The figure shows a velocity-time graph for a car.

The problem states that the distance travelled in the first four seconds is 32 meters. This corresponds to the area of the triangle from $t=0$ s to $t=4$ s.
The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

$$\text{Distance} = \frac{1}{2} \times \text{time} \times V_m$$ $$32 \text{ m} = \frac{1}{2} \times 4 \text{ s} \times V_m$$ $$32 = 2 V_m$$ $$V_m = \frac{32}{2}$$ $$V_m = 16 \text{ m/s}$$

The total distance is the area of the entire trapezium.
The area of a trapezium is given by $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$.
In this graph:

• The parallel sides are the time durations at the top and bottom of the trapezium.
• The top parallel side (constant velocity duration) is from $t=4$ s to $t=10$ s, so its length is $10 - 4 = 6$ s.
• The bottom parallel side (total time duration) is from $t=0$ s to $t=12$ s, so its length is $12 - 0 = 12$ s.
• The height of the trapezium is the maximum velocity $V_m = 16$ m/s.

$$\text{Total distance} = \frac{1}{2} \times (6 \text{ s} + 12 \text{ s}) \times 16 \text{ m/s}$$ $$\text{Total distance} = \frac{1}{2} \times 18 \text{ s} \times 16 \text{ m/s}$$ $$\text{Total distance} = 9 \text{ s} \times 16 \text{ m/s}$$ $$\text{Total distance} = 144 \text{ m}$$

Average velocity is defined as total distance divided by total time.

$$\text{Average velocity} = \frac{\text{Total distance}}{\text{Total time}}$$ $$\text{Average velocity} = \frac{144 \text{ m}}{12 \text{ s}}$$ $$\text{Average velocity} = 12 \text{ m/s}$$

The average velocity for the whole journey is $\text{12 m/s}$.