Answer
\[ \begin{align*} a &= -\mathbf{j}, \\ a_T &= -3\mathbf{j}, \\ a_N &= \mathbf{i} \end{align*} \]
Work Step by Step
Step 1 Given: \[ \begin{align*} v &= -4\mathbf{j}, \\ a &= 2\mathbf{i} + 3\mathbf{j} \end{align*} \] Step 2 Since the scalar tangential and normal components of acceleration given by: \[ \begin{align*} a_T &= \frac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\|} = -\frac{12}{4} = -3, \\ a_N &= \frac{\|\mathbf{v}\| (\mathbf{v} \times \mathbf{a})}{\|\mathbf{v}\|} = \frac{1}{4} \left\| \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -4 & 0 \\ 2 & 3 & 0 \end{vmatrix} \right\| = \frac{1}{4} \|8\mathbf{k}\| = 2. \end{align*} \] Step 3 Since: \[ a = \frac{\mathbf{v}}{\|\mathbf{v}\|} = -\mathbf{j} \] Then: \[ a_T = 3\mathbf{j} \] To find the normal vector component of acceleration, we rewrite: \[ a = a_T - a_N \quad \text{as} \quad a_N = a - a_T \] The normal vector component of acceleration is: \[ a_N = 2\mathbf{i} + 3\mathbf{j} - 3\mathbf{j} = \mathbf{i} \] Result \[ \begin{align*} a &= -\mathbf{j}, \\ a_T &= -3\mathbf{j}, \\ a_N &= \mathbf{i} \end{align*} \]
