2. A particle moves along the curve \[ \begin{array}{l}x=2 t^{2}, \quad y=t^{2}-4 t \quad \text { and } \quad z=3 t-5 \\ \text { where } t \text { is the time. Find the components of its velocity and acceleration at time } t=1 \text {, in the } \\ \text { direction } \hat{i}-3 \hat{j}+2 \hat{k} \text {. } \\ \text { (Nagpur, Summer 200l) Ans. } \frac{8 \sqrt{14}}{7},-\frac{\sqrt{14}}{7}\end{array} \]

To determine the components of the particle's velocity and acceleration at time \( t = 1 \) in the direction of the vector \( \mathbf{d} = \mathbf{\hat{i}} - 3\mathbf{\hat{j}} + 2\mathbf{\hat{k}} \), follow these steps: ### 1. **Determine the Position Vector** Given the parametric equations: \[ \begin{cases} x = 2t^2 \\ y = t^2 - 4t \\ z = 3t - 5 \end{cases} \] The position vector \( \mathbf{r}(t) \) is: \[ \mathbf{r}(t) = 2t^2\,\mathbf{\hat{i}} + (t^2 - 4t)\,\mathbf{\hat{j}} + (3t - 5)\,\mathbf{\hat{k}} \] ### 2. **Find the Velocity Vector \( \mathbf{v}(t) \)** Differentiate \( \mathbf{r}(t) \) with respect to time \( t \): \[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(2t^2)\,\mathbf{\hat{i}} + \frac{d}{dt}(t^2 - 4t)\,\mathbf{\hat{j}} + \frac{d}{dt}(3t - 5)\,\mathbf{\hat{k}} \] \[ \mathbf{v}(t) = 4t\,\mathbf{\hat{i}} + (2t - 4)\,\mathbf{\hat{j}} + 3\,\mathbf{\hat{k}} \] At \( t = 1 \): \[ \mathbf{v}(1) = 4(1)\,\mathbf{\hat{i}} + (2(1) - 4)\,\mathbf{\hat{j}} + 3\,\mathbf{\hat{k}} = 4\,\mathbf{\hat{i}} - 2\,\mathbf{\hat{j}} + 3\,\mathbf{\hat{k}} \] ### 3. **Find the Acceleration Vector \( \mathbf{a}(t) \)** Differentiate \( \mathbf{v}(t) \) with respect to time \( t \): \[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(4t)\,\mathbf{\hat{i}} + \frac{d}{dt}(2t - 4)\,\mathbf{\hat{j}} + \frac{d}{dt}(3)\,\mathbf{\hat{k}} \] \[ \mathbf{a}(t) = 4\,\mathbf{\hat{i}} + 2\,\mathbf{\hat{j}} + 0\,\mathbf{\hat{k}} = 4\,\mathbf{\hat{i}} + 2\,\mathbf{\hat{j}} \] At \( t = 1 \): \[ \mathbf{a}(1) = 4\,\mathbf{\hat{i}} + 2\,\mathbf{\hat{j}} \] ### 4. **Compute the Components in the Direction of \( \mathbf{d} = \mathbf{\hat{i}} - 3\mathbf{\hat{j}} + 2\mathbf{\hat{k}} \)** First, find the unit vector in the direction of \( \mathbf{d} \): \[ \|\mathbf{d}\| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] \[ \mathbf{u} = \frac{\mathbf{d}}{\|\mathbf{d}\|} = \frac{1}{\sqrt{14}}\,\mathbf{\hat{i}} - \frac{3}{\sqrt{14}}\,\mathbf{\hat{j}} + \frac{2}{\sqrt{14}}\,\mathbf{\hat{k}} \] **a. Component of Velocity \( \mathbf{v}(1) \) in the Direction of \( \mathbf{d} \):** \[ \text{Component} = \mathbf{v}(1) \cdot \mathbf{u} = \left(4\,\mathbf{\hat{i}} - 2\,\mathbf{\hat{j}} + 3\,\mathbf{\hat{k}}\right) \cdot \left(\frac{1}{\sqrt{14}}\,\mathbf{\hat{i}} - \frac{3}{\sqrt{14}}\,\mathbf{\hat{j}} + \frac{2}{\sqrt{14}}\,\mathbf{\hat{k}}\right) \] \[ = \frac{4 \times 1 + (-2) \times (-3) + 3 \times 2}{\sqrt{14}} = \frac{4 + 6 + 6}{\sqrt{14}} = \frac{16}{\sqrt{14}} = \frac{16\sqrt{14}}{14} = \frac{8\sqrt{14}}{7} \] **b. Component of Acceleration \( \mathbf{a}(1) \) in the Direction of \( \mathbf{d} \):** \[ \text{Component} = \mathbf{a}(1) \cdot \mathbf{u} = \left(4\,\mathbf{\hat{i}} + 2\,\mathbf{\hat{j}}\right) \cdot \left(\frac{1}{\sqrt{14}}\,\mathbf{\hat{i}} - \frac{3}{\sqrt{14}}\,\mathbf{\hat{j}} + \frac{2}{\sqrt{14}}\,\mathbf{\hat{k}}\right) \] \[ = \frac{4 \times 1 + 2 \times (-3) + 0 \times 2}{\sqrt{14}} = \frac{4 - 6 + 0}{\sqrt{14}} = \frac{-2}{\sqrt{14}} = \frac{-2\sqrt{14}}{14} = -\frac{\sqrt{14}}{7} \] ### **Final Answer** - **Velocity component in the direction \( \mathbf{d} \):** \( \dfrac{8\sqrt{14}}{7} \) - **Acceleration component in the direction \( \mathbf{d} \):** \( -\dfrac{\sqrt{14}}{7} \) These results match the provided answer: \[ \frac{8 \sqrt{14}}{7}, \quad -\frac{\sqrt{14}}{7} \]