Chapter 5, Problem 2P

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?

To determine

A jet plane traveling pulls out of a drive-by moving in an arc of radius $5.20\text{km}$ what is the plane’s acceleration in g.

Answer to Problem 2P

Solution:

Acceleration of jet plane moving in a circular arc $\text{=5}\text{.41g}\text{.}$

Explanation of Solution

Given:

The speed of a jet plane $\text{=525m/s}$

The radiusof the circular arc $\text{=5}{\text{.20×10}}^{\text{3}}\text{m}$

Concept used:

Centripetal acceleration $\text{=}\left(\frac{v^2}{r}\right)$

Where v : velocity of a body moving on a circular path

r : radius of the circular path.

Calculation:

Plane’s acceleration $=\frac{v^2}{r}$

  $\begin{array}{l}\text{=}\frac{{\left(\text{525m/s}\right)}^{\text{2}}}{\text{5}{\text{.20×10}}^{\text{3}}}\\ \text{=53}{\text{.00m/s}}^{\text{2}}\end{array}$

  $\left(\text{1g}\text{=}\text{9}{\text{.8m/s}}^{\text{2}}\right)$

  $\begin{array}{l}\text{=}\frac{\text{53}{\text{.00m/s}}^{\text{2}}}{\text{9}{\text{.8m/s}}^{\text{2}}}\text{g}\\ \text{=5}\text{.41g}\text{.}\end{array}$

Conclusion:

The jet plane is moving in a circulararc, so its centripetal acceleration is $\frac{v^2}{r}$ and it is equal to $5.41\text{g}$ .