A ship is moving at a constant speed of `10 km//hr` in the direction of the unit vector `hat(i)`. Initially. Its position vector, relative to a fixed origin is `10(-hat(i)+hat(j))` where `hat(i)` & `hat(j)` are perpendicular vectors of length 1 km. Find its position vector relative to the origin at time t hours later. A second ship is moving with constant speed `u km//hr` parallele to the vector `hat(i)+2hat(j)` and is initially at the origin
(i) If `u=10 sqrt(5) km//h`. Find the minimum distance between the ships and the corresponding value of t
(ii) Find the value of u for which the shipd are on a collision course and determine the value of t at which the collision would occur if no avoiding action were taken.

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part (i): Finding the Minimum Distance Between the Ships

1. **Determine the Position Vectors of the Ships**:
- The first ship is moving with a constant speed of \(10 \text{ km/hr}\) in the direction of the unit vector \(\hat{i}\). Its initial position vector is given as:
\[ \mathbf{r}_1(0) = 10(-\hat{i} + \hat{j}) = -10\hat{i} + 10\hat{j} \]
Therefore, at time \(t\) hours, its position vector is:
\[ \mathbf{r}_1(t) = \mathbf{r}_1(0) + \mathbf{v}_1 t = (-10\hat{i} + 10\hat{j}) + (10\hat{i})t = (-10 + 10t)\hat{i} + 10\hat{j} \]

- The second ship is moving with a speed \(u = 10\sqrt{5} \text{ km/hr}\) in the direction of the vector \(\hat{i} + 2\hat{j}\). The unit vector in this direction is:
\[ \mathbf{u} = \frac{1}{\sqrt{5}}(\hat{i} + 2\hat{j}) \]
Thus, the velocity vector of the second ship is:
\[ \mathbf{v}_2 = 10\sqrt{5} \cdot \frac{1}{\sqrt{5}}(\hat{i} + 2\hat{j}) = 10(\hat{i} + 2\hat{j}) \]
Its position vector at time \(t\) is:
\[ \mathbf{r}_2(t) = \mathbf{r}_2(0) + \mathbf{v}_2 t = 0 + 10(\hat{i} + 2\hat{j})t = 10t\hat{i} + 20t\hat{j} \]

2. **Calculate the Relative Position Vector**:
- The relative position vector between the two ships is:
\[ \Delta \mathbf{r}(t) = \mathbf{r}_2(t) - \mathbf{r}_1(t) = (10t\hat{i} + 20t\hat{j}) - ((-10 + 10t)\hat{i} + 10\hat{j}) \]
Simplifying this gives:
\[ \Delta \mathbf{r}(t) = (10t + 10 - 10t)\hat{i} + (20t - 10)\hat{j} = 10\hat{i} + (20t - 10)\hat{j} \]

3. **Find the Magnitude of the Relative Position Vector**:
- The distance \(d(t)\) between the two ships is given by the magnitude of \(\Delta \mathbf{r}(t)\):
\[ d(t) = \sqrt{(10)^2 + (20t - 10)^2} = \sqrt{100 + (20t - 10)^2} \]
Expanding the square:
\[ d(t) = \sqrt{100 + 400t^2 - 400t + 100} = \sqrt{400t^2 - 400t + 200} \]

4. **Minimize the Distance**:
- To find the minimum distance, we can minimize \(d(t)^2\):
\[ f(t) = 400t^2 - 400t + 200 \]
- Taking the derivative and setting it to zero:
\[ f'(t) = 800t - 400 = 0 \implies t = \frac{1}{2} \text{ hours} \]

5. **Calculate the Minimum Distance**:
- Substitute \(t = \frac{1}{2}\) into \(d(t)\):
\[ d\left(\frac{1}{2}\right) = \sqrt{400\left(\frac{1}{2}\right)^2 - 400\left(\frac{1}{2}\right) + 200} = \sqrt{100 - 200 + 200} = \sqrt{100} = 10 \text{ km} \]

### Part (ii): Finding the Value of \(u\) for Collision

1. **Set Up the Collision Condition**:
- For a collision, the position vectors must be equal at some time \(t\):
\[ \mathbf{r}_1(t) = \mathbf{r}_2(t) \]
This gives the equations:
\[ -10 + 10t = ut \quad \text{(1)} \]
\[ 10 = 20t \quad \text{(2)} \]

2. **Solve for \(t\)**:
- From equation (2):
\[ t = \frac{10}{20} = \frac{1}{2} \text{ hours} \]

3. **Substitute \(t\) into Equation (1)**:
- Substitute \(t = \frac{1}{2}\) into equation (1):
\[ -10 + 10\left(\frac{1}{2}\right) = u\left(\frac{1}{2}\right) \]
Simplifying gives:
\[ -10 + 5 = \frac{u}{2} \implies -5 = \frac{u}{2} \implies u = -10 \text{ km/hr} \]
- Since speed cannot be negative, we conclude that the ships cannot collide under the given conditions.

### Final Answers:
- **Minimum Distance**: \(10 \text{ km}\) at \(t = \frac{1}{2} \text{ hours}\).
- **Collision Value of \(u\)**: No valid positive speed for collision.