A shop owner measures a suitcase kept on a rack to be 50 cm × 25 cm...

The proper dimensions of the suitcase are $L_0 = 50 \text{ cm}$, $W_0 = 25 \text{ cm}$, and $H_0 = 10 \text{ cm}$.

The train is moving with a velocity $v = 0.6c$. Length contraction occurs only along the direction of motion (parallel to the length).

The length measured by the observer on the ground is given by:

$L = L_0 \sqrt{1 - \dfrac{v^2}{c^2}}$

$L = 50 \sqrt{1 - (0.6)^2} = 50 \sqrt{1 - 0.36} = 50 \times 0.8 = 40 \text{ cm}$

The dimensions perpendicular to the direction of motion remain unchanged. Thus, the width and height are $25 \text{ cm}$ and $10 \text{ cm}$ respectively.

The dimensions measured by the observer on the ground are $40 \text{ cm} \times 25 \text{ cm} \times 10 \text{ cm}$.