Daily Challenge #267 - Braking Speed

#challenge

Jul 9, 2020 ・2 min read

Braking distance d1 is the distance a vehicle will go from the point when it brakes to when it comes to a complete stop. It depends on the original speed v and on the coefficient of friction mu between the tires and the road surface.

Braking distance is just one of two principal components of the total stopping distance. The other component is the reaction distance, which is the product of the speed and the perception-reaction time of the driver. We can just assume that the reaction time is constant and equal to 1 second.

The kinetic energy E can be found using the formula 0.5*m*v**2, the work W given by braking is mu*m*g*d1. Finding E and W gives the braking distance: d1 = v*v / 2*mu*g where g is the gravity of Earth and m the vehicle's mass.

So, to complete this challenge, there are two tasks you need to complete.

Calculate the total stopping distance in meters given v, mu (and the reaction time t = 1).
dist(v, mu) -> d = total stopping distance
Calculate v in km per hour knowing d in meters and mu. The reaction time is still t = 1.
speed(d, mu) -> v such that dist(v, mu) = d

Examples

dist(100, 0.7) -> 83.9598760937531
speed(83.9598760937531, 0.7) -> 100.0

Tests

dist(144, 0.3)
dist(92, 0.5)

speed(159, 0.8)
speed(153, 0.7)

Good luck!

This challenge comes from g964 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Discussion (3)

Valts Liepiņš • Jul 10 '20

First part of the challenge is almost written out as is, but the second part requires solving quadratic equation.

Solution in Haskell:

dist :: Double -> Double -> Double
dist vKmh mu = d_reaction + d_braking
  where
    d_reaction = v*t_reaction
    d_braking = (v*v) / (2*mu*g)
    v = fromKmh vKmh

speed :: Double -> Double -> Double
speed d mu = toKmh v
  where
    -- Solve using quadratic formula
    v = (-b + (sqrt $ b*b - 4*a*c)) / (2*a)
    a = 1 / (2*mu*g)
    b = 1 / t_reaction
    c = -d

fromKmh :: Double -> Double
fromKmh v = v * 1000 / 3600

toKmh :: Double -> Double
toKmh v = v * 3600 / 1000

g :: Double
g = 9.81

t_reaction :: Double
t_reaction = 1

Tests:

> dist 144 0.3
311.83146449201496

> dist 92 0.5
92.12909477605366

> speed 159 0.8
153.79671564846308

> speed 153 0.7
142.14404997566152

Hayden Mankin • Jul 9 '20

Is there something about this problem I'm missing? This seems like it's just a math problem, is there some trick I'm missing that makes it faster? Or not lose precision? Or am I missing an edge case some how?

anyways, here's a javascript solution.

const g = 9.81; // gravity in m/s
const k = 3600 / 1000; // conversion coefficient: m/s to km/hr

const dist = (v, mu, t=1) => {
  v = v / k; 
  return v * t + v * v / (mu * 2 * g); 
}

const speed = (d, mu, t=1) => {
  return k * (Math.sqrt(g * mu * (2 * d + g * t * t * mu)) - g * t * mu);
}

console.log(dist(144, 0.3)) // 311.83146449201496
console.log(dist(92, 0.5)) // 92.12909477605365
console.log(speed(159, 0.8)) // 153.79671564846308
console.log(speed(153, 0.7)) // 142.14404997566152

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Daily Challenge #267 - Braking Speed

Examples

Tests

Discussion (3)

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