Twice as Fast to Stop

solved UniformAcceleration

Q. Braking just as hard $a_{f} (b2) = a_{c} (b1)$ , how much farther does a car need to stop $v_{c} (b1₁) = 0$ m/s, $v_{f} (b2₁) = 0$ m/s from twice the speed $v_{f} (b2₀) = 2 \cdot v_{c} (b1₀)$ ?

How it's solved

Solved with UniformAcceleration

Given

$v_{c} (b1₁) = 0$ m/s from “to stop”
$v_{f} (b2₁) = 0$ m/s from “to stop”
$v_{f} (b2₀) = 2 \cdot v_{c} (b1₀)$ from “twice the speed”
$a_{f} (b2) = a_{c} (b1)$ from “Braking just as hard”

Assumed

$x_{c} (b1₀) = 0$ m
$x_{f} (b2₀) = 0$ m

Find $\frac{x_{f} (b2₁)}{x_{c} (b1₁)}$ .

Working

velocity after a distance(braking) ${v_{c} (b1₁)}^{2} = {v_{c} (b1₀)}^{2} + 2 \cdot a_{c} (b1) \cdot (x_{c} (b1₁) - x_{c} (b1₀))$
$x_{c} (b1₁) = \frac{{v_{c} (b1₀)}^{2}}{- 2 \cdot a_{c} (b1)}$
velocity after a distance(braking too) ${v_{f} (b2₁)}^{2} = {v_{f} (b2₀)}^{2} + 2 \cdot a_{f} (b2) \cdot (x_{f} (b2₁) - x_{f} (b2₀))$
$x_{f} (b2₁) = \frac{{(2 \cdot v_{c} (b1₀))}^{2}}{- 2 \cdot a_{c} (b1)}$
$\frac{x_{f} (b2₁)}{x_{c} (b1₁)} = 4$

Conclusion

Stopping distance goes as the square of the speed, because a car going twice as fast takes twice as long to stop and covers ground twice as fast all that time.

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 :twice_as_fast_to_stop do4  kind :scq5  describes "Braking just as hard, how much farther does a car need to stop from twice the speed?"67  option "Twice as far", as: :twice8  option "Four times as far", as: :four9  option "Eight times as far", as: :eight10  option "Just as far", as: :same11end

solution.rb

1require_relative "../questions/twice_as_fast_to_stop"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :twice_as_fast_to_stop, using: UniformAcceleration do5  objects :car, :faster_car6  car.goes(:braking)7  faster_car.goes(:braking_too)89  deduce_from "to stop", car.final_velocity => "0 m/s", faster_car.final_velocity => "0 m/s"10  deduce_from "twice the speed", faster_car.initial_velocity => 2 * car.initial_velocity11  deduce_from "Braking just as hard", faster_car.acceleration => car.acceleration12  assume car.initial_position => "0 m", faster_car.initial_position => "0 m"13  take twice: 2, four: 4, eight: 8, same: 11415  answer find_relation(faster_car.final_position / car.final_position)1617  conclusion "Stopping distance goes as the square of the speed, because a car going twice as " \18             "fast takes twice as long to stop and covers ground twice as fast all that time."19end

