A Ship A steams down north at a speed of 8 kmph. and ship B due west at a speed of 6 kmph. The velocity of A w.r.t. B is.

Options

A. 10 kmphm, `tan^(-1) (4/3)` N of E
B. 10 kmph, `tan^(-1) (4/3)` E of N
C. 10 kmph NE
D. 2 kmph, `tan^(-1)(4/3)`N of E

Answer

To find the velocity of ship A with respect to ship B, we can follow these steps:

### Step 1: Define the velocities of ships A and B
- Ship A is moving north at a speed of 8 km/h. We can represent this velocity as a vector:
\[ \vec{V_A} = 0 \hat{i} + 8 \hat{j} \text{ km/h} \]
- Ship B is moving west at a speed of 6 km/h. We can represent this velocity as a vector:
\[ \vec{V_B} = -6 \hat{i} + 0 \hat{j} \text{ km/h} \]

### Step 2: Calculate the velocity of A with respect to B
The velocity of A with respect to B (\(\vec{V_{A/B}}\)) is given by the formula:
\[ \vec{V_{A/B}} = \vec{V_A} - \vec{V_B} \]
Substituting the values we found:
\[ \vec{V_{A/B}} = (0 \hat{i} + 8 \hat{j}) - (-6 \hat{i} + 0 \hat{j}) \]
This simplifies to:
\[ \vec{V_{A/B}} = 0 \hat{i} + 8 \hat{j} + 6 \hat{i} = 6 \hat{i} + 8 \hat{j} \text{ km/h} \]

### Step 3: Calculate the magnitude of the resultant velocity
To find the magnitude of the resultant velocity vector \(\vec{V_{A/B}}\):
\[ |\vec{V_{A/B}}| = \sqrt{(6)^2 + (8)^2} \]
Calculating this gives:
\[ |\vec{V_{A/B}}| = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ km/h} \]

### Step 4: Determine the direction of the velocity
To find the angle \(\theta\) that the resultant vector makes with the positive x-axis (east direction), we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{6} \]
Calculating the angle:
\[ \theta = \tan^{-1}\left(\frac{8}{6}\right) \approx 53.13^\circ \]
This angle is measured from the east towards the north.

### Final Result
The velocity of ship A with respect to ship B is:
\[ 10 \text{ km/h at } 53^\circ \text{ north of east} \]