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How many g's are experienced, on average, by the driver?

Acceleration is sometimes expressed in multiples of g, where g=9.8 m/s^2 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?

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Benjamin B. | Experienced and engaging math/physics tutorExperienced and engaging math/physics tu...
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When someone expresses something in terms of g's, it is the acceleration divided by g (9.8m/s^2). So if you experienced 980 m/s^2 of acceleration, you would be experiencing 980/9.8 = 100 g's. In this case you want to assume the acceleration if constant from 26m/s to 0, which we can do because the question says average g. The acceleration is the change in velocity over time (this is why the units are m/s^2). We find the change in velocity, 26-0 = 26m/s. The change is 0.15 seconds. So we divide 26m/s by 0.15s to get 173.33 m/s^2. Now we divide by 9.8m/s^2 to get the number of g's. Notice how the units are the same (although technically the conversion is 9.8m/s^2 per g). This results 17.69 g's.
To give some perspective, roller coasters are usually about 5g's. Thankfully the bus driver is probably still alive, is likely unconscious. However, sustained exposure (several seconds) to such high g's is less than recommend.
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