91Ó°ÊÓ

Kinematics is a branch of physics that describes the motion of objects without considering the causes of this motion. One important aspect of kinematics is the use of formulas to predict various parameters of motion. In this specific scenario, we use a kinematic equation to find acceleration. The formula applied here is: \[ v^2 = u^2 + 2as \] where:

• \( v \) is the final velocity. In our exercise, the passenger comes to rest, so \( v = 0 \text{ m/s} \).
• \( u \) is the initial velocity. After converting the initial speed of 63 km/h to meters per second, we have \( u = 17.5 \text{ m/s} \).
• \( a \) is the acceleration we need to find.
• \( s \) is the distance over which the passenger stops, which is 0.65 meters, converted from 65 centimeters.

When substituting these values into the equation, you can calculate the acceleration needed to determine the force acting upon the passenger.

Acceleration is the rate at which an object changes its velocity. In this exercise, the acceleration indicates how quickly the passenger's speed decreases to zero due to the impact. Calculating acceleration involves solving the kinematic equation for \( a \) where:\[ 0 = (17.5)^2 + 2 \cdot a \cdot 0.65 \] To isolate \( a \), rearrange the equation:\[ 0 = 306.25 + 1.3a \] Next, subtract 306.25 from both sides to simplify:\[ -306.25 = 1.3a \] Finally, divide each side by 1.3:\[ a = \frac{-306.25}{1.3} \approx -235.58 \, \text{m/s}^2 \] This negative sign represents a deceleration, meaning the passenger is rapidly slowing down. Acceleration is a foundational concept in physics that connects various physical situations through such equations.

Force is a vector quantity that causes an object to change its state of motion or shape. According to Newton's Second Law, the force \( F \) acting on an object is the product of its mass \( m \) and acceleration \( a \). In this exercise, to find the force exerted on the passenger's torso, apply Newton's formula:\[ F = ma \] With the passenger's torso mass \( m = 41 \, \text{kg} \), and the previously calculated acceleration \( a = -235.58 \, \text{m/s}^2 \), compute the force:\[ F = 41 \times (-235.58) \approx -9668.78 \, \text{N} \] The negative sign indicates that the force direction is opposite to the initial movement direction, consistent with a stopping force. However, when expressing the magnitude of force, it's positive: \( 9668.78 \, \text{N} \). Understanding this interaction helps in designing safety measures, such as airbags, which aim to minimize forces exerted on passengers during sudden stops.