There and Back: Speed

Q. A man walks 2.5 km to a market in 30 min, finds it shut, and walks straight home in 20 min. What is his average speed over the whole walk, in km/h? expects 6 km/h

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

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}

The solution

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 Ruby

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

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

1require_relative "../solutions/there_and_back_speed"23play_with :there_and_back_speed do4  title "There and Back: Speed"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
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)