Question

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. How far did the bike travel during this period? Figure your answer to the nearest tenth of a meter.

Answer 1

Answer:

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 = $\frac{v-u}{t}$

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

→ a = $\frac{11.4-3.5}{4.2}$ = 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