Braking to a Stop

solved UniformAcceleration

Q. A car doing 126 km/h $v_{0} = 126$ km/h brakes steadily and comes to rest $v = 0$ m/s in 200 m $x = 200$ m. What is its acceleration, in m/s²?

Inputs

speed

v_{0}

126 km/h

distance

x

200.00 m

Outputs

acceleration

a

How it's solved

Solved with UniformAcceleration

Given

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

Assumed

$x_{0} = 0$ m

Find $a$ .

Working

$v_{0} = 126$ km/h $= 35$ m/s
velocity after a distance $v^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x - x_{0})$
$0^{2} = 35^{2} + 2 \cdot a \cdot (200 - 0)$
$a = −3.062$ m/s²
$a \approx −3.06$ m/s²to 3 significant figures

Conclusion

Slowing down along the direction of motion is an acceleration against it, so it comes out negative.

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 :braking_to_a_stop do4  kind :numerical5  describes "A car doing %{speed} brakes steadily and comes to rest in %{distance}. %{acceleration}, in m/s²?"6  given speed: "126 km/h"7  given distance: "200 m"8  asks acceleration: "What is its acceleration"9end

solution.rb

1require_relative "../questions/braking_to_a_stop"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :braking_to_a_stop, using: UniformAcceleration do5  take speed: :v_0, distance: :x6  deduce_from "comes to rest", v: "0 m/s"7  assume x_0: "0 m"8  take acceleration: :a9  answer find_value(:acceleration, in: "m/s2", figures: 3)1011  conclusion "Slowing down along the direction of motion is an acceleration against it, so it " \12             "comes out negative."13end

playground.rb

1require_relative "../solutions/braking_to_a_stop"23play_with :braking_to_a_stop do4  title "Braking to a 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

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 - x_{0}

change of position

v_{av} = \frac{Δ x}{Δ t}

displacement a time

c_{av} = \frac{d}{Δ t}

path a time

Δ v = v - v_{0}

velocity gained

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time

v = v_{0} + a \cdot t

velocity after a time

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

position after a time

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

velocity after a distance

x - x_{0} = \frac{(v + v_{0}) \cdot t}{2}

mean of the velocities