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In an automobile collision, a 44-kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.10-second time interval. What is the magnitude of the average force exerted on the passenger during this time?

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abidemiokin

Answer:

6,600N

Explanation:

According to second law of motion, Force = mass × acceleration

If acceleration = change in velocity/time = 15/0.10

Acceleration = 150m/s²

Given mass = 44kg

Force = 44× 150

Force = 6,600N

Magnitude of the average force exerted on the passenger during this time is 6,600N

The airbag will apply Impulsive Force on the passenger.

What is Impulsive Force?

When a force is applied in a small period of time then the force exerted is known as  Impulsive Force. It is given by the formula,

Impulsive Force = Force x time taken

[tex]J = F \times \triangle t[/tex]

The force exerted by the airbag on the passenger is 660 N.

Given to us

  • Mass of the passenger = 44 kg
  • Velocity = 15 m/sec
  • Time duration taken by the airbag = 0.10 second

Acceleration of the passenger

Acceleration

[tex]a = \dfrac{v}{t}\\\\a = \dfrac{15}{0.10}\\\\a = 150m/s^2[/tex]

Force exerted by the passenger on the airbag

Force

[tex]F = mass \times acceleration\\F = 44 \times 150\\F = 6,600\ N[/tex]

Impulsive Force

As the airbag takes minimum time to apply the force, therefore, the force applied by the airbag will be impulsive.

Impulsive Force

[tex]J = F \times \triangle t\\J = 6,600 \times 0.10\\J = 660\ N[/tex]

Hence, the force exerted by the airbag on the passenger is 660 N.

Learn more about Impulsive Force:

https://brainly.com/question/1219455

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