Galileo's Odd Numbers

Q. A body falls from rest. In successive equal intervals of time, the distances it falls are in the ratio expects A

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The web of concepts

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The equations it used

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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

The playground

The solution

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 Ruby

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

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

1require_relative "../solutions/galileos_odd_numbers"23play_with :galileos_odd_numbers do4  title "Galileo's Odd Numbers"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
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)