1. Use the relation between stopping distance and speed

When a car is stopped by brakes with the same braking force, the retardation is constant.

Using the kinematic equation:

$v^2 = u^2 + 2as$

For stopping,

• final speed: $v = 0$
• initial speed: $u$
• retardation: $a$
• stopping distance: $s$

So,

$0 = u^2 + 2as$

$s = -\frac{u^2}{2a}$

Since $a$ is constant, we get:

$s \propto u^2$

Thus, stopping distance is proportional to the square of speed.

2. Compare the two cases

First speed:

$u_1 = 50\ \text{km/h}, \quad s_1 = 6\ \text{m}$

Second speed:

$u_2 = 100\ \text{km/h} = 2u_1$

Therefore,

$\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2 = 2^2 = 4$

So,

$s_2 = 4s_1 = 4 \times 6 = 24\ \text{m}$

3. Check options

• A: $12\ \text{m}$ ❌
• B: $18\ \text{m}$ ❌
• C: $24\ \text{m}$ ✅
• D: $6\ \text{m}$ ❌

4. Final Answer

The minimum stopping distance is

$\boxed{24\ \text{m}}$

So the correct option is C.