1. **Stating the problem:** We have a velocity-time graph of a particle with three segments. The particle starts at velocity $5$ m/s at $t=0$ and accelerates at $0.75$ m/s² for $20$ s. Then it decelerates to velocity $0$ at $t=40$ s. After $t=40$ s, it moves back towards $O$, coming to rest at time $T$. We need to find the value of $T$. 2. **Calculate velocity at $t=20$ s:** Using $v = u + at$, where $u=5$, $a=0.75$, and $t=20$: $$v = 5 + 0.75 \times 20 = 5 + 15 = 20\text{ m/s}$$ 3. **Given info correction:** The problem states velocity reaches 15 m/s at 20 s, but calculation shows 20 m/s. Assuming problem's initial velocity is 5 m/s and acceleration 0.75 m/s², velocity at 20 s is indeed 20 m/s. We proceed with $v=20$ m/s at $t=20$ s. 4. **Deceleration phase from $t=20$ s to $t=40$ s:** At $t=40$, velocity is $0$. Initial velocity in this interval is $20$ m/s at $t=20$. Time interval $20$ s. Deceleration $a_d$: $$a_d = \frac{v - u}{t} = \frac{0 - 20}{20} = -1\text{ m/s}^2$$ 5. **Third phase from $t=40$ s to $t=T$ s:** Particle moves back towards $O$, velocity goes from $0$ at $t=40$ to $0$ at $t=T$ with negative velocity in between. This indicates acceleration reversing velocity back to zero. Velocity-time graph is a straight line from $(40,0)$ to $(T,0)$ passing below time axis. 6. **Calculate displacement during each phase to find $T$:** Displacement is area under velocity-time graph. - First phase displacement $(0 \to 20)$ s: Initial velocity $5$ m/s, final velocity $20$ m/s Area = trapezoid: $$s_1 = \frac{(5 + 20)}{2} \times 20 = 12.5 \times 20 = 250\text{ m}$$ - Second phase displacement $(20 \to 40)$ s: Velocity decreases from $20$ to $0$ Area = triangle: $$s_2 = \frac{1}{2} \times 20 \times 20 = 200\text{ m}$$ Total displacement up to $t=40$ s: $$s_{total} = s_1 + s_2 = 250 + 200 = 450\text{ m}$$ 7. **Third phase displacement $(40 \to T)$ s:** Velocity goes from $0$ to negative and back to $0$ in time $T - 40$. Area under velocity-time graph is negative area (particle moves back). This area equals $-450$ m to return to origin. Graph is triangle with base $T - 40$ and height $v_{min}$ (negative velocity magnitude). 8. **Find acceleration and velocity in third phase:** From $t=40$ to $t=T$, velocity changes from $0$ to negative max then back to $0$, symmetric straight lines. Acceleration magnitude is same for deceleration and acceleration phases (assumed constant). Slope from $t=40$ to $t=50$ is velocity with constant acceleration: The velocity at $t=50$ is given as the minimum negative peak. Compute it based on slope equal magnitude 1 m/s² but negative. Velocity at $t=50$: $$v = 0 + a (t - 40)$$ Assuming acceleration $a = -1$ m/s²: $$v(50) = 0 + (-1)(50 - 40) = -10\text{ m/s}$$ 9. **Calculate area from $40$ to $50$ s (triangle):** $$s_{40-50} = \frac{1}{2} \times 10 \times 10 = 50\text{ m}$$ (negative direction) 10. **From $50$ to $T$ velocity returns to zero:** It takes time $T - 50$ to go from $-10$ m/s to $0$ m/s with positive acceleration: Using $$0 = -10 + a (T - 50)$$ $$a = \frac{10}{T - 50}$$ Acceleration magnitude equals 1 m/s²: $$1 = \frac{10}{T - 50} \Rightarrow T - 50 = 10 \Rightarrow T = 60$$ 11. **Calculate displacement from $50$ to $T$: ** Area triangle with base $10$ s, height $-10$ m/s: $$s_{50-T} = \frac{1}{2} \times 10 \times 10 = 50\text{ m}$$ (negative) 12. **Total displacement for $40$ to $T$ is:** $$s_3 = s_{40-50} + s_{50-T} = -50 - 50 = -100\text{ m}$$ This contradicts displacement needed to return to $O$ (needs $-450$ m). 13. **Conclusion:** The negative displacement needed is $-450$ m, the last negative segment displacement is currently $-100$ m for $T=60$. The velocity magnitude or acceleration must be adjusted. 14. **Correct approach: The final motion is a straight line from $v=0$ at $t=40$ to $v=0$ at $t=T$ below time axis. Let velocity be linear with slope $a$ (negative). Area under velocity-time plot from $40$ to $T$ is: $$s_3 = \frac{1}{2} \times (T - 40) \times V_{min}$$ Where $V_{min}$ is the negative velocity minimum at midpoint $t = \frac{40 + T}{2}$. Velocity changes from $0$ at $t=40$ to $V_{min}$ at midpoint, then back to $0$ at $t=T$. The constant acceleration magnitude: $$a = \frac{V_{min}}{\frac{T - 40}{2}} = \frac{2 V_{min}}{T - 40}$$ 15. **Velocity returns to zero at $T$, so net displacement after $40$ must be $-450$ m:** $$s_3 = \frac{1}{2} (T - 40) V_{min} = -450$$ 16. **Express $a$ in terms of $V_{min}$ and $T$, but $a$ magnitude must equal 1 m/s² (from deceleration phase):** From deceleration phase, acceleration magnitude $=1$, so: $$|a| = \frac{2|V_{min}|}{T - 40} = 1$$ From this: $$|V_{min}| = \frac{T - 40}{2}$$ 17. **Use this to rewrite displacement:** $$-450 = \frac{1}{2} (T - 40)(-\frac{T - 40}{2}) = -\frac{(T - 40)^2}{4}$$ Multiply both sides by $-1$: $$450 = \frac{(T - 40)^2}{4}$$ Multiply both sides by $4$: $$1800 = (T - 40)^2$$ Take square root: $$T - 40 = \sqrt{1800} = 42.426$$ 18. **Find $T$: ** $$T = 40 + 42.426 = 82.426 \approx 82.4\text{ s}$$ **Final answer: ** $$\boxed{T = 82.4 \text{ seconds}}$$