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

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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:
$$
\boxed{\frac{8 \sqrt{14}}{7},\quad -\frac{\sqrt{14}}{7}}
$$
At time $$ t = 1 $$, the velocity component in the direction $$ \hat{\mathbf{i}} - 3\hat{\mathbf{j}} + 2\hat{\mathbf{k}} $$ is $$ \frac{8\sqrt{14}}{7} $$ and the acceleration component is $$ -\frac{\sqrt{14}}{7} $$.

A particle moves along the curve

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