1. **State the problem:** We have a distance-time graph for two particles A and B. Particle A moves along points E(0,0), F(2,15), G(6,15), and H(t,36). Particle B moves along a straight line from P(0,15) to H(t,36). We need to: (a) Find the duration when particle A is stationary. (b) Calculate the speed of particle A in the first 2 seconds. (c) Given the difference in average speeds of particles A and B is $\frac{5}{3}$ ms$^{-1}$, find $t$. 2. **Important formulas and rules:** - Speed = $\frac{\text{distance}}{\text{time}}$ - Particle is stationary when distance does not change over time (horizontal segment on graph). - Average speed = $\frac{\text{total distance}}{\text{total time}}$ 3. **Step (a): Duration particle A is stationary** - Particle A is stationary between points F(2,15) and G(6,15) because distance remains 15 m. - Duration = $6 - 2 = 4$ seconds. 4. **Step (b): Speed of particle A in first 2 seconds** - Distance traveled from E(0,0) to F(2,15) is 15 m. - Time taken = 2 seconds. - Speed = $\frac{15}{2} = 7.5$ ms$^{-1}$. 5. **Step (c): Calculate $t$ given difference in average speeds is $\frac{5}{3}$ ms$^{-1}$** - Average speed of particle A: - Total distance = distance from E to F + F to G + G to H - EF: 15 m, FG: 0 m (stationary), GH: $36 - 15 = 21$ m - Total distance = $15 + 0 + 21 = 36$ m - Total time = $t - 0 = t$ seconds - Average speed of A = $\frac{36}{t}$ - Average speed of particle B: - Distance from P(0,15) to H(t,36) = $36 - 15 = 21$ m - Time = $t$ seconds - Average speed of B = $\frac{21}{t}$ - Given difference in average speeds: $$\left| \frac{36}{t} - \frac{21}{t} \right| = \frac{5}{3}$$ $$\frac{15}{t} = \frac{5}{3}$$ $$t = \frac{15 \times 3}{5} = 9$$ seconds. **Final answers:** (a) Particle A is stationary for 4 seconds. (b) Speed of particle A in first 2 seconds is 7.5 ms$^{-1}$. (c) The value of $t$ is 9 seconds.