The diagram shows an athletics track with two parallel sides and two semi-circular ends Each of the parallel sides is 60 metres, long and the diameter of ea...

From the diagram the track is a rectangle capped by two semicircular ends. The two straight (parallel) sides are each 60 m long, and each semicircular end has diameter 120 m (the marked vertical distance between the two straights).

(a) Distance for one lap, then two laps.

The two semicircular ends together form one complete circle of diameter \(d = 120\) m. So one lap consists of the two straights plus one full circle:

\[ \text{One lap} = 2(60) + \pi d = 120 + \frac{22}{7}\times 120. \]

\[ \frac{22}{7}\times 120 = \frac{2640}{7} = 377.14\text{ m}. \]

\[ \text{One lap} = 120 + \frac{2640}{7} = \frac{840 + 2640}{7} = \frac{3480}{7} = 497.14\text{ m}. \]

Running round the track two times:

\[ D = 2 \times \frac{3480}{7} = \frac{6960}{7} = 994.29\text{ m} \;(\text{to 2 d.p.}). \]

Distance covered \(\approx 994.29\) m.

(b) Speed in km/h.

Time \(= 200\) s. First find the speed in metres per second:

\[ \text{Speed} = \frac{D}{t} = \frac{6960/7}{200} = \frac{6960}{1400} = 4.971\text{ m/s}. \]

Convert to km/h (multiply by \(\tfrac{3600}{1000} = 3.6\)):

\[ \text{Speed} = 4.971 \times 3.6 = 17.90\text{ km/h}. \]

Alternatively: \(D = \dfrac{6960}{7000}\) km \(= 0.99429\) km, \(t = \dfrac{200}{3600}\) h \(= \dfrac{1}{18}\) h, so speed \(= 0.99429 \times 18 = 17.90\) km/h.

Speed \(\approx 17.9\) km/h.