A balloon is at a height of 50 meters, and it is rising at the constant rate of 5m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10m/sec. How fast is the distance between the bicyclist and the balloon increasing 10 seconds later?

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04-10-2016
Mathematics

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A balloon is at a height of 50 meters, and it is rising at the constant rate of 5m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10m/sec. How fast is the distance between the bicyclist and the balloon increasing 10 seconds later?

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HomertheGenius HomertheGenius · Answer 1 · 2016-10-10T22:00:19+02:00

After 10 seconds the balloon will be at the height of 100 meters ( 50 + 5 * 10 = 50 + 50 = 100 m ).
After 10 seconds the bicyclist will travel the distance of: 10 * 10 = 100 m
The distance between the bicyclist and the balloon was at the start 50 m
( at t = 0 s) and after 10 seconds ( at t = 10 s ) it will be:
d² = 100² + 100² = 10,000 + 10,000 = 20,000
d = √20,000 = 100√2 = 141.42 m
The rate:
v = (141.42 - 50)m / 10 s = 9.142 m/s
The distance is increasing at the rate of 9.142 m/s.

scme1702 scme1702 · Answer 2 · 2016-10-10T22:47:03+02:00

The path of the balloon and cyclist form two perpendicular lines and when connected, they form a right angle triangle. The base of this triangle is the distance covered by the cyclist. The height is the height of the balloon. And the hypotenuse is the distance between the two.
At 10 seconds,
The balloon will be at 100 m
The cyclist will be at 100 m
The distance will be (applying Pythagoras Theorem):
√(100² + 100²)
= 141.4 m
The rate of increase is given by the height increase per second of each subject in the same manner.
The height rises 5 m/s
The base increases 10 m/s
Therefore, the distance increase:
√(5² + 10²)
= 11.2 m/s