Question

A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of the loop in which the plane is flying?

Answer 1

The radius of the curve is 9,183.67 m.

Step-by-step explanation:

Given;

velocity of the jet plane, v = 600 m/s

acceleration of the jet plane, a = 4g = 4 x 9.8 m/s² = 39.2 m/s²

The radius of the curve is calculated from centripetal acceleration formula as given below;

`a = (v^2)/(r) \\\\r = (v^2)/(a) \\\\r = (600^2)/(39.2) \\\\r = 9,183.67 \ m`

Therefore, the radius of the curve is 9,183.67 m.