1. **Problem statement:** A car accelerates uniformly from rest for 20 seconds, moves at constant speed $v$ m/s for 40 seconds, then decelerates uniformly to rest in 10 seconds. Total distance traveled is 880 m. Find $v$. 2. **Speed-time graph:** - From 0 to 20 s: speed increases linearly from 0 to $v$ (uniform acceleration). - From 20 to 60 s: speed is constant at $v$. - From 60 to 70 s: speed decreases linearly from $v$ to 0 (uniform deceleration). 3. **Acceleration-time graph:** - From 0 to 20 s: acceleration is constant positive $a$. - From 20 to 60 s: acceleration is 0 (constant speed). - From 60 to 70 s: acceleration is constant negative $-a'$. 4. **Formulas and rules:** - Distance during acceleration: $$s_1 = \frac{1}{2} a t_1^2$$ where $t_1=20$ s. - Speed at end of acceleration: $$v = a t_1$$ - Distance during constant speed: $$s_2 = v t_2$$ where $t_2=40$ s. - Distance during deceleration: $$s_3 = \frac{1}{2} v t_3$$ where $t_3=10$ s (since speed decreases uniformly from $v$ to 0). - Total distance: $$s = s_1 + s_2 + s_3 = 880$$ m. 5. **Calculate distances:** - From $v = a t_1$, we get $a = \frac{v}{20}$. - Substitute into $s_1$: $$s_1 = \frac{1}{2} \times \frac{v}{20} \times 20^2 = \frac{1}{2} \times \frac{v}{20} \times 400 = 10 v$$ - Distance at constant speed: $$s_2 = v \times 40 = 40 v$$ - Distance during deceleration: $$s_3 = \frac{1}{2} v \times 10 = 5 v$$ 6. **Sum distances:** $$s = s_1 + s_2 + s_3 = 10 v + 40 v + 5 v = 55 v$$ 7. **Solve for $v$:** $$55 v = 880$$ $$v = \frac{880}{55} = 16$$ **Final answer:** The speed $v$ is $16$ m/s.