A Three-Wheeler Stops

solved NewtonsSecondLaw

Q. A three-wheeler of 400 kg $m_{t} = 400$ kg with its driver $m = m_{t} + m_{d}$ of 65 kg $m_{d} = 65$ kg is doing 36 km/h $v_{0} = 36$ km/h when the driver sees a child in the road and brings it to rest $v_{1} = 0$ m/s in 4.0 s $Δ t = 4.0$ s. What is the average force that stops it, in newtons?

Inputs

vehicle mass

m_{t}

400 kg

driver mass

m_{d}

65 kg

speed

v_{0}

36 km/h

time

Δ t

4.00 s

Outputs

force

| F_{net} |

How it's solved

Solved with NewtonsSecondLaw

Given

$m_{t} = 400$ kg
$m_{d} = 65$ kg
$v_{0} = 36$ km/h
$Δ t = 4.0$ s
$v_{1} = 0$ m/s from “brings it to rest”
$m = m_{t} + m_{d}$ from “with its driver”

Find $| F_{net} |$ .

Working

$v_{0} = 36$ km/h $= 10$ m/s
with its driver $m = m_{t} + m_{d}$
$m = 400 + 65$
$m = 465$ kg
velocity after a time(stopping, x) $v_{1} = v_{0} + a \cdot Δ t$
$0 = 10 + a \cdot 4$
$a = −2.5$ m/s²
second law(stopping, x) $F_{net} = m \cdot a$
$F_{net} = 465 \cdot (−2.5)$
$F_{net} = −1162.5$ N
magnitude from components(net force on vehicle with driver during stopping) $| F_{net} | = \sqrt{{F_{net}}^{2}}$
$| F_{net} | = \sqrt{{(−1162.5)}^{2}}$
$| F_{net} | = 1162$ N
$| F_{net} | \approx 1.16 \times 10^{3}$ Nto 3 significant figures

Conclusion

Vehicle and driver are stopped together, so the force has to take the speed out of both masses at once.

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 :three_wheeler_stops do4  kind :numerical5  describes "A three-wheeler of %{vehicle_mass} with its driver of %{driver_mass} is doing %{speed} when the driver sees " \6            "a child in the road and brings it to rest in %{time}. %{force}, in newtons?"7  given vehicle_mass: "400 kg"8  given driver_mass: "65 kg"9  given speed: "36 km/h"10  given time: "4.0 s"11  asks force: "What is the average force that stops it"12end

solution.rb

1require_relative "../questions/three_wheeler_stops"2require_relative "../science/laws_of_motion/newtons_second_law"34solve :three_wheeler_stops, using: NewtonsSecondLaw do5  objects :vehicle_with_driver, :three_wheeler, :driver6  axes :x7  vehicle_with_driver.goes(:stopping)89  take vehicle_mass: three_wheeler.mass10  take driver_mass: driver.mass11  take speed: vehicle_with_driver.initial_velocity12  take time: stopping.time_taken1314  deduce_from "with its driver", vehicle_with_driver.mass => three_wheeler.mass + driver.mass15  deduce_from "brings it to rest", vehicle_with_driver.final_velocity => "0 m/s"1617  force = vehicle_with_driver.net_force.magnitude18  answer force:, in: "N", figures: 31920  conclusion "Vehicle and driver are stopped together, so the force has to take the speed out of " \21             "both masses at once."22end

playground.rb

1require_relative "../solutions/three_wheeler_stops"23play_with :three_wheeler_stops do4  title "A Three-Wheeler Stops"5end

