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 solve the problem, we'll determine the components of the particle's velocity and acceleration at time \( t = 1 \) in the direction of the vector \( \mathbf{d} = \hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 2\hat{\mathbf{k}} \). ### Step 1: Determine the Velocity and Acceleration Vectors **Given:** \[ \begin{cases} x = 2t^2 \\ y = t^2 - 4t \\ z = 3t - 5 \end{cases} \] **Position Vector:** \[ \mathbf{r}(t) = 2t^2\,\hat{\mathbf{i}} + (t^2 - 4t)\,\hat{\mathbf{j}} + (3t - 5)\,\hat{\mathbf{k}} \] **Velocity Vector (\( \mathbf{v}(t) \)):** \[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(2t^2)\,\hat{\mathbf{i}} + \frac{d}{dt}(t^2 - 4t)\,\hat{\mathbf{j}} + \frac{d}{dt}(3t - 5)\,\hat{\mathbf{k}} = 4t\,\hat{\mathbf{i}} + (2t - 4)\,\hat{\mathbf{j}} + 3\,\hat{\mathbf{k}} \] **Acceleration Vector (\( \mathbf{a}(t) \)):** \[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(4t)\,\hat{\mathbf{i}} + \frac{d}{dt}(2t - 4)\,\hat{\mathbf{j}} + \frac{d}{dt}(3)\,\hat{\mathbf{k}} = 4\,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}} + 0\,\hat{\mathbf{k}} \] **At \( t = 1 \):** \[ \mathbf{v}(1) = 4(1)\,\hat{\mathbf{i}} + (2(1) - 4)\,\hat{\mathbf{j}} + 3\,\hat{\mathbf{k}} = 4\,\hat{\mathbf{i}} - 2\,\hat{\mathbf{j}} + 3\,\hat{\mathbf{k}} \] \[ \mathbf{a}(1) = 4\,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}} + 0\,\hat{\mathbf{k}} = 4\,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}} \] ### Step 2: Determine the Unit Vector in the Given Direction **Given Direction Vector:** \[ \mathbf{d} = \hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 2\hat{\mathbf{k}} \] **Magnitude of \( \mathbf{d} \):** \[ |\mathbf{d}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] **Unit Vector (\( \hat{\mathbf{u}} \)):** \[ \hat{\mathbf{u}} = \frac{\mathbf{d}}{|\mathbf{d}|} = \frac{1}{\sqrt{14}}\,\hat{\mathbf{i}} - \frac{3}{\sqrt{14}}\,\hat{\mathbf{j}} + \frac{2}{\sqrt{14}}\,\hat{\mathbf{k}} \] ### Step 3: Calculate the Components of Velocity and Acceleration in the Given Direction The component of a vector \( \mathbf{A} \) in the direction of \( \hat{\mathbf{u}} \) is given by the dot product \( \mathbf{A} \cdot \hat{\mathbf{u}} \). **Component of Velocity (\( v_{\text{component}} \)):** \[ \mathbf{v}(1) \cdot \hat{\mathbf{u}} = \left(4\,\hat{\mathbf{i}} - 2\,\hat{\mathbf{j}} + 3\,\hat{\mathbf{k}}\right) \cdot \left(\frac{1}{\sqrt{14}}\,\hat{\mathbf{i}} - \frac{3}{\sqrt{14}}\,\hat{\mathbf{j}} + \frac{2}{\sqrt{14}}\,\hat{\mathbf{k}}\right) \] \[ = \frac{4 \times 1}{\sqrt{14}} + \frac{(-2) \times (-3)}{\sqrt{14}} + \frac{3 \times 2}{\sqrt{14}} = \frac{4 + 6 + 6}{\sqrt{14}} = \frac{16}{\sqrt{14}} = \frac{8\sqrt{14}}{7} \] **Component of Acceleration (\( a_{\text{component}} \)):** \[ \mathbf{a}(1) \cdot \hat{\mathbf{u}} = \left(4\,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}}\right) \cdot \left(\frac{1}{\sqrt{14}}\,\hat{\mathbf{i}} - \frac{3}{\sqrt{14}}\,\hat{\mathbf{j}} + \frac{2}{\sqrt{14}}\,\hat{\mathbf{k}}\right) \] \[ = \frac{4 \times 1}{\sqrt{14}} + \frac{2 \times (-3)}{\sqrt{14}} + \frac{0 \times 2}{\sqrt{14}} = \frac{4 - 6 + 0}{\sqrt{14}} = \frac{-2}{\sqrt{14}} = -\frac{\sqrt{14}}{7} \] ### **Final Answer:** - **Velocity Component:** \( \frac{8\sqrt{14}}{7} \) - **Acceleration Component:** \( -\frac{\sqrt{14}}{7} \) These match the provided answer: \[ \boxed{\frac{8 \sqrt{14}}{7},\quad -\frac{\sqrt{14}}{7}} \]