1. **Problem statement:** An automobile starts from rest, accelerates at $4\ \text{m/s}^2$ to a speed of $40\ \text{m/s}$, travels at this constant speed for some time, then decelerates at $5\ \text{m/s}^2$ to rest. The total distance traveled is $1000$ meters. We need to find: a. Distance covered during acceleration b. Distance traveled at constant speed c. Total time of travel 2. **Formulas and rules:** - Distance during acceleration with constant acceleration $a$ from rest to speed $v$: $$d = \frac{v^2}{2a}$$ - Distance during deceleration from speed $v$ to rest with deceleration $a$: $$d = \frac{v^2}{2a}$$ - Distance at constant speed $v$ for time $t$: $$d = vt$$ - Time during acceleration: $$t = \frac{v}{a}$$ - Time during deceleration: $$t = \frac{v}{a}$$ 3. **Calculate distance during acceleration:** $$d_1 = \frac{v^2}{2a} = \frac{40^2}{2 \times 4} = \frac{1600}{8} = 200\ \text{m}$$ 4. **Calculate distance during deceleration:** $$d_3 = \frac{v^2}{2a} = \frac{40^2}{2 \times 5} = \frac{1600}{10} = 160\ \text{m}$$ 5. **Calculate distance at constant speed:** Total distance $= 1000$ m, so $$d_2 = 1000 - d_1 - d_3 = 1000 - 200 - 160 = 640\ \text{m}$$ 6. **Calculate time during acceleration:** $$t_1 = \frac{v}{a} = \frac{40}{4} = 10\ \text{s}$$ 7. **Calculate time during deceleration:** $$t_3 = \frac{v}{a} = \frac{40}{5} = 8\ \text{s}$$ 8. **Calculate time at constant speed:** $$t_2 = \frac{d_2}{v} = \frac{640}{40} = 16\ \text{s}$$ 9. **Calculate total time:** $$t_{total} = t_1 + t_2 + t_3 = 10 + 16 + 8 = 34\ \text{s}$$ **Final answers:** - Distance during acceleration: $200$ meters - Distance at constant speed: $640$ meters - Total time of travel: $34$ seconds