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
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
[tex](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=\frac{(5.20m/s)^{2}-(8.30m/s)^{2} }{2*(5.10m)}\\a=-4.103 m/s^{2}[/tex]
