play phryzby with physics

Part of eval main bdd945f six, agent version bdd945f

Car Braking To Rest

solved UniformAcceleration

Q. A car doing 126 km/hv0=126km/h brakes steadily and comes to restv1=0m/s in 200 mΔx=200m. How long does it take to stop, in seconds?

x₀ = 0 mv₀ = 35 m/sx = ?v = 0 m/sdisplacement 200 ma = ?v̄ = ?
Inputs
v0
126 km/h
Δx
200.00 m
Outputs
timeΔt?

How it's solved

Solved with UniformAcceleration

Given

  • v0=126km/h
  • Δx=200m
  • v1=0m/s from “comes to rest”

Assumed

  • x0=0m

Find Δt.

Working
  1. v0=126km/h=35m/s
  2. change of position(braking)Δx=x1-x0
  3. 200=x1-0
  4. x1=200m
  5. velocity after a distance(braking)v12=v02+2⋅a⋅(x1-x0)
  6. 02=352+2⋅a⋅(200-0)
  7. a=−3.0625m/s²
  8. velocity after a time(braking)v1=v0+a⋅Δt
  9. 0=35+(−3.0625)⋅Δt
  10. Δt=11.43 s
  11. Δt≈11.4sto 3 significant figures
Conclusion

The car slows down steadily from its initial speed to rest over the given distance, and the time taken is found from the kinematic relation between speed, distance, and time.

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
1question :car_braking_to_rest do2  kind :numerical3  describes "A car doing %{speed} brakes steadily and comes to rest in %{distance}. %{time}, in seconds?"4  given speed: "126 km/h"5  given distance: "200 m"6  asks time: "How long does it take to stop"7end
solution.rb
1solve :car_braking_to_rest, using: UniformAcceleration do2  objects :car3  car.goes(:braking)45  take speed: car.initial_velocity, distance: car.displacement(during: braking).magnitude6  deduce_from "comes to rest", car.final_velocity => "0 m/s"7  assume car.initial_position => "0 m"8  time = braking.time_taken9  answer time:, in: "s", figures: 31011  conclusion "The car slows down steadily from its initial speed to rest over the given distance, " \12             "and the time taken is found from the kinematic relation between speed, distance, and time."13end
playground.rb
1play_with :car_braking_to_rest do2  title "Car Braking To Rest"3end
The laws it drew on
quantitywrittenunitswithinfrom
positionxm—Velocity
displacementΔxm—Velocity
path lengthdm—Velocity
time takenΔts0 … 1e+05Velocity
average velocityvavm/s—Velocity
average speedcavm/s—Velocity
starting positionx0m—Velocity
final positionxm—Velocity
velocityvm/s—Acceleration
change of velocityΔvm/s—Acceleration
average accelerationaavm/s²—Acceleration
starting velocityv0m/s—Acceleration
final velocityvm/s—Acceleration
accelerationam/s²—UniformAcceleration
gravitygm/s²—UniformAcceleration
timets0 … 1e+05UniformAcceleration
displacement of car during brakingΔxm—
position of car at end of brakingx1m—
position of car at start of brakingx0m—
average velocity of car during brakingvavm/s—
time taken of brakingΔts0 … 1e+05
average speed of car during brakingcavm/s—
path length of car during brakingdm—
change of velocity of car during brakingΔvm/s—
velocity of car at end of braking from groundv1m/s—
velocity of car at start of braking from groundv0m/s—
average acceleration of car during brakingaavm/s²—
acceleration of car during brakingam/s²—
x0≡xi(start of motion)
starting position
x≡xi(end of motion)
final position
v0≡vi(start of motion)
starting velocity
v≡vi(end of motion)
final velocity
t≡Δtm
time
di(whole)=di(part 1)+di(part 2)
path in parts
Δtw=Δtp+Δtp
time in parts
ai(part 1)=ai(whole)
steady through its parts
Δx=x1-x0
change of position(braking)
vav=ΔxΔt
displacement a time(braking)
cav=dΔt
path a time(braking)
Δv=v1-v0
velocity gained(braking)
aav=ΔvΔt
velocity gained a time(braking)
v1=v0+a⋅Δt
velocity after a time(braking)
x1=x0+v0⋅Δt+a⋅Δt22
position after a time(braking)
v12=v02+2⋅a⋅(x1-x0)
velocity after a distance(braking)
x1-x0=(v1+v0)⋅Δt2
mean of the velocities(braking)