A traditional Bugghi (a luxury cart drawn by horses) features a...

Let the mass of the bugghi be $M = 200 \text{ kg}$ and its initial velocity be $v_1 = 10 \text{ km/h}$.

Let the mass of the boy be $m = 25 \text{ kg}$ and his initial velocity be $v_2 = 4 \text{ km/h}$.

Since no external force acts on the system in the horizontal direction during the process of the boy sitting on the bugghi, the linear momentum of the system is conserved.

Applying the principle of conservation of linear momentum:

$M v_1 + m v_2 = (M + m) v_{\text{new}}$

$200 \times 10 + 25 \times 4 = (200 + 25) v_{\text{new}}$

$2000 + 100 = 225 v_{\text{new}}$

$2100 = 225 v_{\text{new}}$

$v_{\text{new}} = \dfrac{2100}{225}$

$v_{\text{new}} = \dfrac{28}{3} \text{ km/h}$