 Solution

If $S^{\prime }$ moves at $+v$ relative to $S$, the Lorentz transformations areA rod at rest in $S$ has length $L$. To measure it in $S^{\prime }$, select endpoint events simultaneous there: $\Delta t^{\prime }=0$, so $\Delta t=vL/c^2$ and $\Delta x^{\prime }=\gamma (L-v\Delta t)=L/\gamma$. This is length contraction.

Now let a clock at rest in $S^{\prime }$ cross the two rod endpoints. These are different events, with $\Delta x=L$, $\Delta t=L/v$ and $\Delta x^{\prime }=0$. Their transformed time interval isThusThe moving clock measures the proper time $\Delta t^{\prime }$. The rod's contracted length divided by its speed in that clock's frame gives exactly the same crossing time; time dilation and length contraction are consistent consequences of the same transformation. Measuring a length and measuring a traversal interval require different simultaneity conditions.