1. **State the problem:** We have a car with initial velocity $u = 20$ m/s. - It accelerates at $a = 2.5$ m/s² for $4$ seconds. - Then it travels at constant velocity $V$ for $7$ seconds. - Finally, it decelerates to rest over time $T - 11$ seconds. - Total journey time is $40$ seconds. - Total distance traveled is $1090$ m. We need to: (i) Sketch the velocity-time graph. (ii) Find $V$. (iii) Find $T$. 2. **Velocity after acceleration:** Using formula $v = u + at$ for the first 4 seconds: $$v = 20 + 2.5 \times 4 = 20 + 10 = 30 \text{ m/s}$$ So, $V = 30$ m/s (velocity during constant speed phase). 3. **Total time and deceleration time:** Total time is $40$ seconds. Time at constant velocity is $7$ seconds (from 4 to 11 seconds). Deceleration starts at $t=11$ seconds and ends at $t=T$ seconds. Since total time is $40$ seconds, deceleration time is: $$T - 11 = 40 - 11 = 29 \text{ seconds}$$ So, $T = 40$ seconds. 4. **Distance traveled during each phase:** - Phase 1 (acceleration): Using $s = ut + \frac{1}{2}at^2$ for $t=4$ s: $$s_1 = 20 \times 4 + \frac{1}{2} \times 2.5 \times 4^2 = 80 + 20 = 100 \text{ m}$$ - Phase 2 (constant velocity): $$s_2 = V \times 7 = 30 \times 7 = 210 \text{ m}$$ - Phase 3 (deceleration): Initial velocity $u = 30$ m/s, final velocity $0$ m/s, time $29$ s. Deceleration $a_d$ is: $$a_d = \frac{0 - 30}{29} = -\frac{30}{29} \approx -1.0345 \text{ m/s}^2$$ Distance during deceleration: Using $s = ut + \frac{1}{2}at^2$: $$s_3 = 30 \times 29 + \frac{1}{2} \times (-1.0345) \times 29^2$$ Calculate: $$s_3 = 870 - 0.51725 \times 841 = 870 - 435 = 435 \text{ m (approx)}$$ 5. **Check total distance:** $$s_{total} = s_1 + s_2 + s_3 = 100 + 210 + 435 = 745 \text{ m}$$ This is less than $1090$ m, so we must re-examine the problem. 6. **Recalculate with unknown $T$:** Let deceleration time be $t_d = T - 11$. Total time $T = 40$ s, so $t_d = 29$ s. Distance during deceleration: $$s_3 = V t_d - \frac{1}{2} \times \frac{V}{t_d} \times t_d^2 = V t_d - \frac{1}{2} V t_d = \frac{1}{2} V t_d$$ Since deceleration is uniform from $V$ to $0$ in $t_d$ seconds. 7. **Total distance formula:** $$s_{total} = s_1 + s_2 + s_3 = 100 + 7V + \frac{1}{2} V (T - 11)$$ Given $s_{total} = 1090$ m and $T = 40$ s: $$1090 = 100 + 7V + \frac{1}{2} V (40 - 11) = 100 + 7V + \frac{1}{2} V \times 29 = 100 + 7V + 14.5V = 100 + 21.5V$$ 8. **Solve for $V$:** $$1090 - 100 = 21.5V$$ $$990 = 21.5V$$ $$V = \frac{990}{21.5} \approx 46.05 \text{ m/s}$$ 9. **Recalculate acceleration phase velocity:** Velocity after acceleration phase is: $$v = 20 + 2.5 \times 4 = 30 \text{ m/s}$$ But $V$ is $46.05$ m/s, so the constant velocity phase must be $46.05$ m/s. 10. **Recalculate acceleration time to reach $V$:** Since acceleration is $2.5$ m/s², time to reach $V$ from $20$ m/s: $$t_a = \frac{V - 20}{2.5} = \frac{46.05 - 20}{2.5} = 10.42 \text{ s}$$ But problem states acceleration lasts 4 seconds, so acceleration phase is fixed. 11. **Adjust problem understanding:** Acceleration phase is fixed 4 seconds, velocity at end is 30 m/s. Constant velocity phase is 7 seconds at $V$ m/s. Deceleration phase lasts $T - 11$ seconds. Total time is 40 seconds, so $T = 40$. 12. **Find $V$ from total distance:** Distance during acceleration: $$s_1 = 20 \times 4 + \frac{1}{2} \times 2.5 \times 4^2 = 100 \text{ m}$$ Distance during constant velocity: $$s_2 = V \times 7$$ Distance during deceleration: Deceleration $a_d = -\frac{V}{T-11} = -\frac{V}{29}$ Distance: $$s_3 = V \times 29 + \frac{1}{2} \times a_d \times 29^2 = 29V - \frac{1}{2} \times \frac{V}{29} \times 29^2 = 29V - \frac{1}{2} V \times 29 = 29V - 14.5V = 14.5V$$ Total distance: $$1090 = 100 + 7V + 14.5V = 100 + 21.5V$$ Solve for $V$: $$21.5V = 990$$ $$V = 46.05 \text{ m/s}$$ 13. **Find $T$:** Given total time is 40 seconds, so $T = 40$ seconds. 14. **Summary:** - Velocity after acceleration phase is 30 m/s. - Constant velocity $V = 46.05$ m/s. - Deceleration time $T - 11 = 29$ seconds. - Total time $T = 40$ seconds. 15. **Velocity-time graph:** - From $0$ to $4$ s: velocity increases linearly from $20$ to $30$ m/s. - From $4$ to $11$ s: velocity is constant at $46.05$ m/s. - From $11$ to $40$ s: velocity decreases linearly from $46.05$ m/s to $0$. **Final answers:** - $V = 46.05$ m/s - $T = 40$ seconds