For a car to move safely around a curved road without skidding, the centripetal force needed is provided by the frictional force between the tires and the road. The relationship governing this scenario can be expressed as:

$$ f_{\text{friction}} = f_{\text{centripetal}} $$

The frictional force is given by:

$$ f_{\text{friction}} = \mu \cdot m \cdot g $$

And the centripetal force is given by:

$$ f_{\text{centripetal}} = \frac{m \cdot v^2}{r} $$

Where:

$\mu$ is the coefficient of friction,

$m$ is the mass of the car,

$g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$),

$v$ is the speed of the car,

$r$ is the radius of the curve.

Since the frictional force equals the centripetal force, we have:

$$ \mu \cdot m \cdot g = \frac{m \cdot v^2}{r} $$

Cancelling $m$ from both sides (assuming $m \neq 0$) gives:

$$ \mu \cdot g = \frac{v^2}{r} $$

Rewriting for $v$ gives the speed:

$$ v = \sqrt{\mu \cdot g \cdot r} $$

For the first scenario with $r = 75 \, \text{m}$ and $v = 30 \, \text{m/s}$:

$$ v_1^2 = \mu \cdot g \cdot r_1 $$

Thus,

$$ \mu \cdot g = \frac{v_1^2}{r_1} = \frac{30^2}{75} = \frac{900}{75} = 12 $$

For the second scenario with $r_2 = 48 \, \text{m}$, the new speed $v_2$ is:

$$ v_2 = \sqrt{\mu \cdot g \cdot r_2} = \sqrt{12 \times 48} $$

Calculating inside the square root:

$$ 12 \times 48 = 576 $$

So,

$$ v_2 = \sqrt{576} = 24 $$

Hence, the maximum allowed speed for a curve with a radius of $48 \, \text{m}$ is 24 m/s.