )

Final distance xf=5.10 m

To find

Acceleration

Solution

From the kinetic equation

`(v_(f))^(2)=(v_(i) )^(2)+2a(x_(f) -x_(i) )\\  (5.20 m/s)^(2)=(8.30m/s)^(2)+2a(5.10m-0m)\\a=((5.20m/s)^(2)-(8.30m/s)^(2)  )/(2*(5.10m))\\a=-4.103 m/s^(2)`