If a dragster accelerates at a rate of 39.2 m/s^2, how many g's (using g = 9.80 m/s^2) does the driver experience?
A1.5 g
B4.0 g
C10.5 g
D24.5 g
Answer & Solution
Correct answer: B. 4.0 g
1. One g of acceleration equals 9.80 m/s^2.
2. Number of g's = 39.2 m/s^2 / 9.80 m/s^2.
3. That gives 4.0 g.
_Source: OpenStax Physics (CC BY 4.0), Ch 3 "Acceleration", section 3.2 Representing Acceleration with Equations and Graphs_
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