1. **Problem statement:** Calculate the total displacement traveled by the object over the entire journey given the velocity-time graph. 2. **Formula and rules:** Displacement is the area under the velocity-time graph. 3. **Step-by-step solution:** - The graph consists of several segments: - From 0 to 2 s: velocity increases linearly from 0 to 15 m/s (triangle area). - From 2 to 7 s: velocity constant at 15 m/s (rectangle area). - From 7 to 8 s: velocity increases linearly from 15 to 30 m/s (triangle area). - From 8 to 10 s: velocity decreases linearly from 30 to 10 m/s (trapezoid area). - From 10 to 12 s: velocity constant at 10 m/s (rectangle area). - From 12 to 14 s: velocity decreases linearly from 10 to 0 m/s (triangle area). Calculate each area: - Area 1 (0 to 2 s): $$\frac{1}{2} \times 2 \times 15 = 15\,m$$ - Area 2 (2 to 7 s): $$5 \times 15 = 75\,m$$ - Area 3 (7 to 8 s): $$\frac{1}{2} \times 1 \times (30 - 15) = 7.5\,m$$ - Area 4 (8 to 10 s): $$\frac{1}{2} \times (30 + 10) \times 2 = 40\,m$$ - Area 5 (10 to 12 s): $$2 \times 10 = 20\,m$$ - Area 6 (12 to 14 s): $$\frac{1}{2} \times 2 \times 10 = 10\,m$$ Total displacement = 15 + 75 + 7.5 + 40 + 20 + 10 = 167.5 m **Note:** The problem states the total displacement is 190 m, which suggests the graph or values might have slight differences or rounding. Based on the given data, the calculated displacement is 167.5 m. --- Since the user specifically asked for parts b ii and iii: b) ii) **Acceleration at t = 4 s:** - At 4 s, velocity is constant at 15 m/s (flat segment from 2 to 7 s). - Acceleration $$a = \frac{\Delta v}{\Delta t} = 0\,m/s^2$$ b) iii) **Deceleration at t = 9 s:** - Between 8 s and 10 s, velocity drops from 30 m/s to 10 m/s. - Time interval $$\Delta t = 2 s$$ - Change in velocity $$\Delta v = 10 - 30 = -20 m/s$$ - Acceleration $$a = \frac{-20}{2} = -10\,m/s^2$$ (negative indicates deceleration) **Final answers:** - Total displacement: approximately 167.5 m (based on given graph data) - Acceleration at 4 s: 0 m/s^2 - Deceleration at 9 s: -10 m/s^2