Stopping distance S= u^2/ 2a. ( from v^2-u^2=2aS.)...as 'S' is directly proportional to u^2. since, if speed of car is half then stopping distance become S/4.

Explanation:

Given, velocity of the car is constant and the minimum distance over which it can be stopped is Y. Let the initial velocity of the car be V. Let the mass of the car be m and the coefficient of friction between the tyres and the road be μ.

The formula for the minimum stopping distance of a car is given by:

Y = V²/2μg

Where g is the acceleration due to gravity.


When the speed of the car is halved:

Let the new speed of the car be V/2. The minimum stopping distance of the car can be calculated using the same formula:

Y' = (V/2)²/2μg

Y' = V²/8μg

Hence, when the speed of the car is halved, the minimum stopping distance of the car is divided by 4.


Explanation of the result:

The stopping distance of a car is proportional to the square of its initial velocity. When the speed of the car is halved, the stopping distance is reduced to one-fourth of its initial value. This is because the kinetic energy of the car, which is proportional to the square of its velocity, is reduced to one-fourth of its initial value. Therefore, the work done by the frictional force to bring the car to rest is also reduced to one-fourth of its initial value, resulting in a reduction in the minimum stopping distance of the car.


Conclusion:

The minimum stopping distance of a car is proportional to the square of its initial velocity. When the speed of the car is halved, the minimum stopping distance of the car is divided by 4. This is because the kinetic energy of the car is reduced to one-fourth of its initial value, resulting in a reduction in the work done by the frictional force to bring the car to rest.