A Position Rule — phryzby

Q. A body's position is x = a + bt², where a is 8.5 m and b is 2.5 m/s². How fast is it moving at t = 2.0 s, in m/s? expects 10.0 m/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 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?

The playground

Inputs

x_{0}

8.50 m

time

t

2.00 s

Outputs

speed

v

The solution

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 Ruby

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

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

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