1. **Problem 12 (i): Sketch the velocity-time graph for the car's motion.** - From $t=0$ to $t=30$ s, velocity is constant at 10 m/s. - From $t=30$ s to $t=42$ s, velocity decreases uniformly from 10 m/s to 0 m/s. 2. **Problem 12 (ii): Find the distance from A to B.** - The distance is the area under the velocity-time graph. - Area from $0$ to $30$ s (rectangle): $10 \times 30 = 300$ m. - Area from $30$ to $42$ s (triangle): $\frac{1}{2} \times 12 \times 10 = 60$ m. - Total distance $= 300 + 60 = 360$ m. **Final answer for problem 12:** Distance from A to B is $360$ m. 3. **Problem 13 (i): Sketch the velocity-time graph for the particle's motion.** - Accelerates from 0 to 20 m/s in 15 s (linear increase). - Moves at constant velocity 20 m/s for time $x$. - Decelerates uniformly to rest in time $t_d$. - Given $x = 4 t_d$. 4. **Problem 13 (ii): Find the total time for which the particle is moving.** - Displacement $= 480$ m. - Displacement during acceleration: $$s_1 = \frac{1}{2} \times 15 \times 20 = 150 \text{ m}$$ - Displacement during constant velocity: $$s_2 = 20 \times x = 20 \times 4 t_d = 80 t_d$$ - Displacement during deceleration: $$s_3 = \frac{1}{2} \times t_d \times 20 = 10 t_d$$ - Total displacement: $$s_1 + s_2 + s_3 = 150 + 80 t_d + 10 t_d = 150 + 90 t_d = 480$$ - Solve for $t_d$: $$90 t_d = 330 \Rightarrow t_d = \frac{330}{90} = \frac{11}{3} = 3.67 \text{ s}$$ - Total time: $$15 + x + t_d = 15 + 4 t_d + t_d = 15 + 5 t_d = 15 + 5 \times 3.67 = 15 + 18.33 = 33.33 \text{ s}$$ **Final answer for problem 13:** Total time is approximately $33.33$ s. 5. **Problem 14 (i): Sketch velocity-time graphs for M and C.** - M accelerates from 0 at $3$ m/s$^2$ for $8$ s, reaching velocity: $$v_M = 3 \times 8 = 24 \text{ m/s}$$ - Then M maintains velocity $24$ m/s. - C accelerates from 0 to 30 m/s in 20 s, then maintains 30 m/s. 6. **Problem 14 (ii): Find the distance of the pedestrian from the road junction where C passes M.** - Let time when C passes M be $t > 8$ s. - Position of M at time $t$: $$s_M = \text{distance during acceleration} + \text{distance at constant velocity}$$ $$= \frac{1}{2} \times 3 \times 8^2 + 24 (t - 8) = 96 + 24 (t - 8)$$ - Position of C at time $t$: $$s_C = \text{distance during acceleration} + \text{distance at constant velocity}$$ $$= \frac{1}{2} \times \frac{30}{20} \times 20^2 + 30 (t - 20) = 300 + 30 (t - 20)$$ - Set $s_M = s_C$: $$96 + 24 (t - 8) = 300 + 30 (t - 20)$$ $$96 + 24 t - 192 = 300 + 30 t - 600$$ $$24 t - 96 = 30 t - 300$$ $$-96 + 300 = 30 t - 24 t$$ $$204 = 6 t \Rightarrow t = 34 \text{ s}$$ - Distance from junction: $$s_M = 96 + 24 (34 - 8) = 96 + 24 \times 26 = 96 + 624 = 720 \text{ m}$$ **Final answer for problem 14:** The pedestrian is 720 m from the road junction. 7. **Example 2: Find velocity from displacement function $S = t^2 + 2 \sin t$.** - Velocity is derivative of displacement: $$v = \frac{dS}{dt} = 2 t + 2 \cos t$$ - Initial velocity at $t=0$: $$v(0) = 2 \times 0 + 2 \times \cos 0 = 0 + 2 = 2 \text{ m/s}$$ - Velocity at $t = \frac{\pi}{3}$: $$v\left(\frac{\pi}{3}\right) = 2 \times \frac{\pi}{3} + 2 \times \cos \frac{\pi}{3} = \frac{2 \pi}{3} + 2 \times \frac{1}{2} = \frac{2 \pi}{3} + 1 \approx 3.09 + 1 = 4.09 \text{ m/s}$$