(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.
(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and (pi) = 3.142].
Explanation
(a) (P propto Q) and (P propto frac{1}{R^{2}}).
(P propto frac{Q}{R^{2}} implies P = frac{kQ}{R^{2}})
When P = 1, Q = 8, R = 2
(1 = frac{8k}{2^{2}} implies 8k = 4)
(k = frac{4}{8} = 0.5)
(P = frac{Q}{2R^{2}})
When P = 3, R = 5, Q = ?
(3 = frac{Q}{2(5^{2})})
(3 = frac{Q}{50} implies Q = 150).
(b)
Angular difference : 60° – 20° = 40°
(frac{theta}{360} times 2pi r )
(r = Rcos theta)
(frac{40}{360} times 2 times 3.142 times 6400 cos 20 = frac{1511687.278}{360})
= (4199.13 km)
(Speed = frac{Distance}{Time})
= (frac{4199.13}{6})
= (699.85 km/hr)
(approxeq 700 km/hr) (to 3 significant figures)
(ii) (D = frac{theta}{360} times 2pi r)
(420 = frac{theta}{360} times 2 times 3.142 times 6400)
(theta = frac{420 times 360}{2 times 3.142 times 6400})
(theta = frac{151200}{40217.6} = 3.76°)
Latitude of C : 20° + 3.76° = 23.76°
Note that due North implies same longitude but different latitude.