t₁=250/9 – 250/7 = 63.5 s
t₂ = 500/8 = 62.5 s
Δt=t₁-t₂=63.5 – 62.5 =1 s.
The second runner wants to save energy, so he runs with v₃=8 m/s for the whole time.
If the distance of the race is L=500 m, by what time in seconds does the second runner beat the first?
Moral of the story: slightly slower but steadier wins the race.
t₂ = 500/8 = 62.5 s
Δt=t₁-t₂=63.5 – 62.5 =1 s.
T1 = 125/9 + 125/7 = 31.75 seconds
Second runner's time =
T2 = 250 m/8 m/s = 31.25 s
T2 - T1 = 0.50 seconds
Time = Distance / Speed
Let's calculate the time for each runner:
For the first runner:
Time₁ = (250 m) / (9 m/s) + (250 m) / (7 m/s)
Time₁ = 27.78 s + 35.71 s
Time₁ ≈ 63.49 s
For the second runner:
Time₂ = (500 m) / (8 m/s)
Time₂ = 62.5 s
To determine by how much the second runner beats the first runner, we subtract the time of the first runner from the time of the second runner:
Time difference = Time₂ - Time₁
Time difference = 62.5 s - 63.49 s
Time difference ≈ -0.99 s
The negative value indicates that the first runner actually finishes the race slightly faster than the second runner. Therefore, the moral of the story "slightly slower but steadier wins the race" does not apply in this scenario.