1. **Problem Statement:** Describe the motion of a car shown on a position-time graph where the car starts at position 5 m at time 0 s, rapidly moves to position 1 m by about 2 s, stays nearly constant between 1 and 2 m until 4 s, then accelerates linearly back to position 5 m by about 15 s. 2. **Understanding the Graph:** - The vertical axis is position in meters. - The horizontal axis is time in seconds. - The car's position decreases quickly from 5 m to 1 m in the first 2 seconds. - From 2 s to 4 s, the position remains almost constant near 1-2 m. - From 4 s to 15 s, the position increases linearly back to 5 m. 3. **Velocity and Motion Analysis:** - Velocity is the slope of the position-time graph: $v = \frac{\Delta x}{\Delta t}$. 4. **Step 1: Calculate initial velocity (0 to 2 s):** $$v_1 = \frac{1 - 5}{2 - 0} = \frac{-4}{2} = -2\ \text{m/s}$$ The negative velocity means the car is moving backward or slowing down. 5. **Step 2: Velocity from 2 s to 4 s:** Position is nearly constant, so velocity is approximately zero: $$v_2 \approx 0\ \text{m/s}$$ The car is nearly stopped. 6. **Step 3: Velocity from 4 s to 15 s:** $$v_3 = \frac{5 - 1}{15 - 4} = \frac{4}{11} \approx 0.36\ \text{m/s}$$ The positive velocity indicates the car is moving forward and accelerating steadily. 7. **Summary:** - The car starts moving backward at about $-2$ m/s, slowing down. - It stops and remains nearly stationary between 2 s and 4 s. - Then it accelerates forward at about $0.36$ m/s until 15 s. This matches the description: the car slows down before a stop light, stops, then accelerates when the light turns green.