Question

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continues on at 6.5 m/s. What is her acceleration on the rough ice?

Answer 1

Acceleration,
`a=-2.48\ m/s^2`

Step-by-step explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

`v^2-u^2=2as`

`a=(v^2-u^2)/(2s)`

`a=((6.5)^2-(8.4)^2)/(2* 5.7)`

`a=-2.48\ m/s^2`

So, the acceleration on the rough ice
`-2.48\ m/s^2` and negative sign shows deceleration.