Acceleration is sometimes expressed in multiples of g, where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ is the acceleration of an object due to the earth's gravity. In a car crash, the car's forward velocity may go from $29 \mathrm {~m} / \mathrm { s }$ to $0 \mathrm {~m} / \mathrm { s }$ in $0.15 \mathrm {~s}$ How many g's are experienced, on average, by the driver?
A) 20 g
B) 14 g
C) 24 g
D) 26 g

Question

Quizplus · Accepted Answer

The average acceleration is:$$a = \frac{\Delta v}{\Delta t} = \frac{0 \mathrm {~m} / \mathrm { s } - 29 \mathrm {~m} / \mathrm { s }}{0.15 \mathrm {~s}} = -193.33 \mathrm {~m} / \mathrm { s } ^ { 2 }$$The acceleration in g's is:$$a_g = \frac{a}{g} = \frac{-193.33 \mathrm {~m} / \mathrm { s } ^ { 2 }}{9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }} = -19.73$$The average acceleration experienced by the driver is -19.73 g's, which is closest to 20 g.

Acceleration Is Sometimes Expressed in Multiples of G, Where g=9.8 m/s2g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }g=9.8 m/s2

Acceleration Is Sometimes Expressed in Multiples of G, Where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$