Question

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

B) 650 m

C) 9200 m

D) 1200 m

Answer 1

Answer:

Option C

Solution:

As per the question:

Velocity of the jet, v = 600 m/s

Acceleration, a = 4.0 g

Now,

To calculate the radius of curvature:

The necessary centripetal force on the jet is given by the force:

$F = F_{c}$                                               (1)

where

$F_{c} = \frac{mv^{2}}{R}$ = centripetal force

where

m = mass of the jet

R = radius of curvature of the path

Using eqn (1):

$ma = \frac{mv^{2}}{R}$

Thus

$a = \frac{v^{2}}{R}$

$4.0 g = \frac{600^{2}}{R}$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

$4.0\times 9.8 = \frac{600^{2}}{R}$

$R = \frac{600^{2}}{4.0\times 9.8}$

R = 9183.67 m ≈ 9200 m