1. **State the problem:** An object starts at position $s_0 = -75$ m with an initial velocity $v_0 = 20$ m/s and constant acceleration $a = -2$ m/s$^2$. We want to sketch the position-time graph for the first 20 seconds. 2. **Write the position function:** The position as a function of time $t$ under constant acceleration is given by: $$s(t) = s_0 + v_0 t + \frac{1}{2} a t^2$$ Substitute the values: $$s(t) = -75 + 20t + \frac{1}{2}(-2) t^2 = -75 + 20t - t^2$$ 3. **Analyze the function:** - Initial position at $t=0$ is $s(0) = -75$ m. - The velocity function is $v(t) = v_0 + at = 20 - 2t$. - The object moves upward initially because velocity is positive at $t=0$. - The velocity becomes zero at $t = 10$ s, which is the time when the object reaches its maximum position. 4. **Find the maximum position:** Set $v(t) = 0$: $$20 - 2t = 0 \implies t = 10$$ Calculate $s(10)$: $$s(10) = -75 + 20(10) - (10)^2 = -75 + 200 - 100 = 25$$ So, the maximum position is 25 m at 10 seconds. 5. **Find position at $t=20$ s:** $$s(20) = -75 + 20(20) - (20)^2 = -75 + 400 - 400 = -75$$ The object returns to the initial position after 20 seconds. 6. **Summary:** - The position-time graph is a downward-opening parabola starting at $-75$ m. - It rises to a maximum of 25 m at 10 s. - Then it falls back to $-75$ m at 20 s. This matches the description of the graph starting at $-75$ m, curving upward then downward due to negative acceleration.