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Kinematic equations are essential tools in physics, used to describe the motion of objects. They relate the initial and final velocities, acceleration, time, and displacement. In our problem, we use the kinematic equation: ewline ewline \[ v^2 = u^2 + 2ad\] ewline ewline Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(d\) is the displacement or stopping distance we are trying to find. After evaluating, we can consider stopping distance as the displacement factor when the automobile comes to a halt.

Newton's second law of motion states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: ewline ewline \[ F = ma \] ewline ewline In our scenario, the braking process includes a frictional force working against the car's motion. The mass of the automobile can be computed from its weight and the gravitational force. Once we know the mass, we can use the frictional force to find the deceleration. ewline ewline This law helps us understand the relationship between force, mass, and acceleration, critical for determining how quickly the car comes to a stop.

Deceleration is the reduction in the speed or velocity of an object. It is simply negative acceleration. With Newton's second law, the deceleration can be calculated by rearranging the formula: ewline ewline \[ a = \frac{F}{m}\] ewline ewline In the context of our problem, where the frictional force opposes the motion of the automobile, the deceleration is computed as: ewline ewline \[ a = \frac{8230 \text{ N}}{1673.47 \text{ kg}} = 4.92 \text{ m/s}^2 \text{ (negative in our case)}\] ewline ewline Always remember, deceleration means the car is slowing down. It's crucial in finding out how long it will take or the distance it will cover before stopping.

Unit conversion is important for ensuring consistency in physics calculations. In our problem, we convert the car's speed from km/h to m/s using the factor \[ \frac{5}{18}\] since 1 km/h is equivalent to \[ \frac{1000}{3600} = \frac{5}{18} \text{ m/s}\]. ewline ewline For instance: ewline ewline \[ 113 \text{ km/h} \times \frac{5}{18} \text{ m/s} = 31.39 \text{ m/s}\] ewline ewline Understanding unit conversions ensures that our calculations are precise and we avoid errors stemming from inconsistent units. This precision is critical when working towards solving the problem accurately.

Automobile stopping distance is the total distance a car travels before coming to a complete stop after the brakes are applied. This distance depends on several factors, such as: ewline ewline

• Initial velocity of the automobile
• Magnitude of the braking force (frictional force)
• Mass of the automobile
• Deceleration rate

ewline ewlineBy using the kinematic equation \[ v^2 = u^2 + 2ad\], where the final velocity \(v\) becomes zero upon stopping, initial velocity \(u\) is given in m/s, and deceleration \(a\) is calculated from frictional force over mass, we can solve for stopping distance \(d\). In our problem: ewline ewline \[ d = \frac{(31.39)^2}{2 \times 4.92} = 100.34 \text{ meters}\] ewlineewline Understanding these components gives us a clear picture of how the car behaves under braking and allows us to predict the distance needed for a safe stop.