/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 134 A drunkard walking in a narrow l... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 134

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is \(1 \mathrm{~m}\) long and requires \(1 \mathrm{~s}\). How long will it take for the drunkard to fall in a pit \(13 \mathrm{~m}\) away from the start. (A) \(13 \mathrm{~s}\) (B) \(16 \mathrm{~s}\) (C) \(24 \mathrm{~s}\) (D) \(32 \mathrm{~s}\)

Short Answer

Expert verified
The drunkard will take \(49 \mathrm{~s}\) to fall into the pit. Note that answer is not among the provided options, suggesting an error in the question.

Step by step solution

01

Calculate effective steps per cycle

In a cycle, the drunkard takes 5 steps forward and 3 steps backward, implying an effective movement of \(5 \mathrm{~m}\) - \(3 \mathrm{~m} = 2 \mathrm{~m}\) forward.
02

Calculate number of cycles

Knowing the drunkard effectively moves \(2 \mathrm{~m}\) per cycle, and the pit is \(13 \mathrm{~m}\) away, the number of full cycles needed is \(13 \mathrm{~m}\) / \(2 \mathrm{~m}\) per cycle = \(6.5\) cycles. Since the drunkard cannot perform half a cycle, the number of cycles he will complete is 6.
03

Calculate time spent in full cycles

A full cycle takes \(5 \mathrm{~s}\) (for \5 steps forward) + \(3 \mathrm{~s}\) (for 3 steps backward) = \(8 \mathrm{~s}\). So for 6 complete cycles, it will take \(6 \times 8 \mathrm{~s} = 48 \mathrm{~s}\).
04

Calculate steps and time in the last incomplete cycle

After the 6 complete cycles, the drunkard is 2 meters away from the pit (\(13 - 6 \times 2 = 1 \mathrm{~m}\)). So, he will take 1 step forward to fall into the pit, which will take \(1 \mathrm{~s}\).
05

Sum the time of full cycles and the last incomplete cycle

The total time taken to fall into the pit will then be the time for all the full cycles plus the time taken in the last cycle. So, \(48 \mathrm{~s}\) + \(1 \mathrm{~s}\) = \(49 \mathrm{~s}\).

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Most popular questions from this chapter

A man wants to reach point \(C\) on the opposite bank of river. He can swim in still water with speed \(5 \mathrm{~m} / \mathrm{s}\) and can walk on ground with \(10 \mathrm{~m} / \mathrm{s}\). Velocity of river is \(3 \mathrm{~m} / \mathrm{s}\) and given \(A B=B C=500 \mathrm{~m} .\) $$ \begin{aligned} &\text { Column-I } & \text { Column-II } \\ &\hline \text { (A) The time to reach point } C \text { if } & \text { (1) } 125 \mathrm{~s} \\ &\text { he crosses the river by shortest } \\ &\text { path is } \\ &\text { (B) The time to reach point } C \text { if } & \text { (2) } 175 \mathrm{~s} \\ &\text { he crosses the river in shortest } \\ &\text { time. } \\ &\text { (C) The time to reach point } C \text { if } & \text { (3) } 120 \mathrm{~s} \\ &\text { he crosses the river along } & \text { (4) } 106 \mathrm{~s} \\ &\text { path } A C \text { (approximate value). } & \\ &\text { (D) Time to reach point } B \\ &\text { with zero drift } \end{aligned} $$

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On a two-lane road, car A is travelling with a speed of \(36 \mathrm{~km} \mathrm{~h}^{-1}\). Two cars \(\mathrm{B}\) and \(\mathrm{C}\) approach car \(\mathrm{A}\) in opposite directions with a speed of \(54 \mathrm{~km} \mathrm{~h}^{-1}\) each. At a certain instant, when the distance \(\mathrm{AB}\) is equal to \(\mathrm{AC}\), both being \(1 \mathrm{~km} . \mathrm{B}\) decides to overtake \(\mathrm{A}\) before C does. What minimum acceleration of car \(\mathrm{B}\) is required to avoid an accident? (A) \(3 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(2 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(1 \mathrm{~m} / \mathrm{s}^{2}\) (D) None

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