Question

a jet fighter flying at 300 m/s (just below the speed of sound) makes a turn of radius 2.30 km. (a) what is its centripetal acceleration in gs? (b) suppose the pilot makes an emergency turn to avoid an approaching missile, subjecting himself to a centripetal acceleration of 10 gs, while flying at 405 m/s (supersonic). what is the radius, in km, of his turn? (this must be short - lived because fighter planes can only briefly endure such large accelerations without serious damage, and the pilot will soon black out at 10 gs.)

Explanation:

Step1: Recall centripetal acceleration formula

The centripetal acceleration formula is \(a_{c}=\frac{v^{2}}{r}\). For part (a), \(v = 300\ m/s\) and \(r=2.30\times10^{3}\ m\). Also, \(g = 9.8\ m/s^{2}\).

Step2: Calculate centripetal acceleration in \(g\)s for part (a)

First, calculate \(a_{c}=\frac{v^{2}}{r}=\frac{300^{2}}{2.30\times 10^{3}}\).

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Then, divide by \(g = 9.8\ m/s^{2}\) to get in \(g\)s: \(n=\frac{a_{c}}{g}=\frac{900/(23)}{9.8}\).

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Step3: Rearrange formula for radius in part (b)

From \(a_{c}=\frac{v^{2}}{r}\), we can solve for \(r\): \(r=\frac{v^{2}}{a_{c}}\). Given \(a_{c}=10\times9.8\ m/s^{2}\) and \(v = 405\ m/s\).

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Answer:

a) \(4.0\ g\)
b) \(1.67\ km\)