1. **Problem Statement:** We analyze the motion of a girl walking between two classrooms with given velocities and times, and answer questions about her displacement, distance, average velocity, and average speed. 2. **Understanding the scenario:** - She starts at Mrs. Gallant's classroom (position 0 m). - She waits there for 3 minutes (180 s), but since she is stationary, position does not change during this time. - She then moves towards Mr. Murphy's classroom (East) but immediately realizes she forgot something and hurries back West towards Mrs. Gallant's classroom. - She moves back at 4 m/s for 6 s, then slows to 2 m/s for another 6 s until she arrives at Mrs. Gallant's classroom. 3. **Position vs. Time Graph (a):** - From 0 to 180 s: position is constant at 0 m. - From 180 s to 186 s: moving East (positive direction) to Mr. Murphy's classroom. However, the problem states she begins her next class there, so we assume she reached Mr. Murphy's classroom at 180 s (start of return trip). - From 180 s to 186 s: moving West at 4 m/s for 6 s, so displacement is $\Delta x_1 = -4 \times 6 = -24$ m. - From 186 s to 192 s: moving West at 2 m/s for 6 s, so displacement is $\Delta x_2 = -2 \times 6 = -12$ m. - Total displacement from Mr. Murphy's classroom back to Mrs. Gallant's classroom is $-24 - 12 = -36$ m. 4. **Calculating displacement for entire trip (b):** - She starts at 0 m, goes East to Mr. Murphy's classroom (unknown distance $d$), then returns 36 m West. - Since she ends at Mrs. Gallant's classroom (starting point), total displacement is $0$ m. 5. **Calculating distance travelled (c):** - Distance to Mr. Murphy's classroom: $d$ m (unknown, but can be inferred from return trip distance). - Distance returning: $24 + 12 = 36$ m. - Since she returns to start, $d = 36$ m. - Total distance travelled = $36 + 36 = 72$ m. 6. **Average velocity (d):** - Total displacement = 0 m. - Total time = waiting time + moving time = $180 + 6 + 6 = 192$ s. - Average velocity $= \frac{0}{192} = 0$ m/s. 7. **Average speed (e):** - Total distance = 72 m. - Total time = 192 s. - Average speed $= \frac{72}{192} = 0.375$ m/s. **Final answers:** - (b) Displacement = $0$ m. - (c) Distance travelled = $72$ m. - (d) Average velocity = $0$ m/s. - (e) Average speed = $0.375$ m/s.