A Position Rule — phryzby

solved UniformAcceleration

Q. A body's position is x = a + bt² $v_{0} = 0$ m/s, where a is 8.5 m $x_{0} = 8.5$ m and b is 2.5 m/s² $a = 5.0$ m/s². How fast is it moving at t = 2.0 s $t = 2.0$ s, in m/s?

Inputs

x_{0}

8.50 m

time

t

2.00 s

Outputs

speed

v

How it's solved

Solved with UniformAcceleration

Given

$x_{0} = 8.5$ m
$t = 2.0$ s
$v_{0} = 0$ m/s from “x = a + bt²”
$a = 5.0$ m/s²

Find $v$ .

Working

velocity after a time $v = v_{0} + a \cdot t$
$v = 0 + 5 \cdot 2$
$v = 10.0 m/s$

Conclusion

A position that grows as the square of the time is a body starting from rest with a steady acceleration of twice b.

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 :a_position_rule do4  kind :numerical5  describes "A body's position is x = a + bt², where a is %{a} and b is 2.5 m/s². %{speed} at " \6          "t = %{t}, in m/s?"7  given a: "8.5 m"8  given t: "2.0 s"9  asks speed: "How fast is it moving"10end

solution.rb

1require_relative "../questions/a_position_rule"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :a_position_rule, using: UniformAcceleration do5  take a: :x_0, t: :t6  deduce_from "x = a + bt²", v_0: "0 m/s"7  deduce_from "b is 2.5 m/s²", a: "5.0 m/s2"8  take speed: :v9  answer find_value(:speed, in: "m/s")1011  conclusion "A position that grows as the square of the time is a body starting from rest with a " \12             "steady acceleration of twice b."13end

playground.rb

1require_relative "../solutions/a_position_rule"23play_with :a_position_rule do4  title "A Position Rule"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