Forty Minutes Out: Speed

Q. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. What is the average speed of the man over the interval of time 0 to 40 min, in km/h? expects 5.625 km/h

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Q. A man walks on a straight road $v_{av} (to) = c_{av} (to)$ from his home $x (start) = 0$ km to a market 2.5 km $x (market) = 2.5$ km away with a speed of 5 km/h $c_{av} (to) = 5$ km/h. Finding the market closed, he instantly turns and walks back home $v_{av} (tu) = - c_{av} (tu)$ with a speed of 7.5 km/h $c_{av} (tu) = 7.5$ km/h. What is the average speed of the man over the interval of time 0 to 40 min $Δ t_{w} = 40$ min, in km/h?

The playground

Inputs

distance

x

2.5 km

speed there

c_{av}

5 km/h

speed back

c_{av}

7.5 km/h

interval

Δ t_{w}

40 min

Outputs

average speed

c_{av}

The solution

Solved with Velocity

Given

$x (market) = 2.5$ km
$c_{av} (to) = 5$ km/h
$c_{av} (tu) = 7.5$ km/h
$Δ t_{w} = 40$ min
$x (start) = 0$ km from “from his home”
$v_{av} (to) = c_{av} (to)$ from “on a straight road”
$v_{av} (tu) = - c_{av} (tu)$ from “walks back home”

Find $c_{av} (w)$ .

Working

$x (market) = 2.5$ km $= 2500$ m
$c_{av} (to) = 5$ km/h $= 1.38888889$ m/s
$c_{av} (tu) = 7.5$ km/h $= 2.08333333$ m/s
$Δ t_{w} = 40$ min $= 2400$ s
$x (start) = 0$ km $= 0$ m
on a straight road $v_{av} (to) = c_{av} (to)$
$v_{av} (to) = 1.38888889$
$v_{av} (to) = 1.38888889$ m/s
change of position(to market) $Δ x (to) = x (market) - x (start)$
$Δ x (to) = 2500 - 0$
$Δ x (to) = 2500$ m
displacement a time(to market) $v_{av} (to) = \frac{Δ x (to)}{Δ t_{to}}$
$1.38888889 = \frac{2500}{Δ t_{to}}$
$Δ t_{to} = 1800$ s
path a time(to market) $c_{av} (to) = \frac{d (to)}{Δ t_{to}}$
$1.38888889 = \frac{d (to)}{1800}$
$d (to) = 2500$ m
time in parts(walking) $Δ t_{w} = Δ t_{to} + Δ t_{tu}$
$2400 = 1800 + Δ t_{tu}$
$Δ t_{tu} = 600$ s
path a time(turning back) $c_{av} (tu) = \frac{d (tu)}{Δ t_{tu}}$
$2.08333333 = \frac{d (tu)}{600}$
$d (tu) = 1250$ m
path in parts(walking) $d (w) = d (to) + d (tu)$
$d (w) = 2500 + 1250$
$d (w) = 3750$ m
path a time(walking) $c_{av} (w) = \frac{d (w)}{Δ t_{w}}$
$c_{av} (w) = \frac{3750}{2400}$
$c_{av} (w) = 5.625$ km/h
$c_{av} (w) = 1.5625$ m/s $= 5.625$ km/h
$c_{av} (w) \approx 5.625$ km/hto 4 significant figures

Conclusion

Average speed counts the whole path, out to the market and part of the way back, so it is more than the size of the average velocity over the same time.

The Ruby

1require_relative "../physics"23question :forty_minutes_out_speed do4  kind :numerical5  describes "A man walks on a straight road from his home to a market %{distance} away with a speed of " \6            "%{speed_there}. Finding the market closed, he instantly turns and walks back home with a speed of " \7            "%{speed_back}. %{average_speed} over the interval of time 0 to %{interval}, in km/h?"8  given distance: "2.5 km", speed_there: "5 km/h", speed_back: "7.5 km/h", interval: "40 min"9  asks average_speed: "What is the average speed of the man"10end

1require_relative "../questions/forty_minutes_out_speed"2require_relative "../science/motion_in_a_straight_line/velocity"34solve :forty_minutes_out_speed, using: Velocity do5  objects :man6  instants :start, :market, :later7  man.goes(:to_market, during: start..market)8  man.goes(:turning_back, during: market..later)9  man.goes(:walking, during: start..later)10  take distance: man.position(at: market), speed_there: to_market.average_speed,11       speed_back: turning_back.average_speed, interval: walking.time_taken12  deduce_from "from his home", man.position(at: start) => "0 km"13  deduce_from "on a straight road", to_market.average_velocity => to_market.average_speed14  deduce_from "walks back home", turning_back.average_velocity => -turning_back.average_speed15  average_speed = walking.average_speed16  answer average_speed:, in: "km/h", figures: 417  conclusion "Average speed counts the whole path, out to the market and part of the way back, so it is " \18             "more than the size of the average velocity over the same time."19end

1require_relative "../solutions/forty_minutes_out_speed"23play_with :forty_minutes_out_speed do4  title "Forty Minutes Out: 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	$Δ x (to)$	m	—
position of man at market	$x (market)$	m	—
position of man at start	$x (start)$	m	—
displacement of man during turning back	$Δ x (tu)$	m	—
position of man at later	$x (later)$	m	—
displacement of man during walking	$Δ x (w)$	m	—
average velocity of man during to market	$v_{av} (to)$	m/s	—
time taken of to market	$Δ t_{to}$	s	0 … 1e+05
average velocity of man during turning back	$v_{av} (tu)$	m/s	—
time taken of turning back	$Δ t_{tu}$	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	$c_{av} (to)$	m/s	—
path length of man during to market	$d (to)$	m	—
average speed of man during turning back	$c_{av} (tu)$	m/s	—
path length of man during turning back	$d (tu)$	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 (to) = x (market) - x (start)

change of position(to market)

Δ x (tu) = x (later) - x (market)

change of position(turning back)

Δ x (w) = x (later) - x (start)

change of position(walking)

v_{av} (to) = \frac{Δ x (to)}{Δ t_{to}}

displacement a time(to market)

v_{av} (tu) = \frac{Δ x (tu)}{Δ t_{tu}}

displacement a time(turning back)

v_{av} (w) = \frac{Δ x (w)}{Δ t_{w}}

displacement a time(walking)

c_{av} (to) = \frac{d (to)}{Δ t_{to}}

path a time(to market)

c_{av} (tu) = \frac{d (tu)}{Δ t_{tu}}

path a time(turning back)

c_{av} (w) = \frac{d (w)}{Δ t_{w}}

path a time(walking)

d (w) = d (to) + d (tu)

path in parts(walking)

Δ t_{w} = Δ t_{to} + Δ t_{tu}

time in parts(walking)

v_{av} (to) = c_{av} (to)

on a straight road

v_{av} (tu) = - c_{av} (tu)

walks back home