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 = [tex]\frac{v-u}{t}[/tex]
Substitute the values above in the rule to find the acceleration a
→ a = [tex]\frac{11.4-3.5}{4.2}[/tex] = 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
