What is her acceleration on the rough ice?

Question

A speed skater moving across frictionless ice at 5.0m/s hits a 3.0-m-wide patch of rough ice, she slows steadily, then continues on at 3.7m/s. What is her acceleration on the rough ice? Use 2 significant figures.

Answer 1

v1 = speed at start of slide on rough ice = 5 m/s

 

v2 = speed at end of slide on rough ice = 3.7 m/s

 

t = time spent crossing the rough ice

 

d = length of slide on rough ice = 3 m

 

a = constant acceleration on the rough ice

 

d = v1*t + a*t^2 / 2

 

Solve for a and get:

 

a = 2(d - v1*t) / t^2

 

But by definition of constant acceleration,

 

a = (v2 - v1) / t

 

This gives us two expressions for a; set them equal and solve for t:

 

2(d - v1*t) / t^2 = (v2 - v1) / t

 

2(d - v1*t) = (v2 - v1)*t

 

t = 2d / (v1 + v2) = 2(3 m) / [(5 + 3.7) m/s] = 0.690 s

 

Now plug this t back into a = (v2 - v1) / t and get:

 

a = [(3.7 - 5) / (0.690)] m/s^2 = -1.88 m/s^2