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A-ship-leaves-a-port-P-which-lies-in-latitude-20-N-It-sails-due-east-through-30-of-longitude-and-then-through-south-to-Q-which-lies-on-the-equator-Calculate-the-distance-it-has-travelled-Take-th

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Question Number 10653 by Saham last updated on 21/Feb/17
A ship leaves a port P which lies in latitude 20°N. It sails due east through 30° of longitude and then through south to Q which lies on the equator. Calculate the distance it has travelled, (Take the cicumference of the earth to be 40,000 km). on the return jouney it sails due west through 30° of longitude and then due north back to P. Show that the difference in length between the outward and return jouney is approximately 201 kilometers. Using this value of 201 km and taking 1 knot to be 1.852 km/hr . Calculate the difference in time between the two journey assuming that on each journey the ship sails at an average speed of 25 knots. Really need help here sirs. God will always increase your wisdom.
$$\mathrm{A}\:\mathrm{ship}\:\mathrm{leaves}\:\mathrm{a}\:\mathrm{port}\:\mathrm{P}\:\mathrm{which}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{latitude}\:\mathrm{20}°\mathrm{N}.\:\mathrm{It}\:\mathrm{sails}\:\mathrm{due}\:\mathrm{east} \\ $$$$\mathrm{through}\:\mathrm{30}°\:\mathrm{of}\:\:\mathrm{longitude}\:\mathrm{and}\:\mathrm{then}\:\mathrm{through}\:\mathrm{south}\:\mathrm{to}\:\mathrm{Q}\:\mathrm{which}\:\mathrm{lies} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{equator}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{it}\:\mathrm{has}\:\mathrm{travelled}, \\ $$$$\left(\mathrm{Take}\:\mathrm{the}\:\mathrm{cicumference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}\:\mathrm{to}\:\mathrm{be}\:\mathrm{40},\mathrm{000}\:\mathrm{km}\right). \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{return}\:\mathrm{jouney}\:\mathrm{it}\:\mathrm{sails}\:\mathrm{due}\:\mathrm{west}\:\mathrm{through}\:\mathrm{30}°\:\mathrm{of}\:\mathrm{longitude}\:\mathrm{and} \\ $$$$\mathrm{then}\:\mathrm{due}\:\mathrm{north}\:\mathrm{back}\:\mathrm{to}\:\mathrm{P}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{in}\:\mathrm{length}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{outward}\:\mathrm{and}\:\mathrm{return}\:\mathrm{jouney}\:\mathrm{is}\:\mathrm{approximately}\:\mathrm{201}\:\mathrm{kilometers}. \\ $$$$\mathrm{Using}\:\mathrm{this}\:\mathrm{value}\:\mathrm{of}\:\mathrm{201}\:\mathrm{km}\:\mathrm{and}\:\mathrm{taking}\:\mathrm{1}\:\mathrm{knot}\:\mathrm{to}\:\mathrm{be}\:\mathrm{1}.\mathrm{852}\:\mathrm{km}/\mathrm{hr}\:. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{in}\:\mathrm{time}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{journey}\:\mathrm{assuming} \\ $$$$\mathrm{that}\:\mathrm{on}\:\mathrm{each}\:\mathrm{journey}\:\mathrm{the}\:\:\mathrm{ship}\:\mathrm{sails}\:\mathrm{at}\:\mathrm{an}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{of}\:\:\mathrm{25}\:\mathrm{knots}. \\ $$$$ \\ $$$$\mathrm{Really}\:\mathrm{need}\:\mathrm{help}\:\mathrm{here}\:\mathrm{sirs}.\:\mathrm{God}\:\mathrm{will}\:\mathrm{always}\:\mathrm{increase}\: \\ $$$$\mathrm{your}\:\mathrm{wisdom}. \\ $$
