1. **Problem statement:** Calculate the acceleration of the car in the first 10 seconds and find the time $T$ when the car travels 700 m. 2. **Acceleration calculation:** Acceleration $a$ is the change in speed divided by the change in time: $$a = \frac{\Delta v}{\Delta t}$$ From the graph, speed changes from 0 m/s to 20 m/s in 10 seconds. So, $$a = \frac{20 - 0}{10 - 0} = \frac{20}{10} = 2\ \text{m/s}^2$$ 3. **Distance calculation:** The car's speed-time graph consists of two parts: - From 0 to 10 seconds, speed increases linearly from 0 to 20 m/s (triangle area under graph). - From 10 to $T$ seconds, speed is constant at 20 m/s (rectangle area under graph). 4. **Distance traveled in first 10 seconds:** Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$ $$= \frac{1}{2} \times 10 \times 20 = 100\ \text{m}$$ 5. **Distance traveled from 10 seconds to $T$ seconds:** Speed is constant at 20 m/s, so distance = speed $\times$ time $$= 20 \times (T - 10)$$ 6. **Total distance traveled is 700 m:** Sum of distances: $$100 + 20(T - 10) = 700$$ 7. **Solve for $T$:** $$20(T - 10) = 700 - 100$$ $$20(T - 10) = 600$$ $$T - 10 = \frac{600}{20} = 30$$ $$T = 30 + 10 = 40\ \text{seconds}$$ **Final answers:** - Acceleration in first 10 seconds: $2\ \text{m/s}^2$ - Time $T$ when car travels 700 m: $40$ seconds