Answer:
Centripetal acceleration of this aircraft: approximately [tex]6.52\; {\rm m \cdot s^{-2}}[/tex].
Distance covered during the turn: approximately [tex]63.2\; {\rm km}[/tex].
Time required for the turn: approximately [tex]0.0242\; \text{hours}[/tex] (approximately [tex]87.2\; {\rm s}[/tex].)
Explanation:
Convert velocity and radius to standard units:
[tex]\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}[/tex].
[tex]\begin{aligned} r = 80.5\; {\rm km} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 \times 10^{4}\; {\rm m}\end{aligned}[/tex].
Hence, the centripetal acceleration of this aircraft:
[tex]\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}[/tex].
The trajectory of the turn is an arc with a radius of [tex]r = 80.5\; {\rm km}[/tex] and a central angle of [tex]\theta = 90^{\circ} = (\pi / 4)[/tex]. The length of this arc would be:
[tex]\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} \times (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}[/tex].
The time required to travel [tex]63.2\; {\rm km}[/tex] at a speed of [tex]2610\; {\rm km \cdot h^{-1}}[/tex] would be:
[tex]\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}[/tex].
