A car traveling at 71 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 63 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 47 kg?

force*.63=1/2 47*v^2

for v, change 71km/hr to m/s

To find the magnitude of force acting on the passenger's upper torso, we can utilize Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the acceleration is the deceleration experienced by the passenger as the car comes to a stop.

The first step is to convert the speed of the car from km/h to m/s, as the SI unit for acceleration is meters per second squared (m/s^2). To do this, we need to divide 71 km/h by 3.6 to get the speed in m/s. Therefore, the speed of the car is approximately 19.7 m/s.

Next, we need to calculate the deceleration of the passenger. To do this, we can use the kinematic equation that relates displacement (d), initial velocity (v), final velocity (u), and acceleration (a):

(v^2 - u^2) = 2ad

Since the final velocity (u) is 0 m/s (the passenger comes to rest), the equation simplifies to:

v^2 = 2ad

Plugging in the values we have, with v = 0 m/s, u = 19.7 m/s, and d = 0.63 m (since 63 cm is equal to 0.63 m), we can solve for the acceleration:

0^2 = 2a * 0.63

0 = 1.26a

a = 0 m/s^2

From the calculation, we find that the acceleration experienced by the passenger is 0 m/s^2. This implies that the passenger's upper torso stayed at a constant velocity during the impact, and the airbag absorbed all the force. Therefore, the magnitude of force acting on the passenger's upper torso is 0 Newtons (N).