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**Hint:**The third equation of motion provides the final velocity of an object under uniform acceleration given the initial velocity and the distance traveled. By using the third equation of motion formula, apply the given values to find the minimum distance. When acceleration is constant, Velocity is proportional to time and Displacement is proportional to the square of time and Displacement is proportional to the square of the velocity.

**Formula Used:**

We will be using the formula of the third equation of motion i.e.,${v^2} - {u^2} = 2as$.

**Complete step by step answer:**

Given: The speed of the car $50km{h^{ - 1}}$

When brakes are applied final velocity becomes zero.

So we apply the third equation of motion

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

Where $u$ is the initial speed of the car $50km{h^{ - 1}}$

$v$ is the final speed of the car that is zero

$a$ is the acceleration

\[s\] is the distance that is $6m$

We know that, Distance $ = $average velocity $ \times $Time

Therefore, for constant acceleration

Average velocity$ = $\[\dfrac{{{\text{final velocity}} \times {\text{initial velocity}}}}{2}\]

Average velocity $ = \dfrac{{(v + u)}}{2}$

Hence Distance $s = \dfrac{{({v^2} - {u^2})}}{{2a}}$

Applying the value in the third equation of motion,

${v^2} - 50 = 2a \times 6$

The final velocity of the car is $v = 0$

Substituting the velocity of the car $v = 0$, and we can calculate the acceleration a is given by

\[a = \dfrac{{ - 2500}}{{2 \times 6}}\]

$a = \dfrac{{ - 2500}}{{12}}$

To find the minimum stopping distance

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

${100^2} = 2 \times (\dfrac{{ - 2500}}{{12}})s$

$s = 10000 \times \dfrac{6}{{2500}}$

$s = 24m$

Therefore the distance s is calculated

**Hence the minimum stopping distance of the car is option (D), $24m$.**

**Notes:**Dynamics and Kinematics are the two main descriptions of motion. Equation of motion can be used to derive the components like acceleration, velocity, and time. The relation between the components such as displacement, velocity, acceleration, speed, time, and distance are called Equations of motion. The third equation of motion is also called Laws of constant acceleration.

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