1. **State the problem:** We need to find the displacement of a runner after 10.6 seconds given the velocity-time graph. 2. **Understand the graph:** The velocity decreases linearly from 8 m/s at 0 s to 0 m/s at 16 s, forming a right triangle under the line. 3. **Formula for displacement:** Displacement is the area under the velocity-time graph. 4. **Calculate the area of the triangle for 10.6 s:** The velocity at 10.6 s is found by linear interpolation: $$v = 8 - \frac{8}{16} \times 10.6 = 8 - 0.5 \times 10.6 = 8 - 5.3 = 2.7\, m/s$$ The area under the graph from 0 to 10.6 s is a trapezoid with bases 8 m/s and 2.7 m/s and height 10.6 s: $$\text{Displacement} = \frac{(8 + 2.7)}{2} \times 10.6 = \frac{10.7}{2} \times 10.6$$ 5. **Calculate displacement:** $$\text{Displacement} = 5.35 \times 10.6 = 56.71\, m$$ 6. **Answer:** The runner's displacement after 10.6 seconds is approximately **56.71 meters**.