Question

I To withstand "g-forces" of up to$10\text{g}$ 's, caused by suddenly BI0 pulling out of a steep dive, fighter jet pilots train on a "human centrifuge." $10\text{g}$'s is an acceleration of $98\text{m}/{\text{s}}^2$. If the length of the centrifuge arm is $12\text{m}$, at what speed is the rider moving when she experiences 10 g's?

Short answer

The speed of the rider is$34.29\text{m}/\text{s}$.

Step-by-step solution

 Step 1. Given information

The centrifugal acceleration of the fighter jet pilot is 10 g that is $98\text{m}/{\text{s}}^2$, the length of the centrifuge arm is $12\text{m}$.

Step 2. Explanation

The centrifugal acceleration of the pilot of the fighter jet is,

-$a_c=\frac{v^2}{r}$

Here,$a_c$ is the centrifugal acceleration of the pilot, $v$is the linear speed of the pilot, and $r$is the length of the centrifuge arm.

$\left(98\text{m}/{\text{s}}^2\right)=\frac{v^2}{(12\text{m})}$

$\begin{array}{l}v=\sqrt{\left(98\text{m}/{\text{s}}^2\right)\times (12\text{m})}\\ =34.29\text{m}/\text{s}\end{array}$

Thus, the linear speed of the rider is $34.29\text{m}/\text{s}$.