Twice as Fast to Stop

Q. Braking just as hard, how much farther does a car need to stop from twice the speed? expects B

What the draft establishes

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The web of concepts

given read from the words equation asked

The equations it used

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Check an answer

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 $v_{f} (b2₀) = 2 \cdot v_{c} (b1₀)$ ?

The playground

The solution

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 Ruby

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

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

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

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