Aspeed bike tops a hill at 3.50 m/s and accelerates steadily down the hill until reaching a speed of 11.4 m/s after 4.20 seconds. how far did the bike travel during this period? figure your answer to the nearest tenth of a meter.

The bike traveled 31.3 meters during this period

Explanation:

A speed bike tops a hill at 3.50 m/s and accelerates steadily down the

hill until reaching a speed of 11.4 m/s after 4.20 seconds

→ The initial speed u = 3.5 m/s

→ The final speed v = 11.4 m/s

→ The time of acceleration t = 4.2 seconds

→ a =

Substitute the values above in the rule to find the acceleration a

→ a = = 1.88 m/s²

We need to find the distance that the bike traveled during the

acceleration

→ v² = u² + 2 a s, where s is the distance

→ (11.4)² = (3.5)² + 2 (1.88)(s)

→ 129.96 = 12.25 + 3.76 s

Subtract 12.25 from both sides

→ 117.71 = 3.76 s

Divide both sides by 3.76

→ s = 31.3 meters

The bike traveled 31.3 meters during this period

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