Question

A jet plane traveling 1890 km/h (525 m/s) pulls out of a dive by moving in an arc of radius 5.20 km. What is the plane's acceleration in g's?

Answer 1

given,

jet plane is traveling = 525 m/s

radius = 5.2 Km

a_c = ?

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

`a_c = (525^2)/(5200)`

`a_c = 53 m/s^2`

we know

g = 9.8 m/s²

`(a_c)/(g)=(53)/(9.8)`

`(a_c)/(g) = 5.41`

`a_c = 5.41 g`

Answer 2

Acceleration of the plane, a = 5.4 g

Step-by-step explanation:

It is given that,

Speed of the jet plane, v = 1890 km/h = 525 m/s

Radius of the arc, r = 5.20 km = 5200 m

The plane is moving in the circular path, the centripetal acceleration will act on it. It is given by :

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

`a=((525\ m/s)^2)/(5200\ m)`

`a=53.004\ m/s^2`

We know that, the value of g is,
`g=9.8\ m/s^2`

`(a)/(g)=(53.004\ m/s^2)/(9.8\ m/s^2)=5.4`

`a=5.4* g`

So, the acceleration of the plane is 5.4 g. Hence, this is the required solution.