Acceleration is sometimes expressed in multiples of g,where $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ is the acceleration of an object when it is in the state of free fall.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 calculated using the formula $a = \frac{\Delta v}{\Delta t} = \frac{29 \, \mathrm{m/s}}{0.15 \, \mathrm{s}} = 193.33 \, \mathrm{m/s^2}$. To express this in g's, divide by $9.8 \, \mathrm{m/s^2}$: $ \frac{193.33}{9.8} \approx 19.74 $, which rounds 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 $g = 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$