A rectangular billboard in space has the dimensions 10 meters by 20meters.
How fast and in what direction with respect to the billboard would a space traveler have to pass for the billboard to appear square?
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Let the length of the billboard `L =20 m` be in the x direction and the width of the billboard `l =10 m` be in the y direction.
The special theory of relativity finds that if we have two inertial systems S and S' where S' is moving relative to S with the speed V in the X direction the longitudinal (along x) dimension is contracting and the perpendicular dimensions stay the same if we measure the dimensions in the same system (S or S').
Therefore if a traveller (system S') is moving with a speed V parallel to the billboard length (in the positive x direction) then the billboard (system S) as seen by the traveller will have
where `C =3*10^8 m/s` is the speed of light in vacuum.
From first relation above we have
`1 -V^2/C^2 = (L')/L`
`V^2/C^2 =1-10/20 =0.5`
`V = C*sqrt(0.5) =0.707*C =0.707*3*10^8 =2.121*10^8 m/s`
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