A Three-Wheeler Stops

Q. A three-wheeler of 400 kg with its driver of 65 kg is doing 36 km/h when the driver sees a child in the road and brings it to rest in 4.0 s. What is the average force that stops it, in newtons? expects 1.16 × 10³ N

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The web of concepts

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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?

The playground

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} |

The solution

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 Ruby

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

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

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

The laws

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)