Kibet ran around a track five times. The track is a rectangle with two semicircles at its ends. The length of the straight sections is 110 m and the diameter of each semicircle is 70 m. Calculate the

The length of the straight sections is $110 \text{ m}$.
The diameter of each semicircle (which is also the width of the rectangular part) is $70 \text{ m}$.

The perimeter of the track consists of two straight sections and the circumference of two semicircles (which combine to form one full circle).
Length of two straight sections = $2 \times 110 \text{ m} = 220 \text{ m}$.
Circumference of one full circle (from the two semicircles) = $\pi \times \text{diameter}$.
Using the approximation $\pi = \frac{22}{7}$:
Circumference = $\frac{22}{7} \times 70 \text{ m} = 22 \times 10 \text{ m} = 220 \text{ m}$.
Perimeter of one lap = (Length of straight sections) + (Circumference of full circle)
Perimeter of one lap = $220 \text{ m} + 220 \text{ m} = 440 \text{ m}$.

Total distance = Perimeter of one lap $\times$ Number of laps
Total distance = $440 \text{ m} \times 5 = 2200 \text{ m}$.

Since $1 \text{ km} = 1000 \text{ m}$:
Total distance in kilometers = $\frac{2200 \text{ m}}{1000 \text{ m/km}} = 2.2 \text{ km}$.

Kibet covered a total distance of $\text{2.2 km}$.