Question

A speed skater moving 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

Acceleration on rough ice, `a=-2.8ms^-2`
`-ve` sign shows that it is retardation

Explanation:

We need to consider the time, distance and speed when she hit the rough patch of ice and when she left that rough patch.

Speed when she hits the patch of ice, initial speed `u=8.0m//s`
Speed when she left the patch of ice final speed `v=6.0m//s`
Width of patch of ice `s=5.0m`

Acceleration `a` due to patch of ice can be found with the relation

`v^2-u^2=2as`, Inserting the given numbers

`6^2-8^2=2xxaxx5`
`=>a=(36-64)/10`
or `a=-2.8ms^-2`
`-ve` sign shows that it is retardation, as we know that speed decreased due to patch of ice.