Q.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 $1\ \text{m}$ long and requires $1\ \text{s}$. Plot the $x$-$t$ graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit $13\ \text{m}$ away from the start.

Concept: Periodic 1-D motion — the pit can be reached mid-surge, not just at the end of a full forward-backward cycle.

Step 1. Each cycle (5 steps forward + 3 steps back) takes $8\ \text{s}$ and gives a net advance of $5-3=2\ \text{m}$. But checking only the position at the end of each cycle misses the fact that during a forward surge, the drunkard reaches a peak $5\ \text{m}$ above the previous cycle's floor — higher than his net position.

Step 2. Position at the end of each full cycle:

Setting up the motion

Each step is 1 m and takes 1 s.

• 5 steps forward → +5 m in 5 s.
• 3 steps backward → −3 m in 3 s.

So one full cycle takes 8 s and gives a net displacement of +2 m.

Reaching the pit

The pit is 13 m from the start. The key point is that during each forward surge the drunkard climbs to a peak higher than his net position, so he can reach the pit mid-cycle.

Position at the end of each completed cycle:

After 4 cycles he is at 8 m at t = 32 s. On the very next forward surge he steps forward one metre at a time:

reaching 13 m after 5 more steps (5 s):

$t = 32 + 5 = 37\ \text{s}$ …

Concept: A Closed-Form Position Function for Periodic Motion

Method: Algebraic Piecewise Formula (solve directly, no cycle-by-cycle table)

Instead of building a table of the drunkard's position at the end of each completed 8-second cycle and then checking the next partial cycle by hand, this method writes a single general formula for his position during the $k$-th cycle's forward phase, and solves it directly for when $x=13\ \text{m}$.

Steps

1. Characterise one cycle. Each cycle: 5 forward steps (5 m in 5 s), then 3 backward steps (3 m in 3 s) — total duration $8\ \text{s}$, net displacement $+2\ \text{m}$ per cycle.

2. Position at the start of cycle $k$ (for $k = 0, 1, 2, \ldots$, cycles counted from zero), i.e. at $t = 8k$, after $k$ complete cycles of net $+2\ \text{m}$ each:

3. Write position during the forward phase of cycle $k$ — for $8k \le t \le 8k+5$, he is walking steadily forward from $x=2k$:

(Position is increasing throughout this window, so it's the only phase where the pit — a fixed forward threshold — can be reached for the first time.)

4. Find the smallest $k$ for which this forward phase reaches $x=13$. The maximum $x$ reached in cycle $k$'s forward phase is $2k+5$; we need $2k+5 \ge 13$:

Check $k=3$: max reach $=2(3)+5=11 < 13$ — not far enough. So $k=4$ is the first cycle whose forward phase can reach the pit.

5. Solve $x(t)=13$ within cycle $k=4$'s forward phase, $32 \le t \le 37$: $2(4) + (t-32) = 13 \;\Rightarrow\; 8 + t - 32 = 13 \;\Rightarrow\; t = 37$ …