With a reckless disregard for safety and the law, you set your high-performance rocket cycle on course to streak through an intersection at top speed. approaching the intersection, you observe green (540 nm) light from the traffic signal. after passing through, you look back to observe red (650 nm) light. actually, the traffic signal never changed color - it didn't have time! what is the top speed of your rocket cycle, and what was the color fo the traffic signal (according to an appaled bystander)?

v = 0.277 10⁸ m / s  and λs = 448.4 nm

Explanation:

This is a relativistic Doppler effect problem, which is described by the equation.

    f₀ = √ [(1-v / c) / (1 + v / c)]

The speed (v) is positive when the emitter (traffic light) and the observer move away

When approaching

    f₀₁ = √ [(1 - (- v) / c) / (1 + (- v) / c)]

    f₀₁ = √ [(1+ v / c) / (1-v / c)]

To get away

   f₀₂ = √ [(1-v / c) / (1 + v / c)]

as they give the wavelengths let's use the relationship

   c = λ f

   f = c / λ

   c /λ₀₁ = √ [(1+ v / c) / (1-v / c)]

   c /λ₀₂ = √ [(1-v / c) / (1 + v / c)]

Now we have to do some algebra to find the speed, let's write the system of equations,

    c / λ₀₁ = √ [(1+ v / c) / (1-v / c)]

    c / λ₀₂ = √ [(1-v / c) / (1 + v / c)]

let's divide the two expressions

 λ₀₂ / λ₀₁ = √ ([(1+ v / c) 2 / (1-v / c) 2]

  λ₀₂ /λ₀₁ = ([(1+ v / c) / (1-v / c)]

  (1+ v / c) =λ₀₂ / λ₀₁ (1-v / c)

  v / c (1 + λ₀₂ / λ₀₁) = λ₀₂ / λ₀₁ - 1

  v = c (λ₀₂ /λ₀₁ - 1) / (1 +λ₀₂ / λ₀₁)

  v = 3 10⁸ (650/540 -1) / 1 + 650/540)

  v = 3,10⁸ 0.2037 / 2.2037

  v = 0.277 10⁸ m / s

To calculate and the emitted frequency we substitute the Doppler equation

   f₀ = fs √ [(1-v / c) / (1 + v / c)]

    c / λ₀ = c / λs √ [(1-v / c) / (1 + v / c)]

     λs = λ₀ √ [(1-v / c) / (1 + v / c)]

    λs = 650 √ (1- 0.277 10⁸/3 10⁸) / (1 + 0.277 10⁸ /3 10⁸8)

   λs = 650 √ (0.9077 / 1.9077)

   λs = 448.4 nm

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