1. **Problem statement:** A particle starts from rest and travels 64 m in 4 s under constant acceleration. We need to find: - The time to travel half the total distance (32 m). - The velocity at that time. - The distance traveled in half the total time (2 s). - The velocity at that time. 2. **Known values:** - Initial velocity $u = 0$ (starts from rest) - Total distance $s = 64$ m - Total time $t = 4$ s 3. **Formula for distance under constant acceleration:** $$ s = ut + \frac{1}{2}at^2 $$ Since $u=0$, this simplifies to: $$ s = \frac{1}{2}at^2 $$ 4. **Find acceleration $a$:** $$ 64 = \frac{1}{2}a(4)^2 $$ $$ 64 = 8a $$ $$ a = 8 \text{ m/s}^2 $$ 5. **Find time to travel half the distance (32 m):** Using the same formula: $$ 32 = \frac{1}{2} \times 8 \times t^2 $$ $$ 32 = 4t^2 $$ $$ t^2 = 8 $$ $$ t = \sqrt{8} = 2\sqrt{2} \approx 2.83 \text{ s} $$ 6. **Velocity at $t = 2.83$ s:** Velocity under constant acceleration: $$ v = u + at = 0 + 8 \times 2.83 = 22.64 \text{ m/s} $$ 7. **Distance traveled in half the total time (2 s):** $$ s = \frac{1}{2} \times 8 \times (2)^2 = 4 \times 4 = 16 \text{ m} $$ 8. **Velocity at $t = 2$ s:** $$ v = 0 + 8 \times 2 = 16 \text{ m/s} $$ **Final answers:** - Time to travel 32 m: $2\sqrt{2} \approx 2.83$ s - Velocity at 32 m: 22.64 m/s - Distance in 2 s: 16 m - Velocity at 2 s: 16 m/s