Acceleration is sometimes expressed in multiples of g,where g = 9.8 m/s² is the acceleration due to the earth's gravity.In a car crash,the car's velocity may go from 26 m/s to 0 m/s in 0.15 s.How many g's are experienced,on average,by the driver? A)18 g B)13 g C)22 g D)23 g

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Quizplus · Accepted Answer

The acceleration is:$$a = \frac{v_f - v_i}{t} = \frac{0 - 26 \ m/s}{0.15 \ s} = -173.33 \ m/s^2$$The acceleration in g's is:$$a_g = \frac{a}{g} = \frac{-173.33 \ m/s^2}{9.8 \ m/s^2} = -17.79$$The negative sign indicates that the acceleration is in the opposite direction of the velocity. The average acceleration is 17.79 g's.

Acceleration Is Sometimes Expressed in Multiples of G,where G =