Answer:
(a) -16 cm
(b) 12(√13-4) ≈ -4.733 cm
Explanation:
You want to know the vertical distance Beetle 1 is above Beetle 2 at the time they are 72 cm apart horizontally when (a) they move downslope in opposite directions on a roof with slope ±2/3, and (b) Beetle 2 is on a vertical slope. Beetle 1 travels twice as fast as Beetle 2.
A Same incline
In the time it takes Beetle 2 to move 3 cm horizontally, it has moved 2 cm down. In the same time, Beetle 1 has moved 6 cm horizontally, and 4 cm down. That is Beetle 2 remains above Beetle 1 by 2 cm for each 9 cm they have of horizontal separation.
When their horizontal separation is 8(9 cm) = 72 cm, the height from Beetle 1 to Beetle 2 is 8(-2 cm) = -16 cm.
B Vertical incline
Suppose it takes 1 unit of time for Beetle 1 to travel downslope so its separation from Beetle 2 is 3 cm horizontally. It will have traveled 3 cm horizontally, so a total distance of √(3² +2²) = √13 cm. After 24 units of time, it has separated from Beetle 2 by 72 cm horizontally, and will have traveled ...
24√13 = 24√13 . . . . . down slope
Of course, it will have traveled down a vertical distance of 24·2 = 48 cm.
Meanwhile, Beetle 2 will have traveled half as far, but in the downward direction. It will be (24√13)/2 = 12√13 cm below the starting position.
The difference in height from Beetle 1 to Beetle 2 is ...
(-48 cm) -(-12√13 cm) = 12(√13 -4) cm ≈ -4.733 cm
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Additional comment
The question asks for the altitude Beetle 1 is above Beetle 2. In both cases, Beetle 1 is below Beetle 2, so that altitude is negative.
The first two attachments show the geometry of the problem. The last attachment shows a way the altitude can be calculated using complex numbers.
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