Question
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?
Answers
Elena
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.
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.