Car Braking — phryzby

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

Q. A car doing 126 km/h brakes steadily and comes to rest in 200 m. How long does it take to stop, in seconds? expects 11.4 s

15 checks, 1 agreed

draft 15: agreed

Latest draft

question.rb1 question :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"7 end

solution.rb1 solve :car_braking, using: UniformAcceleration do2   objects :car3   car.goes(:stopping)4 5   take speed: car.initial_velocity6   take distance: stopping.displacement7 8   deduce_from "comes to rest", car.final_velocity => "0 m/s"9 10 11   assume car.initial_position => "0 m"12   deduce_from "comes to rest", car.final_position => distance13 14   time = stopping.time_taken15   answer time:, in: "s", figures: 316 17   conclusion "The car slows down at a steady rate, so the time to stop is the distance over the average speed."18 end

round 15

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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?

The playground

Inputs

speed

v_{0}

126 km/h

distance

Δ x

200.00 m

Outputs

time

Δ t

The solution

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 Ruby

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

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

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

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 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