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 of finding the components of the velocity and acceleration vectors in the direction of the vector $$\mathbf{d} = \hat{i} - 3\hat{j} + 2\hat{k}$$ at time $$t = 1$$, we'll follow these steps:

1. **Determine the Position Vector $$\mathbf{r}(t)$$**
   
   Given:
   $$
   x = 2t^2, \quad y = t^2 - 4t, \quad z = 3t - 5
   $$
   The position vector is:
   $$
   \mathbf{r}(t) = 2t^2\, \hat{i} + (t^2 - 4t)\, \hat{j} + (3t - 5)\, \hat{k}
   $$

2. **Find the Velocity Vector $$\mathbf{v}(t)$$**
   
   The velocity vector is the first derivative of the position vector with respect to time $$t$$:
   $$
   \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = 4t\, \hat{i} + (2t - 4)\, \hat{j} + 3\, \hat{k}
   $$
   
   At $$t = 1$$:
   $$
   \mathbf{v}(1) = 4(1)\, \hat{i} + (2(1) - 4)\, \hat{j} + 3\, \hat{k} = 4\, \hat{i} - 2\, \hat{j} + 3\, \hat{k}
   $$

3. **Find the Acceleration Vector $$\mathbf{a}(t)$$**
   
   The acceleration vector is the derivative of the velocity vector with respect to time $$t$$:
   $$
   \mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = 4\, \hat{i} + 2\, \hat{j} + 0\, \hat{k}
   $$
   
   At $$t = 1$$:
   $$
   \mathbf{a}(1) = 4\, \hat{i} + 2\, \hat{j} + 0\, \hat{k}
   $$

4. **Determine the Unit Vector in the Direction of $$\mathbf{d}$$**
   
   Given $$\mathbf{d} = \hat{i} - 3\hat{j} + 2\hat{k}$$, the magnitude of $$\mathbf{d}$$ is:
   $$
   |\mathbf{d}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}
   $$
   
   Therefore, the unit vector $$\mathbf{\hat{d}}$$ in the direction of $$\mathbf{d}$$ is:
   $$
   \mathbf{\hat{d}} = \frac{1}{\sqrt{14}}\, \hat{i} - \frac{3}{\sqrt{14}}\, \hat{j} + \frac{2}{\sqrt{14}}\, \hat{k}
   $$

5. **Compute the Components of Velocity and Acceleration in the Direction of $$\mathbf{d}$$**
   
   The component of a vector $$\mathbf{A}$$ in the direction of $$\mathbf{\hat{d}}$$ is given by the dot product:
   $$
   \text{Component} = \mathbf{A} \cdot \mathbf{\hat{d}}
   $$
   
   - **Velocity Component:**
     $$
     \mathbf{v}(1) \cdot \mathbf{\hat{d}} = (4)(1) + (-2)(-3) + (3)(2) = 4 + 6 + 6 = 16
     $$
     $$
     \text{Velocity Component} = \frac{16}{\sqrt{14}} = \frac{8\sqrt{14}}{7}
     $$
   
   - **Acceleration Component:**
     $$
     \mathbf{a}(1) \cdot \mathbf{\hat{d}} = (4)(1) + (2)(-3) + (0)(2) = 4 - 6 + 0 = -2
     $$
     $$
     \text{Acceleration Component} = \frac{-2}{\sqrt{14}} = -\frac{\sqrt{14}}{7}
     $$

**Final Answer:**
- **Component of Velocity:** $$\dfrac{8\, \sqrt{14}}{7}$$
- **Component of Acceleration:** $$-\dfrac{\sqrt{14}}{7}$$
At time $$ t = 1 $$, the component of the velocity in the direction of $$\hat{i} - 3\hat{j} + 2\hat{k}$$ is $$\frac{8\sqrt{14}}{7}$$, and the component of the acceleration is $$-\frac{\sqrt{14}}{7}$$.

A particle moves along the curve

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