Respuesta :
Answer:
The correct option is;
[tex]\left ( -2,\dfrac{2 \cdot \pi }{3} \right)[/tex]
Step-by-step explanation:
The given parameters are
The equation of motion of one (the first) object is r = 4·cos(θ)
The equation of motion of the other (the second) object is r = -1 + 2·cos(θ)
Equating both equations gives;
4·cos(θ) = -1 + 2·cos(θ)
4·cos(θ) - 2·cos(θ) = -1
2·cos(θ) = -1
cos(θ) = -1/2
θ = Arccos(-1/2) = 120° = 2·π/3
Therefore, the two equations are equal when θ = 2·π/3 for which we have;
r = 4·cos(2π/3) = -2 and r = -1 + 2·cos(2π/3) = -2
∴ r = 4·cos(2π/3) = -1 + 2·cos(2π/3) = -2
The coordinate that represents a possible collision point of the objects in the form (r, θ) is therefore [tex]\left ( -2,\dfrac{2 \cdot \pi }{3} \right)[/tex]
Hence, The coordinate that represents a possible collision point of the objects in the form (-2, [tex]\frac{2\pi }{3}[/tex])
Given that ,
The first object is moving around the screen r = 4 cos(0)
The second object is moving around the screen r = -1 + 2cos(0)
We have to find ,
The coordinates represent a possible collision point of the objects.
According to the question,
Coordinate represent collision points to object ,
Equating both equations ;
4·cos(θ) = -1 + 2·cos(θ)
4·cos(θ) - 2·cos(θ) = -1
2·cos(θ) = -1
cos(θ) = -[tex]\frac{1}{2}[/tex]
θ = [tex]cos^{-1} (\frac{1}{2} )[/tex] = 120°
[tex]\theta[/tex] = [tex]\frac{-2\pi }{3}[/tex]
Therefore, the two equations are equal when θ = [tex]\frac{2\pi }{3}[/tex]
We have;
r = 4·cos([tex]\frac{2\pi }{3}[/tex]) = -2
And r = -1 + 2·cos([tex]\frac{2\pi }{3}[/tex]) =
r = -2
r = 4·cos([tex]\frac{2\pi }{3}[/tex]) = -1 + 2·cos([tex]\frac{2\pi }{3}[/tex])
r = -2
The coordinate that represents a possible collision point of the objects in the form (r, θ) is therefore r = -2 and [tex]\theta = \frac{2\pi }{3}[/tex] .
Hence, The coordinate that represents a possible collision point of the objects in the form (-2, [tex]\frac{2\pi }{3}[/tex]) .
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