A dog is sitting in front of a plane mirror. At a command it jumps at the mirror with a speed of 0.50 m/s. How fast does the dog approach its image?
At t= 0 the “dog-mirror” distance is “s”. The “dog-image” distance is “2s”.
At t =1 sec
the “dog-mirror” distance is “s-0.5”
The “dog-image” distance is “2(s-0.5) =
=2s -1”.
v = Δs/Δt =[(2s-1) -2s]/1 = 1 m/s.
To find out how fast the dog approaches its image, we need to consider the relative velocity of the dog with respect to the image in the plane mirror.
The image in the plane mirror is a virtual image, which means that it is formed by the reflection of light rays, but there is no actual object at that location. As a result, the image cannot move or have a velocity of its own.
Therefore, the dog's velocity relative to its image is equal to the dog's velocity itself, which is given as 0.50 m/s.
So, the dog approaches its image at a speed of 0.50 m/s.