The POSITION-TIME GRAPH (not drawn to scale) for a cyclist initially moving west on a straight horizontal road is shown below: 1 Why does the position decrease after 60 seconds? 2 Calculate the velocity of the cyclist for the first 30 seconds.

From the graph:
At $t_1 = 0$ s, position $x_1 = 0$ m.
At $t_2 = 30$ s, position $x_2 = 60$ m.

The cyclist is initially moving west, so the velocity is $2 \text{ m}\cdot\text{s}^{-1}$ west.
The velocity of the cyclist for the first 30 seconds is $\boxed{\text{2 m}\cdot\text{s}^{-1} \text{ west}}$.

The position changes from 0 m to 60 m.
Distance$_1 = |60 \text{ m} - 0 \text{ m}| = 60 \text{ m}$.

The position remains constant at 60 m.
Distance$_2 = |60 \text{ m} - 60 \text{ m}| = 0 \text{ m}$.

Given that the magnitude of the velocity is $2.5 \text{ m}\cdot\text{s}^{-1}$ and the time interval is $70 \text{ s} - 60 \text{ s} = 10 \text{ s}$.
Distance$_3 = \text{magnitude of velocity} \times \Delta t$
Distance$_3 = 2.5 \text{ m}\cdot\text{s}^{-1} \times 10 \text{ s} = 25 \text{ m}$.

Total distance = Distance$_1$ + Distance$_2$ + Distance$_3$
Total distance = $60 \text{ m} + 0 \text{ m} + 25 \text{ m}$
Total distance = $85 \text{ m}$.
The total distance covered by the cyclist for the entire 70 second period is $\boxed{\text{85 m}}$.