1. The problem provides values related to motion: acceleration $9.8\ \text{m/s}^2$, initial velocity $0\ \text{m/s}$, final velocity $-19.99\ \text{m/s}$, and displacement $-20.39\ \text{m}$. We want to analyze the motion using these values. 2. We use the kinematic equation relating velocities, acceleration, and displacement: $$v^2 = v_0^2 + 2a\Delta x$$ where $v$ is final velocity, $v_0$ is initial velocity, $a$ is acceleration, and $\Delta x$ is displacement. 3. Substitute the given values: $$(-19.99)^2 = 0^2 + 2 \times 9.8 \times \Delta x$$ 4. Simplify the left side: $$399.6 = 19.6 \times \Delta x$$ 5. Solve for $\Delta x$: $$\Delta x = \frac{399.6}{19.6}$$ 6. Show cancellation: $$\Delta x = \frac{\cancel{399.6}}{\cancel{19.6}} = 20.39\ \text{m}$$ 7. The displacement calculated matches the given displacement $-20.39\ \text{m}$, but with opposite sign. This indicates the object moved in the negative direction, consistent with the negative final velocity. 8. Summary: The object started at rest, accelerated at $9.8\ \text{m/s}^2$ in the negative direction, reaching a velocity of $-19.99\ \text{m/s}$ after moving $-20.39\ \text{m}$. Final answer: The displacement is $-20.39\ \text{m}$, confirming the motion parameters are consistent.