Solution:

Given,
Acceleration produced by brakes, a = -6 ms-2 (negative sign indicates opposite direction to motion)
Time taken to stop the car, t = 2s

We know that,

Final velocity of the car, v = u + at
where u is the initial velocity of the car, which is not given in the problem statement. However, we can assume that the car was initially moving with a constant velocity before the brakes were applied. Therefore, we can assume that the initial velocity of the car, u = 0.

So, v = 0 + (-6) x 2 = -12 m/s
The negative sign indicates that the car is now moving in the opposite direction to its initial motion.

Now, we can use the equation,

Distance travelled by the car, s = ut + 1/2 at2
where u = 0, a = -6 ms-2, and t = 2s

So, s = 0 x 2 + 1/2 x (-6) x (2)2 = -12 m
The negative sign indicates that the car has travelled a distance of 12 m in the opposite direction to its initial motion before coming to a stop.

Therefore, the car travels a distance of 12 m during the time it takes to stop after the application of brakes.

Explanation:

The problem statement involves the concept of acceleration, which is the rate of change of velocity of an object. When the brakes are applied to the car, an acceleration of 6 ms-2 is produced in the opposite direction to the motion. This means that the car's velocity decreases at a rate of 6 ms-2. The time taken for the car to come to a complete stop after the application of brakes is given as 2s.

To calculate the distance travelled by the car during this time, we use the equation for distance travelled with constant acceleration. Since the car's initial velocity is not given, we assume that it was moving with a constant velocity before the brakes were applied. This means that the car's initial velocity, u, is zero.

Using the equation for distance travelled with constant acceleration, we calculate the distance travelled by the car during the time it takes to stop after the application of brakes. The negative sign in the final answer indicates that the car has travelled a distance of 12 m in the opposite direction to its initial motion before coming to a stop.