The pilot of an airplane traveling 180km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The
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Answer:
t₂=6.35min
Explanation:
t₁ = first observed time (=5.1 min)
t₂ = unknown; this is the quantity we want to find
V₁ = observer's initial speed (=0.84c)
V₂ = observer's final speed (=0.90c)
Lorentz factors for V₁ and V₂:
γ₁ = 1/√(1−(V₁/c)²)
γ₂ = 1/√(1−(V₂/c)²)
The "proper time" (the time measured by the person filling her car) is:
t′ = t₁/γ₁
The proper time is stated to be the same for both observations, so we also have:
t′ = t₂/γ₂
Combine those two equations and solve for t₂
t₂ = t₁(γ₂/γ₁)
t₂= t₁√((1−(V₁/c)²)/(1−(V₂/c)²))