There and Back: Speed

solved Velocity

Q. A man walks 2.5 km $d (t) = 2.5$ km to a market in 30 min $Δ t_{t} = 30$ min, finds it shut, and walks straight home $d (b) = 2.5$ km in 20 min $Δ t_{b} = 20$ min. What is his average speed over the whole walk, in km/h?

Inputs

path there

d

2.5 km

time there

Δ t_{t}

30 min

time back

Δ t_{b}

20 min

Outputs

average speed

c_{av}

How it's solved

Solved with Velocity

Given

$d (t) = 2.5$ km
$Δ t_{t} = 30$ min
$Δ t_{b} = 20$ min
$d (b) = 2.5$ km from “walks straight home”

Find $c_{av} (w)$ .

Working

$d (t) = 2.5$ km $= 2500$ m
$Δ t_{t} = 30$ min $= 1800$ s
$Δ t_{b} = 20$ min $= 1200$ s
$d (b) = 2.5$ km $= 2500$ m
path in parts(walking) $d (w) = d (t) + d (b)$
$d (w) = 2500 + 2500$
$d (w) = 5000$ m
time in parts(walking) $Δ t_{w} = Δ t_{t} + Δ t_{b}$
$Δ t_{w} = 1800 + 1200$
$Δ t_{w} = 3000$ s
path a time(walking) $c_{av} (w) = \frac{d (w)}{Δ t_{w}}$
$c_{av} (w) = \frac{5000}{3000}$
$c_{av} (w) = 6$ km/h
$c_{av} (w) = 1.66666667$ m/s $= 6$ km/h

Conclusion

Average speed counts every step of the path, so it is never less than the size of the average velocity, and a tired man is not told he stood still.

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 :there_and_back_speed do4  kind :numerical5  describes "A man walks %{path_there} to a market in %{time_there}, finds it shut, and walks straight home in " \6          "%{time_back}. %{average_speed}, in km/h?"7  given path_there: "2.5 km", time_there: "30 min", time_back: "20 min"8  asks average_speed: "What is his average speed over the whole walk"9end

solution.rb

1require_relative "../questions/there_and_back_speed"2require_relative "../science/motion_in_a_straight_line/velocity"34solve :there_and_back_speed, using: Velocity do5  objects :man6  instants :start, :market, :home7  man.goes(:to_market, during: start..market)8  man.goes(:back_home, during: market..home)9  man.goes(:walking, during: start..home)10  take path_there: to_market.path_length, time_there: to_market.time_taken, time_back: back_home.time_taken11  deduce_from "walks straight home", back_home.path_length => "2.5 km"12  average_speed = walking.average_speed13  answer average_speed:, in: "km/h"14  conclusion "Average speed counts every step of the path, so it is never less than the size of " \15             "the average velocity, and a tired man is not told he stood still."16end

playground.rb

1require_relative "../solutions/there_and_back_speed"23play_with :there_and_back_speed do4  title "There and Back: Speed"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
displacement of man during to market	$Δ x (t)$	m	—
position of man at market	$x (market)$	m	—
position of man at start	$x (start)$	m	—
displacement of man during back home	$Δ x (b)$	m	—
position of man at home	$x (home)$	m	—
displacement of man during walking	$Δ x (w)$	m	—
average velocity of man during to market	$v_{av} (t)$	m/s	—
time taken of to market	$Δ t_{t}$	s	0 … 1e+05
average velocity of man during back home	$v_{av} (b)$	m/s	—
time taken of back home	$Δ t_{b}$	s	0 … 1e+05
average velocity of man during walking	$v_{av} (w)$	m/s	—
time taken of walking	$Δ t_{w}$	s	0 … 1e+05
average speed of man during to market	$c_{av} (t)$	m/s	—
path length of man during to market	$d (t)$	m	—
average speed of man during back home	$c_{av} (b)$	m/s	—
path length of man during back home	$d (b)$	m	—
average speed of man during walking	$c_{av} (w)$	m/s	—
path length of man during walking	$d (w)$	m	—

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 (t) = x (market) - x (start)

change of position(to market)

Δ x (b) = x (home) - x (market)

change of position(back home)

Δ x (w) = x (home) - x (start)

change of position(walking)

v_{av} (t) = \frac{Δ x (t)}{Δ t_{t}}

displacement a time(to market)

v_{av} (b) = \frac{Δ x (b)}{Δ t_{b}}

displacement a time(back home)

v_{av} (w) = \frac{Δ x (w)}{Δ t_{w}}

displacement a time(walking)

c_{av} (t) = \frac{d (t)}{Δ t_{t}}

path a time(to market)

c_{av} (b) = \frac{d (b)}{Δ t_{b}}

path a time(back home)

c_{av} (w) = \frac{d (w)}{Δ t_{w}}

path a time(walking)

d (w) = d (t) + d (b)

path in parts(walking)

Δ t_{w} = Δ t_{t} + Δ t_{b}

time in parts(walking)