The Drunkard's Walk

Q. A drunkard in a narrow lane takes 5 steps forward and 3 back, again and again, each step 1 m long and 1 s long. How long before he falls into a pit 13 m ahead? expects A

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Q. A drunkard in a narrow lane takes 5 steps forward and 3 back, again and again, each step 1 m $Δ x = 1$ m long and 1 s $Δ t = 1$ s long. How long before he falls into a pit 13 m $x = 13$ m ahead?

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The solution

Solved with Velocity

Given

$Δ x = 1$ m
$Δ t = 1$ s
$x = 13$ m
forward = 5
back = 3

Assumed

$x_{0} = 0$ m

Working

$Δ x = 1$ m, $Δ t = 1$ s
displacement a time $v_{av} = \frac{Δ x}{Δ t}$
$v_{av} = 1$ m/s
$Δ x = −1$ m, $Δ t = 1$ s $\Rightarrow$ $v_{av} = −1$ m/s
$x_{0} = 0$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s
displacement a time $v_{av} = \frac{Δ x}{Δ t}$
$Δ x = 1$ m
change of position $Δ x = x - x_{0}$
$x = 1$ m
$x_{0} = 1$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 2$ m
$x_{0} = 2$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 3$ m
$x_{0} = 3$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 4$ m
$x_{0} = 4$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 5$ m
$x_{0} = 5$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 4$ m
$x_{0} = 4$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 3$ m
$x_{0} = 3$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 2$ m
$x_{0} = 2$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 3$ m
$x_{0} = 3$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 4$ m
$x_{0} = 4$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 5$ m
$x_{0} = 5$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 6$ m
$x_{0} = 6$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 7$ m
$x_{0} = 7$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 6$ m
$x_{0} = 6$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 5$ m
$x_{0} = 5$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 4$ m
$x_{0} = 4$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 5$ m
$x_{0} = 5$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 6$ m
$x_{0} = 6$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 7$ m
$x_{0} = 7$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 8$ m
$x_{0} = 8$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 9$ m
$x_{0} = 9$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 8$ m
$x_{0} = 8$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 7$ m
$x_{0} = 7$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 6$ m
$x_{0} = 6$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 7$ m
$x_{0} = 7$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 8$ m
$x_{0} = 8$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 9$ m
$x_{0} = 9$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 10$ m
$x_{0} = 10$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 11$ m
$x_{0} = 11$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 10$ m
$x_{0} = 10$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 9$ m
$x_{0} = 9$ m, $v_{av} = −1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 8$ m
$x_{0} = 8$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 9$ m
$x_{0} = 9$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 10$ m
$x_{0} = 10$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 11$ m
$x_{0} = 11$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 12$ m
$x_{0} = 12$ m, $v_{av} = 1$ m/s, $Δ t = 1$ s $\Rightarrow$ $x = 13$ m

Conclusion

Following him second by second is the point: averaging his progress would miss that he gets there at the end of a forward stretch, before he can step back.

The Ruby

1require_relative "../physics"23question :the_drunkards_walk do4  kind :scq5  describes "A drunkard in a narrow lane takes %{forward} steps forward and %{back} back, again and again, each step " \6          "%{step} long and %{step_time} long. How long before he falls into a pit %{pit} ahead?"7  given forward: 58  given back: 39  given step: "1 m"10  given step_time: "1 s"11  given pit: "13 m"1213  option "37 s"14  option "39 s"15  option "52 s"16  option "65 s"17end

1require_relative "../questions/the_drunkards_walk"2require_relative "../science/motion_in_a_straight_line/velocity"34solve :the_drunkards_walk, using: Velocity do5  take step: :Delta_x, step_time: :Delta_t, pit: :x6  assume x_0: "0 m"7  step = value_of(:Delta_x)8  tick = value_of(:Delta_t)9  pit = value_of(:x)10  forward = find_value :v_av, given: { Delta_x: step, Delta_t: tick }11  back = find_value :v_av, given: { Delta_x: -step, Delta_t: tick }12  rhythm = [ forward ] * value_of(:forward) + [ back ] * value_of(:back)13  x = value_of(:x_0)14  t = 015  rhythm.cycle do |v|16    x = find_value :x, given: { x_0: x, v_av: v, Delta_t: tick }17    t += tick18    break if x >= pit19  end2021  answer t2223  conclusion "Following him second by second is the point: averaging his progress would miss that " \24             "he gets there at the end of a forward stretch, before he can step back."25end

1require_relative "../solutions/the_drunkards_walk"23play_with :the_drunkards_walk do4  title "The Drunkard's Walk"5end

The laws

quantity	written	units	within	from
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

Δ x = x - x_{0}

change of position

v_{av} = \frac{Δ x}{Δ t}

displacement a time

c_{av} = \frac{d}{Δ t}

path a time