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Sagot :
Let D be the total length of the race, and T the time it takes for the tortoise to finish it. Then
D = (0.15 m/s) T
Let t be the time after the start of the race when the hare decides to take a nap. In this time, the tortoise travels a distance
(0.15 m/s) t
and the hare travels
25 (0.15 m/s) t = (3.75 m/s) t
For the next 5.0 min = 300 s, the tortoise keeps moving and covers a distance of
(0.15 m/s) (300 s) = 45 m
while the hare covers no distance.
After these 300 seconds, the remaining time for the tortoise to complete the race is (T - t - 300 s). In this same time, the hare loses the race by 35 cm = 0.35 m, so overall it has covered a total distance of
(3.75 m/s) t + 0 (300 s) + (3.75 m/s) (T - t - 300 s) = D - 0.35 m
Solve for T :
(3.75 m/s) t + 0 + (3.75 m/s) T - (3.75 m/s) t - 1125 m = D - 0.35 m
(3.75 m/s) T - 1125 m = D - 0.35 m
(3.75 m/s) T - 1125 m = (0.15 m/s) T - 0.35 m
(3.60 m/s) T = 1124.65 m
T = 312.403 s
(b) So, the race takes about 312 seconds, or 5.21 minutes.
(a) It also follows that the race has a total length of
D = (0.15 m/s) (312.403 s) ≈ 46.9 m
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