An aeroplane flies from a town P(lat. 40°N, 38°E) to another town Q(lat. 40°N, 22°W). It later flies to a third town T(28°N, 22°W). Calculate the :
(a) distance between P and Q along their parallel of latitude ;
(b) distance between Q and T along their line of longitudes;
(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \(\pi = 3.142\)]
(a) distance between P and Q along their parallel of latitude ;
(b) distance between Q and T along their line of longitudes;
(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \(\pi = 3.142\)]
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Correct Answer: Option n
Explanation:
(a) Distance along arc PQ = \(\frac{\theta}{360°} \times 2 \pi r\)
where \(r = R \cos \theta\)
\(|PQ| = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 6400 \times \cos 40°\)
\(|PQ| = \frac{1}{6} \times 2 \times \frac{22}{7} \times 6400 \times 0.7660\)
= \(5134.74 km \approxeq 5135km\) (to the nearest whole number)
(b) Difference in latitude between Q and T = 40° - 28° = 12°
\(\therefore |QT| \text{along the line of longitude} = \frac{12}{360} \times 2 \times \frac{22}{7} \times 6400\)
\(\frac{1}{30} \times 2 \times \frac{22}{7} \times 6400 = 1340.95km \approxeq 1341km\)
(c) Average speed = \(\frac{\text{Total distance covered}}{\text{Total time taken}}\)
= \(\frac{5134.74 + 1340.95}{12} = 539.64km/hr\)
\(\approxeq 540km/hr\) (to 3 significant figures)
(a) Distance along arc PQ = \(\frac{\theta}{360°} \times 2 \pi r\)
where \(r = R \cos \theta\)
\(|PQ| = \frac{60°}{360°} \times 2 \times \frac{22}{7} \times 6400 \times \cos 40°\)
\(|PQ| = \frac{1}{6} \times 2 \times \frac{22}{7} \times 6400 \times 0.7660\)
= \(5134.74 km \approxeq 5135km\) (to the nearest whole number)
(b) Difference in latitude between Q and T = 40° - 28° = 12°
\(\therefore |QT| \text{along the line of longitude} = \frac{12}{360} \times 2 \times \frac{22}{7} \times 6400\)
\(\frac{1}{30} \times 2 \times \frac{22}{7} \times 6400 = 1340.95km \approxeq 1341km\)
(c) Average speed = \(\frac{\text{Total distance covered}}{\text{Total time taken}}\)
= \(\frac{5134.74 + 1340.95}{12} = 539.64km/hr\)
\(\approxeq 540km/hr\) (to 3 significant figures)