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

Answer

given,

jet plane is traveling = 525 m/s

radius = 5.2 Km

a_c = ?                          

$a_c = \dfrac{v^2}{r}$          

$a_c = \dfrac{525^2}{5200}$

$a_c = 53 m/s^2$                  

we know                                              

g = 9.8 m/s²                                        

$\dfrac{a_c}{g}=\dfrac{53}{9.8}$

$\dfrac{a_c}{g} = 5.41$

$a_c = 5.41 g$

Answer 2

Answer:

Acceleration of the plane, a = 5.4 g

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 :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(525\ m/s)^2}{5200\ m}$

$a=53.004\ m/s^2$

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

$\dfrac{a}{g}=\dfrac{53.004\ m/s^2}{9.8\ m/s^2}=5.4$

$a=5.4\times g$

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