A particle has an initial velocity of ` 3hati + 4hatj` and on acceleration of ` 0.4hatj + 0.3hatj` . The magnitude of its velocity after 10 s is

Options

A. ` 7sqrt2 ` units
B. 7 units
C. 8.5 units
D. 10 units .

Answer

To find the magnitude of the velocity of a particle after 10 seconds, we can follow these steps:

### Step 1: Identify the initial velocity and acceleration
The initial velocity \( \mathbf{u} \) is given as:
\[ \mathbf{u} = 3\hat{i} + 4\hat{j} \]
The acceleration \( \mathbf{a} \) is given as:
\[ \mathbf{a} = 0.4\hat{i} + 0.3\hat{j} \]

### Step 2: Use the formula for final velocity
The formula for the final velocity \( \mathbf{v} \) after time \( t \) is:
\[ \mathbf{v} = \mathbf{u} + \mathbf{a} \cdot t \]
Substituting the values we have:
\[ \mathbf{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \cdot t \]

### Step 3: Substitute \( t = 10 \) seconds
Now, substituting \( t = 10 \) seconds into the equation:
\[ \mathbf{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j}) \cdot 10 \]
Calculating the acceleration part:
\[ (0.4\hat{i} + 0.3\hat{j}) \cdot 10 = 4\hat{i} + 3\hat{j} \]
Thus, we can rewrite \( \mathbf{v} \):
\[ \mathbf{v} = (3 + 4)\hat{i} + (4 + 3)\hat{j} = 7\hat{i} + 7\hat{j} \]

### Step 4: Calculate the magnitude of the velocity
The magnitude of the velocity \( |\mathbf{v}| \) is given by:
\[ |\mathbf{v}| = \sqrt{(v_x)^2 + (v_y)^2} \]
Substituting \( v_x = 7 \) and \( v_y = 7 \):
\[ |\mathbf{v}| = \sqrt{7^2 + 7^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \]

### Final Answer
The magnitude of the velocity after 10 seconds is:
\[ |\mathbf{v}| = 7\sqrt{2} \text{ m/s} \]

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