Question

A frog starts climbing \(30 \mathrm{ft}\) well. Each hour frog climbs \(3 \mathrm{ft}\) and slips back \(2 \mathrm{ft}\). How many hours does it take to reach top and get out?

Short answer

The frog takes 29 hours to climb out of the well.

Step-by-step solution

Understand Daily Progress

Calculate the net distance climbed by the frog in one hour. The frog climbs 3 ft but slips back 2 ft. Therefore, net progress per hour is 3 ft - 2 ft = 1 ft.

Calculate Total Time to Reach Near the Top

To climb 30 ft, the frog needs to make progress of about 28 ft first (so it can reach exactly 30 ft in the final jump without slipping back). Since the frog climbs 1 ft per hour, it takes 28 hours to reach 29 ft, which is near the top and not quite 30 ft.

Calculate Final Hour Progress

Once the frog reaches 29 ft, in the next hour, it will climb 3 ft. So, in the 29th hour, it reaches 29 + 3 = 32 ft. At this point, the frog climbs out of the well after 28 hours and completes this process in the 29th hour.

Key concepts

Climbing Problems

Climbing problems often involve scenarios where progress is made with a certain amount of effort, but obstacles cause setbacks. To solve these problems, understanding the net progress is crucial. Let's consider the example of a frog climbing a well. With each effort (or hour, in our case), the frog climbs 3 feet, but then slips back 2 feet due to a slick well wall.

To determine how much progress the frog actually makes, we calculate the net gain, which is the difference between the ascent and the backslide: 3 feet climbed minus 2 feet slipped equals 1 foot net gain per hour. This simple calculation, however, doesn't immediately tell the entire story of the climb due to the final ascent.

• Initial assumption: Calculate your net progress for small steps first.
• Always account for the setback in your calculations.
• Determine how much net progress is needed before considering the effect of the final push.

Distance and Speed

In problems like these, the concepts of distance and speed play crucial roles. Although there's no vehicle involved, the action of the frog can be likened to a real-world journey with a start, a path, and an end. Here, the distance is the 30-foot well, which can be compared to a specific route to travel.

The speed, in this analogy, is akin to the frog's hourly net movement. Understanding this lets us answer how quickly the frog moves towards its goal per unit of time—in this scenario, 1 foot per hour. But what happens when speed meets resistance, such as the slipping back? We have to adjust our final timing expectations.

• Evaluate how speed changes with effective movement (3 ft/hour, but reset to 1 ft/hour after slips).
• Acknowledge intermittent setbacks, akin to wait times or stops in an actual journey.
• Ensure that the endpoint calculation accounts for favorable last-phase conditions.

Logical Reasoning

Logical reasoning is at the heart of solving climbing and distance problems. By breaking down the overall task into smaller parts, you leverage logic to see the path to the solution. Consider each aspect of the problem separately, then integrate your findings. Here, we divided the frog's journey into two phases: reaching near the top and making the final jump.

First, we reason that since the frog moves net 1 foot per hour, it will require 28 hours to climb 28 feet. By saving the climactic leap out of the well for the end, we bypass slipping back. This logical separation of journey stages ensures accuracy in solutions.

• Use logical blocks to solve segmented parts of the problem.
• Plan for the problem's conclusion in a separate step.
• Verify if assumptions correctly map onto real-world scenarios or outcomes.