runner is jogging at a steady 1 km/hr. When the runner is 0.88 km from the finish line a bird begins flying from the runner to the finish line at 3 km/hr (3 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far did the bird travel?

Question

Farrell

Physics

17 December, 05:47

runner is jogging at a steady 1 km/hr. When the runner is 0.88 km from the finish line a bird begins flying from the runner to the finish line at 3 km/hr (3 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far did the bird travel?

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Waldo · Answer 1

1.548 km i believe. since 60 minutes times 0.88 equals to 52.8 minutes and the bird travels 3 times his speed 52.8 times 3 = 154.8 then you convert that to km and its 1.548 km