3.4 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 1m long and requires 1 s. Plot the xt graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

3.4

The time taken for each step is 1s, he covers 5 m in 5 secs and goes backward by 3 m in 3 secs.

So in 8 s he covers 2 m. To cover 13m distance he needs to complete 8 m (13-5). To cover 8m, he will require 8 * 4 = 32 s. To fall in a pit, he needs to move another 5m in 5s.

So total time taken = 32 +5 = 37 s

Let’s have a look at how we came to this answer. 

We have to calculate how long it will take for the drunkard to fall into a pit located 13 metres from his starting point.

In each cycle of movement, the drunkard takes 5 steps forward and 3 steps backward. We calculate the net displacement per cycle. 

Time for one cycle =

...more

3.4

The time taken for each step is 1s, he covers 5 m in 5 secs and goes backward by 3 m in 3 secs.

So in 8 s he covers 2 m. To cover 13m distance he needs to complete 8 m (13-5). To cover 8m, he will require 8 * 4 = 32 s. To fall in a pit, he needs to move another 5m in 5s.

So total time taken = 32 +5 = 37 s

Let’s have a look at how we came to this answer. 

We have to calculate how long it will take for the drunkard to fall into a pit located 13 metres from his starting point.

In each cycle of movement, the drunkard takes 5 steps forward and 3 steps backward. We calculate the net displacement per cycle. 

Time for one cycle = 5 seconds (forward) + 3 seconds (backward) = 8 seconds.

Net displacement in one cycle = 5 meters (forward) - 3 meters (backward) = 2 meters.

Reaching the pit: The pit is 13 metres away. We can determine the time taken by analysing his position after each cycle.

After 1 cycle (8 s): Position = 2 m

After 2 cycles (16 s): Position = 4 m

After 3 cycles (24 s): Position = 6 m

After 4 cycles (32 s): Position = 8 m

At the 32-second mark, the drunkard is at the 8-metre position. This we get to know when learning about instantaneous velocity. 

In his next set of forward steps, he will cover 5 metres in 5 seconds. This means he will reach the 13-metre mark and fall into the pit. 

Total time calculation: The total time is the time taken to reach the 8-metre mark plus the time to take the final 5 steps.

Total time = 32 seconds + 5 seconds = 37 seconds.

Practice more NCERT Solutions for Motion in a Straight Line chapter.

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<p><strong>3.4</strong></p><div><div><strong><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728557003phpNef5ng_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728557003phpNef5ng.jpeg" alt="" width="298" height="148"></picture></strong></div></div><p>The time taken for each step is 1s, he covers 5 m in 5 secs and goes backward by 3 m in 3 secs.</p><p>So in 8 s he covers 2 m. To cover 13m distance he needs to complete 8 m (13-5). To cover 8m, he will require 8 * 4 = 32 s. To fall in a pit, he needs to move another 5m in 5s.</p><p>So total time taken = 32 +5 = 37 s</p><p>Let’s have a look at how we came to this answer.&nbsp;</p><p dir="ltr">We have to calculate how long it will take for the drunkard to fall into a pit located 13 metres from his starting point.</p><p dir="ltr">In each cycle of movement, the drunkard takes 5 steps forward and 3 steps backward. We calculate the <strong>net displacement per cycle</strong>.&nbsp;</p><p dir="ltr"><strong>Time for one cycle </strong>= 5 seconds (forward) + 3 seconds (backward) = 8 seconds.</p><p dir="ltr"><strong>Net displacement in one cycle </strong>= 5 meters (forward) - 3 meters (backward) = 2 meters.</p><p dir="ltr"><strong>Reaching the pit</strong>: The pit is 13 metres away. We can determine the time taken by analysing his position after each cycle.</p><p dir="ltr">After 1 cycle (8 s): Position = 2 m</p><p dir="ltr">After 2 cycles (16 s): Position = 4 m</p><p dir="ltr">After 3 cycles (24 s): Position = 6 m</p><p dir="ltr">After 4 cycles (32 s): Position = 8 m</p><p dir="ltr">At the 32-second mark, the drunkard is at the 8-metre position. This we get to know when learning about &lt;a href="https://www.shiksha.com/preparation/physics-motion-in-straight-line-instantaneous-velocity-and-speed-5436-tp"&gt;instantaneous velocity&lt;/a&gt;.&nbsp;</p><p dir="ltr">In his next set of forward steps, he will cover 5 metres in 5 seconds. This means he will reach the 13-metre mark and fall into the pit.&nbsp;</p><p dir="ltr"><strong>Total time calculation</strong>: The total time is the time taken to reach the 8-metre mark plus the time to take the final 5 steps.</p><p dir="ltr">Total time = 32 seconds + 5 seconds = 37 seconds.</p><p dir="ltr">Practice more&nbsp;&lt;a href="https://www.shiksha.com/preparation/physics-ncert-solutions-class-11th-motion-in-a-straight-line-4309-ns"&gt;NCERT Solutions for Motion in a Straight Line chapter&lt;/a&gt;.<strong> </strong></p>

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