A man swims across the river which flows with a velocity of 3 km/h due east. If the velocity of man relative to water is 4 km/h due north, then what is his velocity and its direction relative to the shore of the river?

Options

A. 4 km/h, `63^(@)50'` west of north
B. 5 km/h, `36^(@)52'` west of north
C. 4 km/h, `63^(@)50'` east of north
D. 5 km/h, `36^(@)52'` east of north

Answer

To solve the problem of finding the man's velocity relative to the shore of the river, we can break it down into a few steps:

### Step 1: Identify the velocities
- The velocity of the river (V_r) is 3 km/h due east. In vector form, this can be represented as:
\[ V_r = 3 \hat{i} \text{ km/h} \]
- The velocity of the man relative to the water (V_mw) is 4 km/h due north. In vector form, this can be represented as:
\[ V_mw = 4 \hat{j} \text{ km/h} \]

### Step 2: Determine the man's velocity relative to the ground
To find the man's velocity relative to the ground (V_m), we can use vector addition:
\[ V_m = V_mw + V_r \]
Substituting the values:
\[ V_m = 4 \hat{j} + 3 \hat{i} \]

### Step 3: Calculate the magnitude of the resultant velocity
The magnitude of the resultant velocity can be calculated using the Pythagorean theorem:
\[ |V_m| = \sqrt{(V_{mx})^2 + (V_{my})^2} \]
Where \(V_{mx} = 3\) km/h and \(V_{my} = 4\) km/h:
\[ |V_m| = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km/h} \]

### Step 4: Determine the direction of the resultant velocity
To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{V_{my}}{V_{mx}} = \frac{4}{3} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
Calculating this gives:
\[ \theta \approx 53.13^\circ \]
This angle is measured from the east towards the north.

### Step 5: State the final answer
The man's velocity relative to the shore of the river is:
- Magnitude: 5 km/h
- Direction: 53.13° north of east.

### Summary
The final answer is:
- Velocity: 5 km/h
- Direction: 53.13° north of east.