Are distance and displacement always frame independent? I'm aware that distance and displacement, both are independent of reference frames, when the two frames are stationary wrt each other. Because the actual distance (or the shortest distance) between two points remain the same, as long as the two frames are stationary wrt each other, i.e there is no relative velocity between the two frames (Am I correct?) 
But what if the other frame starts moving wrt the other? From what I understand, in this case, displacement and distance are no longer frame independent. Are they? Please do let me know. 
My question is, are distance and displacement frame independent, always? Even when there's a relative motion between the two frames?
 A: The following discussion disregards effects of special relativity and discusses purely Galilean transformations.
If two frames are at rest with respect to each other and oriented parallel to one another, then the transformation between one frame $S$ with coordinates $(x,y,z)$ to the other $S'$ with coordinates $(x',y',z')$ will take the following form, where $x_0, y_0, z_0$ are constants.
$$x' = x + x_0$$
$$y' = y + y_0$$
$$z' = z + z_0$$
Thus, the displacements between two events in $S'$, event 1 $(x_i', y_i', z_i', t_i)$ and event 2 $(x_f', y_f', z_f', t_f)$, in the new frame are given as follows.
$$\Delta x' = x'_f - x_i' = x_f + x_0 - (x_i + x_0) = x_f - x_i = \Delta x$$
$$\Delta y' = y'_f - y_i' = y_f + y_0 - (y_i + y_0) = y_f - y_i = \Delta y$$
$$\Delta z' = z'_f - z_i' = z_f + z_0 - (z_i + z_0) = z_f - z_i = \Delta z$$
Thus, when frames are at rest with respect to each other, distances are preserved. This is true regardless of whether the initial and final positions are measured simultaneously ($t_i = t_f$) or not ($t_i \ne t_f$).
If the frames are moving with respect to each other with a velocity $\mathbf v = (v_x, v_y, v_z)$ and the frames are coincident at $t = 0$, then the transformation between $S$ and $S'$ is as follows.
$$x' = x + v_x t$$
$$y' = y + v_y t$$
$$z' = z + v_z t$$
If you're looking for the distance between the same two events described above in $S'$, we would have the following.
$$\Delta x' = x'_f - x'_i = x_f + v_x t_f - (x_i + v_x t_i) = x_f - x_i + v_x (t_f - t_i) = \Delta x + v_x (t_f - t_i)$$
$$\Delta y' = y'_f - y'_i = y_f + v_y t_f - (y_i + v_y t_i) = y_f - y_i + v_y (t_f - t_i) = \Delta y + v_y (t_f - t_i)$$
$$\Delta z' = z'_f - z'_i = z_f + v_z t_f - (z_i + v_z t_i) = z_f - z_i + v_z (t_f - t_i) = \Delta z + v_z (t_f - t_i)$$
Thus, when frames are moving with respect to each other, distances are preserved only when the events are simultaneous ($t_i = t_f$).
A: Displacement of a body is dependent on the frame
Suppose a ball is moving on the ground. In the frame of ball the displacement of ball is zero and in the frame of ground there is a some finite displacement
Distance between two points or the shortest distance between two points is not dependent on frame even if those points are moving. Distance between two points will always be same as seen by any observer
A: Distance and displacement are both frame-dependent. For example, if I go to Gettysburg, Pennsylvania, I can say that I'm at the same place where the Civil War battle occurred, but in a different frame of reference I can say that the earth has moved since 1863.
