The Drunkard's Walk

solved Velocity

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?

How it's solved

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 web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

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

solution.rb

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

playground.rb

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

The laws it drew on

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