Answer:
a) 20 feet per year
b) 44.1 m
Step-by-step explanation:
Given:
Distance traveled during the observation period = 55.3 meters
Observation period = 9 years
initial velocity of v₀ = 11 m/s
friction coefficient of stone on ice is μ = 0.14
g = 9.81 m/s²
also,
v₀² = 2μgS
Now,
1 m = 3.28084 ft
thus,
Total distance in feet = 55.3 × 3.28084 = 181.430452 ft
Average speed = [tex]\frac{\textup{Distance}}{\textup{Time}}[/tex]
or
Average speed = [tex]\frac{\textup{181.430452}}{\textup{9}}[/tex]
or
Average speed = 20.159 feet/year ≈ 20 feet per year
b) v₀² = 2μgS
substituting the values in the above equation, we get
11² = 2 × 0.14 × 9.81 × s
or
121 = 2.7468 × s
or
s = 44.051 ≈ 44.1 m