Galileo's Odd Numbers

solved UniformAcceleration

Q. A body falls from rest $v_{0} = 0$ m/s. In successive equal intervals of time, the distances it falls are in the ratio

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 0$ m/s from “from rest”

Assumed

$x_{0} = 0$ m

Find how $x_{1}$ goes with $Δ t$ .

Working

position after a time(falling) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} = \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} \propto {Δ t}^{2}$ $a$ held
$Δ t \times 0 \Rightarrow x_{1} \times 0$
$Δ t \times 1 \Rightarrow x_{1} \times 1$
$Δ t \times 2 \Rightarrow x_{1} \times 4$
$Δ t \times 3 \Rightarrow x_{1} \times 9$
$Δ t \times 4 \Rightarrow x_{1} \times 16$
so in each equal interval in turn it falls 1 : 3 : 5 : 7

Conclusion

The distance fallen grows as the square of the time, so what each equal interval adds is the difference of two squares in a row: the odd numbers. Galileo found this before anyone could write the equation down.

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 :galileos_odd_numbers do4  kind :scq5  describes "A body falls from rest. In successive equal intervals of time, the distances it falls are " \6          "in the ratio"78  option "1 : 3 : 5 : 7 …", as: :odd9  option "1 : 2 : 3 : 4 …", as: :counting10  option "1 : 4 : 9 : 16 …", as: :squares11  option "1 : 1 : 1 : 1 …", as: :same12end

solution.rb

1require_relative "../questions/galileos_odd_numbers"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :galileos_odd_numbers, using: UniformAcceleration do5  objects :body6  body.goes(:falling)78  deduce_from "from rest", body.initial_velocity => "0 m/s"9  assume body.initial_position => "0 m"10  take odd: [ 1, 3, 5, 7 ], counting: [ 1, 2, 3, 4 ], squares: [ 1, 4, 9, 16 ], same: [ 1, 1, 1, 1 ]11  rel = find_relation body.final_position, to: falling.time_taken12  fallen = ->(n) { rel.find body.final_position, when: { falling.time_taken => n.times } }13  steps = (1..4).map { |n| fallen[n] - fallen[n - 1] }14  note "so in each equal interval in turn it falls #{steps.join(" : ")}"1516  answer steps1718  conclusion "The distance fallen grows as the square of the time, so what each equal interval " \19             "adds is the difference of two squares in a row: the odd numbers. Galileo found this " \20             "before anyone could write the equation down."21end

playground.rb

1require_relative "../solutions/galileos_odd_numbers"23play_with :galileos_odd_numbers do4  title "Galileo's Odd Numbers"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
velocity	$v$	m/s	—	Acceleration
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
acceleration	$a$	m/s²	—	UniformAcceleration
gravity	$g$	m/s²	—	UniformAcceleration
time	$t$	s	0 … 1e+05	UniformAcceleration
displacement of body during falling	$Δ x$	m	—
position of body at end of falling	$x_{1}$	m	—
position of body at start of falling	$x_{0}$	m	—
average velocity of body during falling	$v_{av}$	m/s	—
time taken of falling	$Δ t$	s	0 … 1e+05
average speed of body during falling	$c_{av}$	m/s	—
path length of body during falling	$d$	m	—
change of velocity of body during falling	$Δ v$	m/s	—
velocity of body at end of falling from ground	$v_{1}$	m/s	—
velocity of body at start of falling from ground	$v_{0}$	m/s	—
average acceleration of body during falling	$a_{av}$	m/s²	—
acceleration of body during falling	$a$	m/s²	—

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

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

Δ x = x_{1} - x_{0}

change of position(falling)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(falling)

c_{av} = \frac{d}{Δ t}

path a time(falling)

Δ v = v_{1} - v_{0}

velocity gained(falling)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(falling)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(falling)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(falling)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(falling)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(falling)