A cyclist travels a distance of 1 km in the first hour, 0.5 km in the second hour and 0.3 km in the third hour. Find the average speed of the cyclist in:

(i) km h^-1

(ii) m s^-1.

(i) total distance travelled = 1 km + 0.5 km + 0.3 km = 1.8 km
total time taken = 1 h + 1 h + 1 h = 3 h

$\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \\[1em] \text{Average Speed} = \dfrac{\text{1.8}}{\text{3}} = 0.6 \text{km h}^{-1}$

So, average speed of the cyclist = 0.6 km h^-1.

(ii) total distance travelled = 1 km + 0.5 km + 0.3 km = 1.8 km
1 km = 1000 m
1.8 km = 1.8 x 1000 = 1800 m

total time taken = 1 h + 1 h + 1 h = 3 h
1 h = 3600 sec
3 h = 3 x 3600 = 10800 sec

$\text{Average Speed} = \dfrac{\text{Total distance travelled}}{\text{total time taken}} \\[1em] \text{Average Speed} = \dfrac{\text{1800}}{\text{10800}} = 0.167 \text{m s}^{-1}$

So, average speed of the cyclist = 0.167 m s^-1.