Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation a(t) \( =v^{\prime}(\mathrm{t})=\mathrm{g} \), where \( \mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^{2} \). a. \( \mathrm{v}(\mathrm{t})=-9.8 \mathrm{t}+33 \) b. \( \mathrm{s}(\mathrm{t})=\square \)

To determine the displacement \( s(t) \) of the object over time, we'll integrate the velocity function provided in part **a**. ### Given: - **Acceleration:** \( a(t) = v'(t) = g = -9.8 \, \text{m/s}^2 \) - **Velocity:** \( v(t) = -9.8t + 33 \, \text{m/s} \) ### Finding Displacement \( s(t) \): The displacement is the integral of the velocity function with respect to time: \[ s(t) = \int v(t) \, dt = \int (-9.8t + 33) \, dt \] Performing the integration: \[ s(t) = -\frac{9.8}{2} t^2 + 33t + C \] \[ s(t) = -4.9t^2 + 33t + C \] ### Determining the Constant \( C \): If the initial position \( s(0) \) is known (let's assume \( s(0) = 0 \) for simplicity), then: \[ s(0) = -4.9(0)^2 + 33(0) + C = 0 \implies C = 0 \] ### Final Displacement Function: \[ s(t) = -4.9t^2 + 33t \] **Answer:** \[ \mathrm{s}(t) = -4.9\, t^{2} +\, 33\, t \]