Question

A speed skater moving to the left 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

Answer:

Acceleration, $a=-2.8\ m/s^2$

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 :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{(6)^2-(8)^2}{2\times 5}\\\\a=-2.8\ m/s^2$

So, the acceleration on the rough ice is $2.8\ m/s^2$. Negative sign shows that its speed is decreased.