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

The bike traveled 31.3 meters during this period

Step-by-step 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 =
`(v-u)/(t)`

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

→ a =
`(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