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padilas
3 months ago
5

During a car accident, a 125 kg driver is moving at 31 m/s and in 1.5 seconds is brought to rest by an inflating air bag. What i

s the magnitude of the change in momentum of the driver?
A. 4.0 kg • m/s
B. 21 kg • m/s
C. 47 kg • m/s
D. 3900 kg • m/s
Physics
2 answers:
inna [3.1K]3 months ago
8 0
The momentum of a moving object can be expressed as:
p=mv
where
m is the object's mass
and v stands for velocity.

Prior to the vehicular mishap, the driver (with a mass of m=125 kg) was traveling at a rate of v=31 m/s, indicating his initial momentum as:
p_i = mv=(125 kg)(31 m/s)=3875 kg m/s \sim 3900 kgm/s

However, after the accident, the driver is at rest: this implies his final velocity is nil, resulting in a final momentum of zero: p_f=0. This indicates the change in momentum equals to his initial momentum:
\Delta p = 3900 kg m/s
thus the accurate response is D.

Sav [3.1K]3 months ago
8 0

Response:

D. 3900 kg • m/s

Explanation:

I completed the test

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