Given an acceleration vector, initial velocity \( \left\langle u_{0}, v_{0}\right\rangle \), and initial position \( \left\langle x_{0}, y_{0}\right\rangle \), find the velocity and position vectors for \( t \geq 0 \). \[ \mathbf{a}(\mathrm{t})=\langle 0,14\rangle,\left\langle u_{0}, v_{0}\right\rangle=\langle 0,7\rangle,\left\langle x_{0}, y_{0}\right\rangle=\langle 4,-1\rangle \] What is the velocity vector? \( \mathbf{v}(t)=\langle\square, \square \) \( \mathbf{r}(\mathrm{t})=\langle\square, \square \)

To find the velocity and position vectors given the acceleration vector, initial velocity, and initial position, we can follow these steps: 1. **Acceleration Vector**: \[ \mathbf{a}(t) = \langle 0, 14 \rangle \] 2. **Initial Velocity**: \[ \mathbf{v}(0) = \langle u_0, v_0 \rangle = \langle 0, 7 \rangle \] 3. **Initial Position**: \[ \mathbf{r}(0) = \langle x_0, y_0 \rangle = \langle 4, -1 \rangle \] ### Step 1: Find the Velocity Vector The velocity vector is found by integrating the acceleration vector with respect to time \( t \). \[ \mathbf{v}(t) = \int \mathbf{a}(t) \, dt + \mathbf{v}(0) \] Integrating the acceleration vector: \[ \mathbf{a}(t) = \langle 0, 14 \rangle \implies \int \langle 0, 14 \rangle \, dt = \langle 0, 14t \rangle + \mathbf{C} \] Where \( \mathbf{C} \) is a constant vector. Since we know the initial velocity: \[ \mathbf{v}(0) = \langle 0, 7 \rangle \] We can find \( \mathbf{C} \): \[ \mathbf{v}(0) = \langle 0, 14(0) \rangle + \mathbf{C} \implies \mathbf{C} = \langle 0, 7 \rangle \] Thus, the velocity vector becomes: \[ \mathbf{v}(t) = \langle 0, 14t \rangle + \langle 0, 7 \rangle = \langle 0, 14t + 7 \rangle \] ### Step 2: Find the Position Vector The position vector is found by integrating the velocity vector with respect to time \( t \). \[ \mathbf{r}(t) = \int \mathbf{v}(t) \, dt + \mathbf{r}(0) \] Integrating the velocity vector: \[ \mathbf{v}(t) = \langle 0, 14t + 7 \rangle \implies \int \langle 0, 14t + 7 \rangle \, dt = \langle 0, 7t + 7t^2/2 \rangle + \mathbf{D} \] Where \( \mathbf{D} \) is a constant vector. Since we know the initial position: \[ \mathbf{r}(0) = \langle 4, -1 \rangle \] We can find \( \mathbf{D} \): \[ \mathbf{r}(0) = \langle 0, 7(0) + \frac{14(0)^2}{2} \rangle + \mathbf{D} \implies \mathbf{D} = \langle 4, -1 \rangle \] Thus, the position vector becomes: \[ \mathbf{r}(t) = \langle 0, 7t + 7t^2/2 \rangle + \langle 4, -1 \rangle = \langle 4, 7t + \frac{14t^2}{2} - 1 \rangle \] ### Final Results The velocity vector is: \[ \mathbf{v}(t) = \langle 0, 14t + 7 \rangle \] The position vector is: \[ \mathbf{r}(t) = \langle 4, 7t + 7t^2 - 1 \rangle \] So, filling in the blanks: \[ \mathbf{v}(t) = \langle 0, 14t + 7 \rangle \] \[ \mathbf{r}(t) = \langle 4, 7t + 7t^2 - 1 \rangle \]