91Ó°ÊÓ

Newton's Second Law of Motion is a fundamental principle that relates the acceleration of an object to the net force acting upon it and its mass. It is expressed mathematically as \( F = ma \), where \( F \) is the force in newtons (N), \( m \) is the mass of the object in kilograms (kg), and \( a \) is the acceleration in meters per second squared (m/s²). This simple yet profound equation tells us that the force acting on an object is directly proportional to the acceleration of that object, with the mass being the constant of proportionality.

In our exercise, we have a 65-kg person who undergoes a deceleration of 30 \( g \), where \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).

Let's break this down:

• We calculated the deceleration \( a \) as \( 30 \times 9.8 = 294 \text{ m/s}^2 \).
• We then applied Newton's Second Law: \( F = 65 \text{ kg} \times 294 \text{ m/s}^2 \).
• The force \( F \) is calculated to be 19110 N.

This high force is what a person would feel during rapid deceleration, emphasizing the importance of understanding motion and safety in vehicles.

Kinematic equations provide a way to describe motion, allowing us to calculate various aspects such as displacement, velocity, and time during motion. They are particularly useful in problems involving constant acceleration, which is common in physics.

One such kinematic equation used here is:\[ v^2 = u^2 + 2as \]where:

• \( v \) is the final velocity (in m/s)
• \( u \) is the initial velocity (in m/s)
• \( a \) is the acceleration (in m/s²)
• \( s \) is the displacement or stopping distance (in meters)

In our exercise:

• The final velocity \( v \) is 0 m/s since the car comes to a stop,
• The initial velocity \( u \) has been converted from 95 km/h to approximately 26.39 m/s,
• The deceleration \( a \) is -294 m/s², indicating a slowing down motion,

By substituting these values into the equation, we find the stopping distance \( s \) to be approximately 1.19 meters. This formula helps us understand how quickly a vehicle can stop under specific conditions, which is crucial for safety.

Stopping distance is a critical concept for understanding how quickly a vehicle can come to a rest from a given speed. It is essential in automotive safety design and driving regulations to minimize the risk of accidents.

Stopping distance depends on factors including:

• The initial speed of the object (or vehicle),
• The rate of deceleration,
• The physical condition of the road, and the vehicle's braking efficiency.

In our problem, when traveling at 95 km/h with a deceleration of 30 \( g \), the vehicle comes to a stop in approximately 1.19 meters.

These details highlight the powerful role of deceleration in reducing stopping distance, underscoring the necessity of proper vehicle design that can withstand such forces without compromising passenger safety. Understanding stopping distance helps drivers recognize safe speeds and maintain enough distance from other vehicles to react effectively under sudden braking conditions.

In real-world driving, it's always safer to assume a longer stopping distance to account for unexpected variables, such as wet or icy roads.