# A car traveling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65 cm (with respect to the road) while being brought to rest by an air bag. What magnitude of foce (assume constant) acts on the passengers's upper torso, which is 41 kg?

I changed 65 cm into .00065 km. I think i need to use F=ma so I need to find a. To find a I used d =1/2 (v0+v)t to find time and I got 2.45X10^-5. then I used a = v/t and i got 2163265.31 as a. Then i pluged m and a into F=ma and I got 88693877.71 for the answer. Is this correct? I wouldn't think the force would be this high.

Why not use

force*distance= 1/2 mass*velocity^2

That is the work done on the bag is equal to the change in KE of the passenger.

You should get the same answer as your technique, but I suspect the teacher was exploring the energy relationships in motion, as I laid out.
For information:

force*distance= 1/2 mass*velocity^2

force*distance= 1/2 mass*(avgveloctity*2)^2

force*distance= 2 mass*(avgveloctity )^2

force*avgvelociy*time= 2 mass*(avgveloctity )^2

force = 2 mass*(avgveloctity )/time

force = mass*(2avgveloctity )/time

force= mass*acceleration as you started from.

16 years ago

8 months ago

## Your approach to finding the force acting on the passenger's upper torso is correct, using Newton's second law, F = ma. However, the steps you took to find the acceleration and ultimately the force seem to have some errors.

To clarify, let's go through the correct steps:

1. Convert the distance moved by the passenger to meters (not kilometers) since the velocity is given in km/h. So, 65 cm becomes 0.65 m.

2. Convert the velocity of the car from km/h to m/s by dividing by 3.6 (since 1 km/h = 0.2778 m/s). Therefore, 53 km/h becomes approximately 14.72 m/s.

3. Use the formula: d = v0t + (1/2)at^2, where d is the distance moved by the passenger (0.65 m), v0 is the initial velocity (14.72 m/s), a is the acceleration, and t is the time taken to stop.

4. Rearrange the formula to find the time: t = (2d - v0t^2) / (2v0). Substituting the values, you obtain a quadratic equation: t^2 - (v0/2a) * t + d/a = 0.

5. Solve the quadratic equation for t using the quadratic formula: t = [-(v0/2a) Ā± sqrt((v0/2a)^2 - 4(d/a))] / 2.

6. Plug in the values of v0, d, and solve for t. Note that you have mistyped the distance as 0.00065 km instead of 0.65 m in your calculations. The correct calculation yields t ā 0.0739 seconds.

7. Use the value of t to find the acceleration, a = v0 / t. Substituting the values, you obtain a ā 14.72 m/s / 0.0739 s ā 199.03 m/sĀ².

8. Finally, calculate the force using F = ma. Substituting the values, you obtain F = 41 kg * 199.03 m/sĀ² ā 8143.23 N.

So, the correct magnitude of force acting on the passenger's upper torso is approximately 8143.23 Newtons. This force may seem high, but it is essential for the airbag to rapidly decelerate the passenger and cushion the impact of the collision.