Car Braking To Rest

Part of eval pr72 v4 branch rerun 3, agent version stated-pi-squared · 5f5be23

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

6 checks, 1 agreed

draft 6: agreed

Latest draft

question.rb1 question :car_braking_to_rest 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_to_rest, using: UniformAcceleration do2   objects :car3   car.goes(:braking)4 5   take speed: car.initial_velocity, distance: car.displacement(during: braking).magnitude6   deduce_from "comes to rest", car.final_velocity => "0 m/s"7   assume car.initial_position => "0 m"8   deduce_from "comes to rest", car.final_position => car.initial_position + distance9   take time: braking.time_taken10 11   answer time:, in: "s", figures: 312 13   conclusion "The car slows down steadily from its initial speed to rest over the given distance, " \14              "and the time taken follows from the kinematic relation between speed, distance and time."15 end

round 6

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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, $x_{1} = x_{0} + Δ x$ 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”
$x_{1} = x_{0} + Δ x$ from “comes to rest”

Assumed

$x_{0} = 0$ m

Find $Δ t$ .

Working

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

Conclusion

The car slows down steadily from its initial speed to rest over the given distance, and the time taken follows from the kinematic relation between speed, distance and time.

The Ruby

1question :car_braking_to_rest 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_to_rest, using: UniformAcceleration do2  objects :car3  car.goes(:braking)45  take speed: car.initial_velocity, distance: car.displacement(during: braking).magnitude6  deduce_from "comes to rest", car.final_velocity => "0 m/s"7  assume car.initial_position => "0 m"8  deduce_from "comes to rest", car.final_position => car.initial_position + distance9  take time: braking.time_taken1011  answer time:, in: "s", figures: 31213  conclusion "The car slows down steadily from its initial speed to rest over the given distance, " \14             "and the time taken follows from the kinematic relation between speed, distance and time."15end

1play_with :car_braking_to_rest do2  title "Car Braking To Rest"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 braking	$Δ x$	m	—
position of car at end of braking	$x_{1}$	m	—
position of car at start of braking	$x_{0}$	m	—
average velocity of car during braking	$v_{av}$	m/s	—
time taken of braking	$Δ t$	s	0 … 1e+05
average speed of car during braking	$c_{av}$	m/s	—
path length of car during braking	$d$	m	—
change of velocity of car during braking	$Δ v$	m/s	—
velocity of car at end of braking from ground	$v_{1}$	m/s	—
velocity of car at start of braking from ground	$v_{0}$	m/s	—
average acceleration of car during braking	$a_{av}$	m/s²	—
acceleration of car during braking	$a$	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(braking)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(braking)

c_{av} = \frac{d}{Δ t}

path a time(braking)

Δ v = v_{1} - v_{0}

velocity gained(braking)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(braking)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(braking)

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

position after a time(braking)

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

velocity after a distance(braking)

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

mean of the velocities(braking)

x_{1} = x_{0} + Δ x

comes to rest