Find the position vector for a particle with acceleration, initial velocity, and initial position given below. \( \vec{a}(t)=\langle 5 t, 4 \sin (t), \cos (2 t)\rangle \) \( \vec{v}(0)=\langle-3,0,0\rangle \) \( \vec{r}(0)=\langle 1,-5,-5\rangle \) \( \vec{r}(t)= \)

To find the position vector \(\vec{r}(t)\), we first need to integrate the acceleration vector \(\vec{a}(t)\) to get the velocity vector \(\vec{v}(t)\). The acceleration vector is given as \(\vec{a}(t)=\langle 5 t, 4 \sin (t), \cos (2 t)\rangle\). Integrating the components of \(\vec{a}(t)\): 1. For the \(x\)-component: \[ v_x(t) = \int 5t \, dt = \frac{5}{2}t^2 + C_1 \] 2. For the \(y\)-component: \[ v_y(t) = \int 4 \sin(t) \, dt = -4 \cos(t) + C_2 \] 3. For the \(z\)-component: \[ v_z(t) = \int \cos(2t) \, dt = \frac{1}{2} \sin(2t) + C_3 \] Now we apply the initial velocity condition, \(\vec{v}(0) = \langle -3, 0, 0 \rangle\): 1. Setting \(t = 0\) in \(v_x(t)\): \[ -3 = \frac{5}{2}(0)^2 + C_1 \implies C_1 = -3 \] Thus, \(v_x(t) = \frac{5}{2}t^2 - 3\). 2. Setting \(t = 0\) in \(v_y(t)\): \[ 0 = -4 \cos(0) + C_2 \implies C_2 = 4 \] Hence, \(v_y(t) = -4 \cos(t) + 4\). 3. Setting \(t = 0\) in \(v_z(t)\): \[ 0 = \frac{1}{2} \sin(0) + C_3 \implies C_3 = 0 \] So, \(v_z(t) = \frac{1}{2} \sin(2t)\). Now, we obtain \(\vec{v}(t) = \left\langle \frac{5}{2} t^2 - 3, -4 \cos(t) + 4, \frac{1}{2} \sin(2t) \right\rangle\). Next, we integrate \(\vec{v}(t)\) to get \(\vec{r}(t)\): 1. Integrating \(v_x(t)\): \[ x(t) = \int \left(\frac{5}{2}t^2 - 3\right) dt = \frac{5}{6}t^3 - 3t + D_1 \] 2. Integrating \(v_y(t)\): \[ y(t) = \int \left(-4 \cos(t) + 4\right) dt = -4 \sin(t) + 4t + D_2 \] 3. Integrating \(v_z(t)\): \[ z(t) = \int \left(\frac{1}{2} \sin(2t)\right) dt = -\frac{1}{4} \cos(2t) + D_3 \] Now apply the initial position \(\vec{r}(0) = \langle 1, -5, -5 \rangle\): 1. At \(t = 0\) for \(x(t)\): \[ 1 = \frac{5}{6}(0)^3 - 3(0) + D_1 \implies D_1 = 1 \] Thus, \(x(t) = \frac{5}{6}t^3 - 3t + 1\). 2. At \(t = 0\) for \(y(t)\): \[ -5 = -4 \sin(0) + 4(0) + D_2 \implies D_2 = -5 \] So, \(y(t) = -4 \sin(t) + 4t - 5\). 3. At \(t = 0\) for \(z(t)\): \[ -5 = -\frac{1}{4} \cos(0) + D_3 \implies D_3 = -5 + \frac{1