1. **State the problem:** A car starts from rest, accelerates to 10 m/s in 10 seconds, travels at this velocity for 20 seconds, then stops in 5 seconds. We need to find the initial acceleration, total distance traveled, and average velocity. 2. **Formulas and rules:** - Acceleration $a = \frac{\Delta v}{\Delta t}$ where $\Delta v$ is change in velocity and $\Delta t$ is change in time. - Distance during acceleration $d = v_i t + \frac{1}{2} a t^2$ where $v_i$ is initial velocity. - Distance at constant velocity $d = v t$. - Average velocity $v_{avg} = \frac{\text{total distance}}{\text{total time}}$. 3. **Calculate initial acceleration:** - Initial velocity $v_i = 0$ m/s (rest). - Final velocity $v_f = 10$ m/s. - Time $t = 10$ s. - Using $a = \frac{v_f - v_i}{t} = \frac{10 - 0}{10} = 1$ m/s$^2$. 4. **Calculate distance during acceleration:** - Using $d = v_i t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 1 \times 10^2 = 50$ m. 5. **Calculate distance at constant velocity:** - Velocity $v = 10$ m/s. - Time $t = 20$ s. - Distance $d = v t = 10 \times 20 = 200$ m. 6. **Calculate deceleration and distance during stopping:** - Initial velocity $v_i = 10$ m/s. - Final velocity $v_f = 0$ m/s. - Time $t = 5$ s. - Deceleration $a = \frac{0 - 10}{5} = -2$ m/s$^2$. - Distance $d = v_i t + \frac{1}{2} a t^2 = 10 \times 5 + \frac{1}{2} \times (-2) \times 5^2 = 50 - 25 = 25$ m. 7. **Calculate total distance:** - Total distance $= 50 + 200 + 25 = 275$ m. 8. **Calculate total time:** - Total time $= 10 + 20 + 5 = 35$ s. 9. **Calculate average velocity:** - $v_{avg} = \frac{275}{35} \approx 7.86$ m/s. **Final answers:** - Initial acceleration: $1$ m/s$^2$. - Total distance traveled: $275$ m. - Average velocity: approximately $7.86$ m/s.