Question

A jet plane flying 600 m/s experiences an acceleration of 4.0 g when pulling out of the circular section of a dive. What is the radius of curvature of this section of the dive?

Answer 1

The radius is
`r = 9183.7 \ m`

Step-by-step explanation:

From the question we are told that

The speed of the jet plane is
`v = 600 \ m/s`

The acceleration is
`a = 4g` here
`g = 9.8 \ m/s^2` so
`a = 4* 9.8 = 39.2 \ m/s^2`

Generally the centripetal acceleration of the jet just before it pulled out of the circular section dive is mathematically represented as

`a = (v^2)/(r)`

=>
`r = (v^2)/(a)`

=>
`r = (600^2)/(39.2)`

=>
`r = 9183.7 \ m`