A car is moving with speed of $$150 \text{ km h}^{-1}$$ and after applying the brake it will move $$27 \text{ m}$$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ______ m distance.

We use the kinematic equation relating velocity, acceleration, and distance:

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

When the car stops, $$v = 0$$, so:

$$0 = u^2 + 2as \implies s = \dfrac{u^2}{2|a|}$$

This shows that the stopping distance is proportional to the square of the initial speed:

$$s \propto u^2$$

Case 1: Speed $$u_1 = 150 \text{ km h}^{-1}$$, stopping distance $$s_1 = 27 \text{ m}$$.

Case 2: Speed $$u_2 = \dfrac{u_1}{3} = 50 \text{ km h}^{-1}$$, stopping distance $$s_2 = ?$$

Since the braking force (and hence deceleration) is the same in both cases:

$$\dfrac{s_2}{s_1} = \dfrac{u_2^2}{u_1^2} = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$$

$$s_2 = \dfrac{27}{9} = 3 \text{ m}$$

Therefore, the car will stop after travelling $$\boxed{3}$$ m.