A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of a circular dive. What is the radius of curvature of the circular part of the path in which the plane is flying?
A) 9200 m
B) 640 m
C) 1200 m
D) 7100 m

Question

Quizplus · Accepted Answer

The radius of curvature (r) can be calculated using the formula for centripetal acceleration: $a = \frac{v^2}{r}$, where $a$ is the acceleration and $v$ is the velocity. Given that the acceleration is 4.0 g (where g = 9.8 m/s$^2$), the actual acceleration is $4.0 \times 9.8 = 39.2$ m/s$^2$. Substituting the given values into the formula gives $39.2 = \frac{600^2}{r}$, solving for $r$ gives $r \approx 9200$ m.

A Jet Plane Flying 600 M/s Experiences an Acceleration of 4.0