1. **Problem 1:** Determine when the object passes through its initial position again using the velocity vs. time graph. 2. The position of the object is the integral (area under the velocity-time curve) of velocity over time. 3. The object starts at position zero at $t=0$. 4. To find when it returns to the initial position, find when the net displacement (area under the velocity curve) is zero again. 5. Calculate the area under the velocity curve from $t=0$ to $t=1$ s: - From $0$ to $0.5$ s, velocity is negative, area is approximately a triangle with base $0.5$ s and height $-1$ m/s, area $= \frac{1}{2} \times 0.5 \times (-1) = -0.25$ m. - From $0.5$ to $1$ s, velocity goes from $-1$ m/s to $0$ m/s, area is another triangle with base $0.5$ s and height $-1$ m/s, area $= \frac{1}{2} \times 0.5 \times (-1) = -0.25$ m. - Total area from $0$ to $1$ s is $-0.25 + (-0.25) = -0.5$ m. 6. From $1$ to $2$ s, velocity increases from $0$ to $2$ m/s, area is a triangle with base $1$ s and height $2$ m/s, area $= \frac{1}{2} \times 1 \times 2 = 1$ m. 7. Net displacement at $t=2$ s is $-0.5 + 1 = 0.5$ m (positive). 8. From $2$ to $3$ s, velocity is constant at $2$ m/s, area is a rectangle with base $1$ s and height $2$ m/s, area $= 2$ m. 9. Net displacement at $t=3$ s is $0.5 + 2 = 2.5$ m. 10. From $3$ to $4$ s, velocity decreases linearly from $2$ m/s to $0$ m/s, area is a triangle with base $1$ s and height $2$ m/s, area $= \frac{1}{2} \times 1 \times 2 = 1$ m. 11. Net displacement at $t=4$ s is $2.5 + 1 = 3.5$ m. 12. Since the net displacement never returns to zero after $t=0$, the object does not pass through its initial position again between $0$ and $4$ s. 13. However, the velocity crosses zero at $t=1$ s, indicating a change in direction. 14. The object passes through its initial position again between $1$ and $2$ seconds because the displacement changes sign. **Answer for Question 1:** B. Between 1 and 2 s --- 15. **Problem 2:** Calculate the distance traveled by a car accelerating uniformly from rest to $20.0$ m/s in $15.0$ s. 16. Use the formula for distance under uniform acceleration starting from rest: $$ d = \frac{1}{2} a t^2 $$ 17. First, find acceleration $a$ using: $$ a = \frac{v}{t} = \frac{20.0}{15.0} = \frac{4}{3} \approx 1.333\, m/s^2 $$ 18. Calculate distance: $$ d = \frac{1}{2} \times 1.333 \times (15.0)^2 = 0.6667 \times 225 = 150 \text{ m} $$ **Answer for Question 2:** A. 150 m