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in an automobile collision, a 45 kg passenger moving at 15 m/s is brought to rest by an air bag during a 0.1 s time interval. What is the magnitude of the average force exerted on the passenger during this time?



Answer :

MathPhys

Answer:

6750 N

Explanation:

There are two ways to solve this problem. One method is to use impulse equations; impulse (J) is equal to the change in momentum (Δp), and is also equal to the average force times the time interval (FΔt). Another method is to use kinematics to find the average acceleration, then use Newton's second law of motion to find the force (F = ma).

Using impulse:

J = Δp = FΔt

mΔv = FΔt

F = mΔv / Δt

F = (45 kg) (0 m/s − 15 m/s) / (0.1 s)

F = -6750 N

Alternatively, we can use kinematics to find acceleration:

v = u + at

0 m/s = 15 m/s + a (0.1 s)

a = -150 m/s²

Applying Newton's second law:

F = ma

F = (45 kg) (-150 m/s²)

F = -6750 N

The magnitude of the force is 6750 N.

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