An airplane is traveling at 250 m/s in level flight. In order to make a change in direction, the airplane travels in a horizontal curved path. To fly in the curved path, the pilot banks the airplane at an angle such that the lift has a horizontal component that provides the horizontal radial acceleration to move in a horizontal circular path. If the airplane is banked at an angle of 15.0 degrees, then the radius of curvature of the curved path of the airplane is
A) 20.1 km.
B) 23.8 km.
C) 25.0 km.
D) 27.5 km.
E) 30.1 km.

Question

Quizplus · Accepted Answer

The radius of curvature (r) of the airplane's path can be found using the formula for centripetal acceleration ($a_c = \frac{v^2}{r}$) and the relationship between the components of the lift force. The lift force provides the centripetal force necessary for circular motion. The horizontal component of the lift force ($L\sin(\theta)$) is responsible for the centripetal acceleration, where $\theta$ is the banking angle. Given that the lift force acts perpendicular to the wings and gravity acts downward, the vertical component of the lift must balance the weight of the airplane for level flight, implying $L\cos(\theta) = mg$. The centripetal force needed for circular motion is provided by the horizontal component of the lift, which can be expressed as $L\sin(\theta) = \frac{mv^2}{r}$, where $m$ is the mass of the airplane, $v$ is its velocity, and $r$ is the radius of the circular path.Given:- $v = 250 \, m/s$- $\theta = 15^\circ$We can rearrange the equation to solve for $r$:$$r = \frac{v^2}{g\tan(\theta)}$$Using $g = 9.8 \, m/s^2$, we can calculate $r$:$$r = \frac{(250)^2}{9.8\tan(15^\circ)} \approx 23800 \, m$$ or $23.8 \, km$.Therefore, the correct answer is 23.8 km, which corresponds to choice B.

An Airplane Is Traveling at 250 M/s in Level Flight