Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 19986 by Tinkutara last updated on 19/Aug/17

$$\mathrm{An}\mathrm{aeroplane}\mathrm{has}\mathrm{to}\mathrm{go}\mathrm{from}\mathrm{a}\mathrm{point}A$$$$\mathrm{to}\mathrm{point}B,500\mathrm{km}\mathrm{away}\mathrm{due}30°\mathrm{east}$$$$\mathrm{of}\mathrm{north}.\mathrm{A}\mathrm{wind}\mathrm{is}\mathrm{blowing}\mathrm{due}\mathrm{north}$$$$\mathrm{at}\mathrm{a}\mathrm{speed}\mathrm{of}20{\mathrm{ms}}^{-1}.\mathrm{The}\mathrm{air}\mathrm{speed}\mathrm{of}$$$$\mathrm{the}\mathrm{plane}\mathrm{is}150{\mathrm{ms}}^{-1}.\mathrm{Find}\mathrm{the}\mathrm{direction}$$$$\mathrm{in}\mathrm{which}\mathrm{the}\mathrm{pilot}\mathrm{should}\mathrm{head}\mathrm{the}$$$$\mathrm{plane}\mathrm{to}\mathrm{reach}\mathrm{point}B.$$

Answered by ajfour last updated on 19/Aug/17

$$\theta ={\sin }^{-1}\left(\frac{1}{15}\right)eastofAB.$$$$whichmeans30°+{\sin }^{-1}\left(\frac{1}{15}\right)east$$$$ofnorth.$$$$airspeediassumetobespeedof$$$$planewhennowindblows.$$

Commented by ajfour last updated on 20/Aug/17

Commented by mrW1 last updated on 20/Aug/17

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{time}\mathrm{the}\mathrm{plane}\mathrm{needs}\mathrm{to}$$$$\mathrm{reach}\mathrm{B}?$$

Commented by ajfour last updated on 20/Aug/17

$$t=\frac{d}{w\cos 30°+v\cos \theta }$$$$=\frac{500km}{(20\left(\frac{\sqrt{3}}{2}\right)+150\sqrt{1-\frac{1}{225}})m/s}$$$$\approx \frac{500}{\frac{18}{5}(10\sqrt{3}+150(1-\frac{1}{450})}h$$$$=\frac{500}{36\sqrt{3}+540-1.2}h$$$$=\frac{500}{601.152}s\approx 50\mathit{m}\mathit{i}\mathit{n}.$$$$withoutthewinditwouldhave$$$$taken{t}_0=\frac{500}{\left(\frac{18}{5}\times 150\right)}=\frac{500}{540}h\approx \frac{25}{27}h$$$$t_0=\frac{25}{3\times 9}\times 60=55.5min.$$

Commented by Tinkutara last updated on 20/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!\mathrm{That}\mathrm{was}$$$$\mathrm{the}\mathrm{next}\mathrm{part}.$$

Commented by Tinkutara last updated on 26/Aug/17

$$\mathrm{Can}\mathrm{you}\mathrm{explain}\mathrm{how}\mathrm{you}\mathrm{get}{\sin }^{-1}\left(\frac{1}{15}\right)?$$

Commented by ajfour last updated on 26/Aug/17

$$thetransversecomponent$$$$\left(\mathrm{\perp }toAB\right)ofwindvelocity=10m/s$$$$needstobecancelledbycomponent$$$$of{plane}^{\prime }svelocity;theplaneheads$$$$forwardinsuchadirection\Rightarrow$$$$\sin \theta =\frac{20\cos 60°}{150}=\frac{1}{15}.$$

Commented by Tinkutara last updated on 26/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com