Car Braking — phryzby

Part of eval multi-llm branch, agent version multi-llm-servers · 2b11b97

solved UniformAcceleration

Q. A car doing 126 km/h $v_{0} = 126$ km/h brakes steadily and comes to rest $v_{1} = 0$ m/s, $x_{1} = Δ x$ in 200 m $Δ x = 200$ m. How long does it take to stop, in seconds?

Inputs

speed

v_{0}

126 km/h

distance

Δ x

200.00 m

Outputs

time

Δ t

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 126$ km/h
$Δ x = 200$ m
$v_{1} = 0$ m/s from “comes to rest”
$x_{1} = Δ x$ from “comes to rest”

Assumed

$x_{0} = 0$ m

Find $Δ t$ .

Working

$v_{0} = 126$ km/h $= 35$ m/s
change of position(stopping) $Δ x = x_{1} - x_{0}$
$200 = x_{1} - 0$
$x_{1} = 200$ m
velocity after a distance(stopping) ${v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})$
$0^{2} = 35^{2} + 2 \cdot a \cdot (200 - 0)$
$a = −3.0625$ m/s²
velocity after a time(stopping) $v_{1} = v_{0} + a \cdot Δ t$
$0 = 35 + (−3.0625) \cdot Δ t$
$Δ t = 11.43 s$
$Δ t \approx 11.4$ sto 3 significant figures

Conclusion

The car slows down at a steady rate, so the time to stop is the distance over the average speed.

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

1question :car_braking do2  kind :numerical3  describes "A car doing %{speed} brakes steadily and comes to rest in %{distance}. %{time}, in seconds?"4  given speed: "126 km/h"5  given distance: "200 m"6  asks time: "How long does it take to stop"7end

solution.rb

1solve :car_braking, using: UniformAcceleration do2  objects :car3  car.goes(:stopping)45  take speed: car.initial_velocity6  take distance: stopping.displacement78  deduce_from "comes to rest", car.final_velocity => "0 m/s"91011  assume car.initial_position => "0 m"12  deduce_from "comes to rest", car.final_position => distance1314  time = stopping.time_taken15  answer time:, in: "s", figures: 31617  conclusion "The car slows down at a steady rate, so the time to stop is the distance over the average speed."18end

playground.rb

1play_with :car_braking do2  title "Car Braking"3end

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 stopping	$Δ x$	m	—
position of car at end of stopping	$x_{1}$	m	—
position of car at start of stopping	$x_{0}$	m	—
average velocity of car during stopping	$v_{av}$	m/s	—
time taken of stopping	$Δ t$	s	0 … 1e+05
average speed of car during stopping	$c_{av}$	m/s	—
path length of car during stopping	$d$	m	—
change of velocity of car during stopping	$Δ v$	m/s	—
velocity of car at end of stopping from ground	$v_{1}$	m/s	—
velocity of car at start of stopping from ground	$v_{0}$	m/s	—
average acceleration of car during stopping	$a_{av}$	m/s²	—
acceleration of car during stopping	$a$	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(stopping)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(stopping)

c_{av} = \frac{d}{Δ t}

path a time(stopping)

Δ v = v_{1} - v_{0}

velocity gained(stopping)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(stopping)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(stopping)

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

position after a time(stopping)

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

velocity after a distance(stopping)

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

mean of the velocities(stopping)

x_{1} = Δ x

comes to rest