Given the initial speeds of two identical cars as $u$ and $4u$, and considering that both cars eventually stop (final speed $v = 0$), we note that both cars decelerate with the same acceleration $-a$.

Using the kinematic equation:

$ v^2 = u^2 - 2as $

Since $v = 0$,

$ 0 = u^2 - 2as $

Hence,

$ u^2 = 2as $

For the first car with speed $u$,

$ u^2 = 2a s_1 \quad \text{...(i)} $

For the second car with speed $4u$,

$ (4u)^2 = 2a s_2 \quad \text{...(ii)} $

Dividing equation (i) by equation (ii),

$ \frac{u^2}{(4u)^2} = \frac{2a s_1}{2a s_2} $

$ \frac{u^2}{16u^2} = \frac{s_1}{s_2} $

$ \frac{1}{16} = \frac{s_1}{s_2} $

Thus, the ratio of the stopping distances of the two cars is $\frac{1}{16}$.