(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 ...
Question 1 Report
(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate. correct to the nearest whole number, the:
(i) distance between cyclist X and Y;
(ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
Answer Details
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
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