Answer:
Acceleration, [tex]a=-2.8\ m/s^2[/tex]
Explanation:
It is given that,
The initial speed of a speed skater, u = 8 m/s
The final speed of a speed skater, v = 6 m/s
Width of patch of rough ice, s = 5 m
We need to find the acceleration on the rough ice. Acceleration can be calculated using third equation of motion as :
[tex]v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{(6)^2-(8)^2}{2\times 5}\\\\a=-2.8\ m/s^2[/tex]
So, the acceleration on the rough ice is [tex]2.8\ m/s^2[/tex]. Negative sign shows that its speed is decreased.
