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This is a related rates problem involving a balloon rising vertically and a bicycle moving horizontally. Here's the breakdown:

### Given:
- The balloon rises vertically at a constant rate of \( \frac{dy}{dt} = 2 \, \text{ft/sec} \).
- The bicycle moves horizontally at a constant rate of \( \frac{dx}{dt} = 17 \, \text{ft/sec} \).
- Initially, the balloon is 124 feet above the ground when the bicycle is directly below it (at time \( t = 0 \)).
- We need to find how fast the distance \( s(t) \) between the balloon and the bicycle is increasing 6 seconds later.

### Step-by-step Solution:

1. **Position of the Balloon**:  
   The balloon is rising at a rate of 2 ft/sec, so its height at time \( t \) is:  
   \[
   y(t) = 124 + 2t
   \]
   After 6 seconds:  
   \[
   y(6) = 124 + 2(6) = 136 \, \text{feet}
   \]

2. **Position of the Bicycle**:  
   The bicycle is moving horizontally at a rate of 17 ft/sec, so its position at time \( t \) is:  
   \[
   x(t) = 17t
   \]
   After 6 seconds:  
   \[
   x(6) = 17(6) = 102 \, \text{feet}
   \]

3. **Distance Between the Balloon and the Bicycle**:  
   By the Pythagorean theorem, the distance \( s(t) \) between the balloon and the bicycle is:
   \[
   s(t) = \sqrt{x(t)^2 + y(t)^2}
   \]
   At \( t = 6 \):
   \[
   s(6) = \sqrt{102^2 + 136^2} = \sqrt{10404 + 18496} = \sqrt{28900} = 170 \, \text{feet}
   \]

4. **Related Rates**:  
   Differentiate \( s(t) \) with respect to \( t \):
   \[
   \frac{ds}{dt} = \frac{1}{2} \cdot \frac{2x(t)\frac{dx}{dt} + 2y(t)\frac{dy}{dt}}{\sqrt{x(t)^2 + y(t)^2}}
   \]
   Simplifying the equation:
   \[
   \frac{ds}{dt} = \frac{x(t) \cdot \frac{dx}{dt} + y(t) \cdot \frac{dy}{dt}}{s(t)}
   \]

   Substituting the values at \( t = 6 \):
   \[
   \frac{ds}{dt} = \frac{102 \cdot 17 + 136 \cdot 2}{170}
   \]
   \[
   \frac{ds}{dt} = \frac{1734 + 272}{170} = \frac{2006}{170} = 11.8 \, \text{ft/sec}
   \]

### Final Answer:
The distance between the balloon and the bicycle is increasing at **11.8 ft/sec** after 6 seconds.

Would you like a more detailed explanation or any clarifications?

Here are 5 related questions:
1. How would the answer change if the balloon were rising at a different rate?
2. What if the bicycle was accelerating instead of moving at a constant speed?
3. How would you calculate the total distance traveled by the bicycle in this scenario?
4. What is the significance of the Pythagorean theorem in related rates problems?
5. How would the problem differ if there were air resistance affecting the balloon?

**Tip:** Always check units and simplify expressions during related rates problems to avoid unnecessary complexity.

A balloon is rising vertically above a level, straight road at a constant rate of 2 ft/sec. Just when the balloon is 124 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 6 seconds later?

Solve this related rates problem where a balloon rises at 2 ft/sec, and a bicycle moves at 17 ft/sec. Learn how to calculate the rate at which the distance between the balloon and the bicycle changes after 6 seconds using derivatives and the Pythagorean theorem.

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