Answered by mrW1 last updated on 22/Feb/17
Earth P=2πR=40000 km Distance between 20°N and equator d_1 d_1 =((20)/(360))P=((20)/(360))×40000=((20000)/9)=2222 km P_(20°N) =2πRcos 20=Pcos 20=40000cos 20=37588 km Distance along 20°N for 30° d_2 d_2 =((30)/(360))P_(20°N) =((30)/(360))×37588=3132 km Distance along equator for 30° d_3 d_3 =((30)/(360))P=((30)/(360))×40000=3333 km Total distance of outward journey: d_2 +d_1 =3132+2222=5354 km Total distance of return journey: d_3 +d_1 =3333+2222=5555 km Difference in distance =d_3 −d_2 =3333−3132=201 km Difference in sailing time =((201)/(25×1.852))=4.34 h=4 h 20 m
$${Earth}\:{P}=\mathrm{2}\pi{R}=\mathrm{40000}\:{km} \\ $$$${Distance}\:{between}\:\mathrm{20}°{N}\:{and}\:{equator}\:{d}_{\mathrm{1}} \\ $$$${d}_{\mathrm{1}} =\frac{\mathrm{20}}{\mathrm{360}}{P}=\frac{\mathrm{20}}{\mathrm{360}}×\mathrm{40000}=\frac{\mathrm{20000}}{\mathrm{9}}=\mathrm{2222}\:{km} \\ $$$$ \\ $$$${P}_{\mathrm{20}°{N}} =\mathrm{2}\pi{R}\mathrm{cos}\:\mathrm{20}={P}\mathrm{cos}\:\mathrm{20}=\mathrm{40000cos}\:\mathrm{20}=\mathrm{37588}\:{km} \\ $$$${Distance}\:{along}\:\mathrm{20}°{N}\:{for}\:\mathrm{30}°\:{d}_{\mathrm{2}} \\ $$$${d}_{\mathrm{2}} =\frac{\mathrm{30}}{\mathrm{360}}{P}_{\mathrm{20}°{N}} =\frac{\mathrm{30}}{\mathrm{360}}×\mathrm{37588}=\mathrm{3132}\:{km} \\ $$$$ \\ $$$${Distance}\:{along}\:{equator}\:{for}\:\mathrm{30}°\:{d}_{\mathrm{3}} \\ $$$${d}_{\mathrm{3}} =\frac{\mathrm{30}}{\mathrm{360}}{P}=\frac{\mathrm{30}}{\mathrm{360}}×\mathrm{40000}=\mathrm{3333}\:{km} \\ $$$$ \\ $$$${Total}\:{distance}\:{of}\:{outward}\:{journey}: \\ $$$${d}_{\mathrm{2}} +{d}_{\mathrm{1}} =\mathrm{3132}+\mathrm{2222}=\mathrm{5354}\:{km} \\ $$$$ \\ $$$${Total}\:{distance}\:{of}\:{return}\:{journey}: \\ $$$${d}_{\mathrm{3}} +{d}_{\mathrm{1}} =\mathrm{3333}+\mathrm{2222}=\mathrm{5555}\:{km} \\ $$$$ \\ $$$${Difference}\:{in}\:{distance}\:={d}_{\mathrm{3}} −{d}_{\mathrm{2}} =\mathrm{3333}−\mathrm{3132}=\mathrm{201}\:{km} \\ $$$$ \\ $$$${Difference}\:{in}\:{sailing}\:{time}\:=\frac{\mathrm{201}}{\mathrm{25}×\mathrm{1}.\mathrm{852}}=\mathrm{4}.\mathrm{34}\:{h}=\mathrm{4}\:{h}\:\mathrm{20}\:{m} \\ $$
Commented by Saham last updated on 22/Feb/17
Wow.. Great.. i have tried it but i did not get it. Now i understand. God bless you sir.
$$\mathrm{Wow}..\:\mathrm{Great}..\:\mathrm{i}\:\mathrm{have}\:\mathrm{tried}\:\mathrm{it}\:\mathrm{but}\:\mathrm{i}\:\mathrm{did}\:\mathrm{not}\:\mathrm{get}\:\mathrm{it}. \\ $$$$\mathrm{Now}\:\mathrm{i}\:\mathrm{understand}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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