Question

A supersonic airplane is flying horizontally at a speed of 2610 km/h.

Answer 1

Answer:

Centripetal acceleration of this aircraft: approximately $6.52\; {\rm m \cdot s^{-2}}$.

Distance covered during the turn: approximately $63.2\; {\rm km}$.

Time required for the turn: approximately $0.0242\; \text{hours}$ (approximately $87.2\; {\rm s}$.)

Explanation:

Convert velocity and radius to standard units:

$\begin{aligned}v &= 2610\; {\rm km \cdot h^{-1}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &= 725\; {\rm m\cdot s^{-1}} \end{aligned}$.

$\begin{aligned} r = 80.5\; {\rm km} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 \times 10^{4}\; {\rm m}\end{aligned}$.

Hence, the centripetal acceleration of this aircraft:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(725\; {\rm m\cdot s^{-1}})^{2} }{8.05 \times 10^{4}\; {\rm m}} \\ &\approx 6.53\; {\rm m \cdot s^{-2}}\end{aligned}$.

The trajectory of the turn is an arc with a radius of $r = 80.5\; {\rm km}$ and a central angle of $\theta = 90^{\circ} = (\pi / 4)$. The length of this arc would be:

$\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} \times (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}$.

The time required to travel $63.2\; {\rm km}$ at a speed of $2610\; {\rm km \cdot h^{-1}}$ would be:

$\begin{aligned}t &= \frac{s}{v} \\ &\approx \frac{63.2\; {\rm km}}{2610\; {\rm km \cdot h^{-1}}} \\ &\approx 0.0242\; {\rm h} \\ &\approx 0.0242 \; {\rm h} \times \frac{3600\; {\rm s}}{1\; {\rm h}} \\ &\approx 87.2\; {\rm s} \end{aligned}$.