playground.rb

1require_relative "../solutions/twice_as_fast_to_stop"23play_with :twice_as_fast_to_stop do4  title "Twice as Fast to Stop"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 car during braking	$Δ x_{c} (b1)$	m	—
position of car at end of braking	$x_{c} (b1₁)$	m	—
position of car at start of braking	$x_{c} (b1₀)$	m	—
displacement of faster car during braking too	$Δ x_{f} (b2)$	m	—
position of faster car at end of braking too	$x_{f} (b2₁)$	m	—
position of faster car at start of braking too	$x_{f} (b2₀)$	m	—
average velocity of car during braking	$v_{av,c} (b1)$	m/s	—
time taken of braking	$Δ t_{b1}$	s	0 … 1e+05
average velocity of faster car during braking too	$v_{av,f} (b2)$	m/s	—
time taken of braking too	$Δ t_{b2}$	s	0 … 1e+05
average speed of car during braking	$c_{av,c} (b1)$	m/s	—
path length of car during braking	$d_{c} (b1)$	m	—
average speed of faster car during braking too	$c_{av,f} (b2)$	m/s	—
path length of faster car during braking too	$d_{f} (b2)$	m	—
change of velocity of car during braking	$Δ v_{c} (b1)$	m/s	—
velocity of car at end of braking from ground	$v_{c} (b1₁)$	m/s	—
velocity of car at start of braking from ground	$v_{c} (b1₀)$	m/s	—
change of velocity of faster car during braking too	$Δ v_{f} (b2)$	m/s	—
velocity of faster car at end of braking too from ground	$v_{f} (b2₁)$	m/s	—
velocity of faster car at start of braking too from ground	$v_{f} (b2₀)$	m/s	—
average acceleration of car during braking	$a_{av,c} (b1)$	m/s²	—
average acceleration of faster car during braking too	$a_{av,f} (b2)$	m/s²	—
acceleration of car during braking	$a_{c} (b1)$	m/s²	—
acceleration of faster car during braking too	$a_{f} (b2)$	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_{c} (b1) = x_{c} (b1₁) - x_{c} (b1₀)

change of position(braking)

Δ x_{f} (b2) = x_{f} (b2₁) - x_{f} (b2₀)

change of position(braking too)

v_{av,c} (b1) = \frac{Δ x_{c} (b1)}{Δ t_{b1}}

displacement a time(braking)

v_{av,f} (b2) = \frac{Δ x_{f} (b2)}{Δ t_{b2}}

displacement a time(braking too)

c_{av,c} (b1) = \frac{d_{c} (b1)}{Δ t_{b1}}

path a time(braking)

c_{av,f} (b2) = \frac{d_{f} (b2)}{Δ t_{b2}}

path a time(braking too)

Δ v_{c} (b1) = v_{c} (b1₁) - v_{c} (b1₀)

velocity gained(braking)

Δ v_{f} (b2) = v_{f} (b2₁) - v_{f} (b2₀)

velocity gained(braking too)

a_{av,c} (b1) = \frac{Δ v_{c} (b1)}{Δ t_{b1}}

velocity gained a time(braking)

a_{av,f} (b2) = \frac{Δ v_{f} (b2)}{Δ t_{b2}}

velocity gained a time(braking too)

v_{c} (b1₁) = v_{c} (b1₀) + a_{c} (b1) \cdot Δ t_{b1}

velocity after a time(braking)

v_{f} (b2₁) = v_{f} (b2₀) + a_{f} (b2) \cdot Δ t_{b2}

velocity after a time(braking too)

x_{c} (b1₁) = x_{c} (b1₀) + v_{c} (b1₀) \cdot Δ t_{b1} + \frac{a_{c} (b1) \cdot {Δ t_{b1}}^{2}}{2}

position after a time(braking)

x_{f} (b2₁) = x_{f} (b2₀) + v_{f} (b2₀) \cdot Δ t_{b2} + \frac{a_{f} (b2) \cdot {Δ t_{b2}}^{2}}{2}

position after a time(braking too)

{v_{c} (b1₁)}^{2} = {v_{c} (b1₀)}^{2} + 2 \cdot a_{c} (b1) \cdot (x_{c} (b1₁) - x_{c} (b1₀))

velocity after a distance(braking)

{v_{f} (b2₁)}^{2} = {v_{f} (b2₀)}^{2} + 2 \cdot a_{f} (b2) \cdot (x_{f} (b2₁) - x_{f} (b2₀))

velocity after a distance(braking too)

x_{c} (b1₁) - x_{c} (b1₀) = \frac{(v_{c} (b1₁) + v_{c} (b1₀)) \cdot Δ t_{b1}}{2}

mean of the velocities(braking)

x_{f} (b2₁) - x_{f} (b2₀) = \frac{(v_{f} (b2₁) + v_{f} (b2₀)) \cdot Δ t_{b2}}{2}

mean of the velocities(braking too)

v_{f} (b2₀) = 2 \cdot v_{c} (b1₀)

twice the speed

a_{f} (b2) = a_{c} (b1)

Braking just as hard