Question

A jet plane traveling$\mathbf{1890}\mathbf{km}\mathbf{/}\mathbf{s}\left(\mathbf{525}\mathbf{m}\mathbf{/}\mathbf{s}\right)$pulls out of a dive by moving in an arc of radius 5.20 km. What is the plane’s acceleration in g’s?

Short answer

The plane’s acceleration in g’s is $5.403g\text{'}\mathrm{s}$.

Step-by-step solution

Step 1. Understanding the centripetal acceleration of the jet plane

The jet plane is travelling in an arc. Then the jet plane experiences a centripetal force.The centripetal acceleration of the jet plane depends on the jet plane's speed and the radius of an arc.

Step 2. Identification of given data 

The given data can be listed below as,

• Theradius of an arc is,$r=5.20\mathrm{km}\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)=5200\mathrm{m}$.

• The speed of the jet plane is, $v=525\mathrm{m}/\mathrm{s}$.
• The acceleration due to gravity is, $g=9.81\mathrm{m}/{\mathrm{s}}^2$.

Step 3. Determination of the centripetal acceleration of the jet plane

The centripetal acceleration can be expressed as,

$a_{\mathrm{c}}=\frac{v^2}{r}$

Here, v is the speed of the jet plane, r is the radius of the merry-go-round.

Substitute the values in the above equation.

$\begin{array}{c}{a}_c=\frac{{\left(525\mathrm{m}/\mathrm{s}\right)}^2}{5200\mathrm{m}}\\ =53\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{g}}{9.81\mathrm{m}/{\mathrm{s}}^2}\right)\\ =5.403g\text{'}\mathrm{s}\end{array}$

Thus, the centripetal acceleration of the jet plane is $5.403g\text{'}\mathrm{s}$.