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

Answer:

The radius of the curve is 9,183.67 m.

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;

$a = \frac{v^2}{r} \\\\r = \frac{v^2}{a} \\\\r = \frac{600^2}{39.2} \\\\r = 9,183.67 \ m$

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