Question

Question:

An airplane experiences a uniform circular motion. The airplane has an initial velocity of \( \vec v \) = [0.500, 0.330] km/s. Two minutes later, the velocity of the airplane is \( - \vec v \). What centripetal acceleration does the pilot experience?

Answer 1

Step 1

Given the following values:

\( \vec v_1 = {\rm (0.500\,km/s)\hat i + (0.330\,km/s)\hat j} \\ \)

\( \vec v_2 = -\vec v_1 = -\left [{\rm (0.500\,km/s)\hat i + (0.330\,km/s)\hat j} \right ] \\ \)

The magnitude of both velocities is:

\( v_1 = v_2 = v = {\rm \sqrt{(0.500\,km/s)^2 + (0.330\,km/s)^2 } = 0.599\,km/s = 599\,m/s }\\ \)

 

Step 2

\( s = \pi\,r \\ \)

\( v = \dfrac{s}{t} = \dfrac{\pi\,r}{t} \\ \)

\( r = \color{blue}{\dfrac{v\,t}{\pi}} \\ \)

 

Step 3

\( a_{\rm\ centripetal} = \dfrac{v^2}{r} = \dfrac{v^2}{ \color{blue}{\dfrac{v\,t}{\pi }} } = \pi\dfrac{v}{t} \\ \)

\( a_{\rm\ centripetal} = \pi\dfrac{v}{t} = {\rm \pi\dfrac{599\,m/s }{120\,s} = \boxed{\rm 15.68\,m/s^2} }\\ \)

The centripetal acceleration experienced by the pilot is about \( \rm 15.68\ m/s^2 \)