1. Problem: Given the velocity-time graph of a car moving on a straight road, calculate the total distance traveled by the car until it stops. The graph shows three intervals: velocity increases linearly from 0 to 10 m/s over 0 to 5 s, remains constant at 10 m/s from 5 to 15 s, then decreases linearly back to 0 m/s from 15 to 30 s. 2. Calculate distance for each time interval using the area under the velocity-time graph (area represents displacement): - From $t=0$ to $t=5$ seconds: velocity increases linearly, so the area is a triangle. $$\text{Area}_1 = \frac{1}{2} \times 5 \times 10 = 25\text{ m}$$ - From $t=5$ to $t=15$ seconds: velocity is constant at 10 m/s, so the area is a rectangle. $$\text{Area}_2 = 10 \times (15 - 5) = 10 \times 10 = 100\text{ m}$$ - From $t=15$ to $t=30$ seconds: velocity decreases linearly back to 0 m/s, so the area is another triangle. $$\text{Area}_3 = \frac{1}{2} \times (30 - 15) \times 10 = \frac{1}{2} \times 15 \times 10 = 75\text{ m}$$ 3. Total distance traveled is the sum of the three areas: $$\text{Total distance} = 25 + 100 + 75 = 200\text{ m}$$ 4. Final answer: The total distance traveled by the car until stopping is \textbf{200 m}.