The diagram below is a speed-time graph for an athlete who took 100 seconds to complete a race. The athlete ran 1,75 km in 100 seconds. Calculate the value of V. and acceleration of the athlete from time 60 seconds to time 100 seconds.

The athlete ran $1.75 \text{ km}$ in $100 \text{ seconds}$.

The graph is a trapezium. The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.
In this speed-time graph, the "height" of the trapezium is the maximum speed $V$. The parallel sides are the time intervals on the x-axis.
The top parallel side (where speed is constant) is from $t=20 \text{ s}$ to $t=60 \text{ s}$, so its length is $60 - 20 = 40 \text{ s}$.
The bottom parallel side (total time) is from $t=0 \text{ s}$ to $t=100 \text{ s}$, so its length is $100 - 0 = 100 \text{ s}$.

The value of $V$ is $\boxed{25 \text{ m/s}}$.

Acceleration is the gradient of the speed-time graph.
At $t_1 = 60 \text{ s}$, the speed is $V_1 = V = 25 \text{ m/s}$.
At $t_2 = 100 \text{ s}$, the speed is $V_2 = 0 \text{ m/s}$.

The acceleration is $\boxed{-0.625 \text{ m/s}^2}$.