How Long to Stop

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

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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. How long does it take to stop, in seconds?

The playground

Inputs

speed

v_{0}

126 km/h

distance

x

200.00 m

Outputs

time

t

The solution

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 $t$ .

Working

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

Conclusion

Under a steady acceleration the average velocity is halfway between the first and the last, so the time is the distance over that average.

The Ruby

1require_relative "../physics"23question :how_long_to_stop do4  kind :numerical5  describes "A car doing %{speed} brakes steadily and comes to rest in %{distance}. %{time}, in seconds?"6  given speed: "126 km/h"7  given distance: "200 m"8  asks time: "How long does it take to stop"9end

1require_relative "../questions/how_long_to_stop"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :how_long_to_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 time: :t9  answer find_value(:time, in: "s", figures: 3)1011  conclusion "Under a steady acceleration the average velocity is halfway between the first and " \12             "the last, so the time is the distance over that average."13end

1require_relative "../solutions/how_long_to_stop"23play_with :how_long_to_stop do4  title "How Long 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

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