Answer:
A.) D = 5.9 + 1.5sin (0.02534t – 51.96) (billions km)
B.) D = 5.87 billions km
Step-by-step explanation:
Assume a sine/cosine wave.
the period (min to max) is 2237–1989 = 248 years
so the first trial is D = M sin ((2πt/248) + k)
where k is the phase and M is the amplitude.
D = M sin (2π(t–1989)/248)
D = M sin ((2πt/248) – (2π1989/248))
at t = 1989, we have sin(0) = 0
at t = 2237, we have sin(56.67534–50.39216) = sin 2π = 0
but we want a min at those points, equivalent to sin 270º or 3π/2, or –π/2, so we have to add in additional phase shift
D = M sin ((2πt/248) – (2π1989/248) – (π/2))
D = M sin (0.0253354t – 50.39216 – 1.570796)
D = M sin (0.0253354t – 51.96296)
at t = 1989,
D = M sin (50.39216 – 51.96296) = M sin 1.5708 = –1 ok
min to max distance is 7.4 – 4.4 = 3
so the equaton is
D = 5.9 + 1.5sin (0.02534t – 51.96) (billions km)
in 2000
D = 5.9 + 1.5 sin (0.02534×2000 – 51.96) (billions km)
D = 5.9 + 1.5 sin (-1.28)
D = 5.9 – 1.5 × 0.02234
d = 5.87 billions km
Answer:
D(t)=-1.5cos(2[tex]pi[/tex]/248(t+11))+5.9
b) 4.89
Step-by-step explanation:
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