If a pull of 210 N accelerates a 40-kg child on ice skates at a rate of 5 m/s², what is the frictional force acting on the skates?

If f is the magnitude of friction, then by Newton's second law the net force on the child is

∑ F = 210 N - f = (40 kg) (5 m/s²)

Solve for f :

f = 210 N - (40 kg) (5 m/s²)

f = 210 N - 200 N

f = 10 N

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day