1. **State the problem:** A car travels along a straight path for 10 minutes with a velocity given by a velocity-time graph. We need to find the distance from the start after 10 minutes. 2. **Understand the velocity-time graph:** The graph shows two semicircles: one above the time axis from 0 to 6 minutes with a peak velocity of 3 km/min at 3 minutes, and one below the time axis from 6 to 10 minutes with a minimum velocity of -2 km/min at 8 minutes. 3. **Formula used:** Distance from start (displacement) is the integral of velocity over time: $$\text{Distance} = \int_0^{10} v(t) \, dt$$ Since the velocity graph is composed of semicircles, the area under the curve corresponds to the displacement. 4. **Calculate the area of the first semicircle (above time axis):** - Radius $r = \frac{6 - 0}{2} = 3$ minutes - Height (max velocity) $= 3$ km/min Area of full circle $= \pi r^2 = \pi \times 3^2 = 9\pi$ Area of semicircle $= \frac{1}{2} \times 9\pi = \frac{9\pi}{2}$ Since the velocity axis is vertical and time horizontal, the area represents km. 5. **Calculate the area of the second semicircle (below time axis):** - Radius $r = \frac{10 - 6}{2} = 2$ minutes - Height (velocity magnitude) $= 2$ km/min Area of full circle $= \pi r^2 = \pi \times 2^2 = 4\pi$ Area of semicircle $= \frac{1}{2} \times 4\pi = 2\pi$ Since this semicircle is below the time axis, its area is negative displacement. 6. **Calculate net displacement:** $$\text{Distance} = \frac{9\pi}{2} - 2\pi = \frac{9\pi}{2} - \frac{4\pi}{2} = \frac{5\pi}{2}$$ 7. **Evaluate numerically:** $$\frac{5\pi}{2} \approx \frac{5 \times 3.1416}{2} = \frac{15.708}{2} = 7.854$$ 8. **Final answer:** The car is approximately **7.854 km** from the start after 10 minutes.