The Long Way Round: Velocity

solved MotionInAPlane

Q. A hotel is 10 km $x = 10$ km from the station along a straight road, but a dishonest driver takes a 23 km route and gets there in 28 min $Δ t = 28$ min. What is the size of the taxi's average velocity, in km/h?

Inputs

distance

x

10 km

time

Δ t

28 min

Outputs

average velocity

v_{av}

How it's solved

Solved with MotionInAPlane

Given

$x = 10$ km
$Δ t = 28$ min

Assumed

$x_{0} = 0$ km

Find $v_{av}$ .

Working

$x = 10$ km $= 10000$ m
$Δ t = 28$ min $= 1680$ s
$x_{0} = 0$ km $= 0$ m
change of position $Δ x = x - x_{0}$
$Δ x = 10000 - 0$
$Δ x = 10000$ m
displacement a time $v_{av} = \frac{Δ x}{Δ t}$
$v_{av} = \frac{10000}{1680}$
$v_{av} = 21.43$ km/h
$v_{av} = 5.95238095$ m/s $= 21.4285714$ km/h
$v_{av} \approx 21.4$ km/hto 3 significant figures

Conclusion

Average velocity asks only how far the hotel is from the station, so the detour makes the taxi slower by this measure, not faster. The two agree only for a straight run.

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_long_way_round_velocity do4  kind :numerical5  describes "A hotel is %{distance} from the station along a straight road, but a dishonest driver takes a " \6          "23 km route and gets there in %{time}. %{average_velocity}, " \7          "in km/h?"8  given distance: "10 km"9  given time: "28 min"10  asks average_velocity: "What is the size of the taxi's average velocity"11end

solution.rb

1require_relative "../questions/the_long_way_round_velocity"2require_relative "../science/motion_in_a_plane/motion_in_a_plane"34solve :the_long_way_round_velocity, using: MotionInAPlane do5  take distance: :x, time: :Delta_t6  assume x_0: "0 km"7  take average_velocity: :v_av8  answer find_value(:average_velocity, in: "km/h", figures: 3)910  conclusion "Average velocity asks only how far the hotel is from the station, so the detour " \11             "makes the taxi slower by this measure, not faster. The two agree only for a " \12             "straight run."13end

playground.rb

1require_relative "../solutions/the_long_way_round_velocity"23play_with :the_long_way_round_velocity do4  title "The Long Way Round: Velocity"5end

The laws it drew on

quantity	written	units	within	from
volume	$volume$	—	—	Vectors
mass	$mass$	—	—	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$work$	—	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$power$	—	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
charge	$charge$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$force$	—	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
gravity	$g$	m/s²	—	UniformAcceleration

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

starting position

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

final position

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

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

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

a_{i} (part 1) = a_{i} (whole)

steady through its parts

v_{o,x} (t) = v_{o} (t) \cdot \cos θ_{o} (t)

parts of a velocity

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

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

Δ v = v - v_{0}

velocity gained

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time

v = v_{0} + a \cdot t

velocity after a time

x = x_{0} + v_{0} \cdot t + \frac{a \cdot t^{2}}{2}

position after a time

v^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x - x_{0})

velocity after a distance

x - x_{0} = \frac{(v + v_{0}) \cdot t}{2}

mean of the velocities