The laws it drew on

quantity	written	units	within	from
volume	$volume$	—	—	Vectors
mass	$m$	kg	0 … ∞	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$work$	—	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$power$	—	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
charge	$charge$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$F$	N	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
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
gravity	$g$	m/s²	—	UniformAcceleration
net force	$F_{net}$	N	—	NewtonsSecondLaw
weight	$W$	N	—	NewtonsSecondLaw
displacement of vehicle with driver during stopping along x	$Δ x$	m	—
position of vehicle with driver at end of stopping along x	$x_{1}$	m	—
position of vehicle with driver at start of stopping along x	$x_{0}$	m	—
average velocity of vehicle with driver during stopping along x	$v_{av}$	m/s	—
time taken of stopping	$Δ t$	s	0 … 1e+05
average speed of vehicle with driver during stopping	$c_{av}$	m/s	—
path length of vehicle with driver during stopping	$d$	m	—
change of velocity of vehicle with driver during stopping along x	$Δ v$	m/s	—
velocity of vehicle with driver at end of stopping from ground along x	$v_{1}$	m/s	—
velocity of vehicle with driver at start of stopping from ground along x	$v_{0}$	m/s	—
average acceleration of vehicle with driver during stopping along x	$a_{av}$	m/s²	—
acceleration of vehicle with driver during stopping along x	$a$	m/s²	—
net force on vehicle with driver during stopping along x	$F_{net}$	N	—
mass of vehicle with driver	$m$	kg	0 … ∞
mass of three wheeler	$m_{t}$	kg	0 … ∞
mass of driver	$m_{d}$	kg	0 … ∞
weight on ground	$\| W_{g} \|$	N	—
weight on vehicle with driver	$\| W \|$	N	—
weight on three wheeler	$\| W_{t} \|$	N	—
weight on driver	$\| W_{d} \|$	N	—
displacement of vehicle with driver during stopping	$\| Δ x \|$	m	—
position of vehicle with driver at end of stopping	$\| x_{1} \|$	m	—
position of vehicle with driver at start of stopping	$\| x_{0} \|$	m	—
average velocity of vehicle with driver during stopping	$\| v_{av} \|$	m/s	—
change of velocity of vehicle with driver during stopping	$\| Δ v \|$	m/s	—
velocity of vehicle with driver at end of stopping from ground	$\| v_{1} \|$	m/s	—
velocity of vehicle with driver at start of stopping from ground	$\| v_{0} \|$	m/s	—
average acceleration of vehicle with driver during stopping	$\| a_{av} \|$	m/s²	—
acceleration of vehicle with driver during stopping	$\| a \|$	m/s²	—
net force on vehicle with driver during stopping	$\| F_{net} \|$	N	—

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

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

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

m_{b} \neq 0

has mass?

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

Δ x = x_{1} - x_{0}

change of position(stopping, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(stopping, x)

c_{av} = \frac{d}{Δ t}

path a time(stopping, x)

Δ v = v_{1} - v_{0}

velocity gained(stopping, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(stopping, x)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(stopping, x)

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

position after a time(stopping, x)

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

velocity after a distance(stopping, x)

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

mean of the velocities(stopping, x)

F_{net} = m \cdot a

second law(stopping, x)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of vehicle with driver during stopping)

| x_{1} | = \sqrt{{x_{1}}^{2}}

magnitude from components(position of vehicle with driver at end of stopping)

| x_{0} | = \sqrt{{x_{0}}^{2}}

magnitude from components(position of vehicle with driver at start of stopping)

| v_{av} | = \sqrt{{v_{av}}^{2}}

magnitude from components(average velocity of vehicle with driver during stopping)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of vehicle with driver during stopping)

| v_{1} | = \sqrt{{v_{1}}^{2}}

magnitude from components(velocity of vehicle with driver at end of stopping from ground)

| v_{0} | = \sqrt{{v_{0}}^{2}}

magnitude from components(velocity of vehicle with driver at start of stopping from ground)

| a_{av} | = \sqrt{{a_{av}}^{2}}

magnitude from components(average acceleration of vehicle with driver during stopping)

| a | = \sqrt{a^{2}}

magnitude from components(acceleration of vehicle with driver during stopping)

| F_{net} | = \sqrt{{F_{net}}^{2}}

magnitude from components(net force on vehicle with driver during stopping)

m = m_{t} + m_{d}

with its driver

m \neq 0

has mass?(vehicle with driver)

m_{t} \neq 0

has mass?(three wheeler)

m_{d} \neq 0

has mass?(driver)

F_{net} = 0

free of forces?(vehicle with driver, stopping)