To solve the problem it is necessary to consider the kinematic equations of motion, distance and speed that would allow finding the solution to this problem.
By definition we know that speed is equivalent to the distance traveled in a certain period of time, that is
[tex]v= \frac{x}{t}[/tex]
Where,
x= distance
t= time
The speed at which the transmission waves travel is equivalent to the speed of light [tex](3 * 10 ^ 8m / s),[/tex] so if a signal takes 270 seconds to go and return to Venus, the distance would be equivalent to
[tex]v = \frac{x}{t}[/tex]
[tex]x = v*t[/tex]
[tex]x = 3*10^8 * 270[/tex]
[tex]x = 8.1*10^{10} m[/tex]
But this is the round trip distance, so the distance (d) would be half the way,
[tex]d= \frac{x}{2} = \frac{8.1*10^{10}}{2}[/tex]
[tex]d= 4.05*10^{10}m[/tex]
Therefore the distance between earth and Venus is [tex]4.05*10^{10}m[/tex]
The Astronomical Units is defined as the distance from the sun to the earth, that is [tex]1.495978707 *10 ^{11} m[/tex]
Therefore the conversion to meters of the distance to Venus would be
[tex]d= 4.05*10^{10}m(\frac{1UA}{1.495978707*10^{11}m})[/tex]
[tex]d = 0.2707UA[/tex]
Therefore the distance between earth and Venus in UA is 0.2707UA
