A force `vec(F)=(2hat(i)+hat(j)+3hat(k))` N acts on a particle of mass 1 kg for t = 2s. If initial velocity of particle is `vec(u)=(2hat(i)+hat(j))ms^(-1)`, then the speed of particle at the t = 2 s will be

Options

A. `4 ms^(-1)`
B. `6 ms^(-1)`
C. `9 ms^(-1)`
D. `12 ms^(-1)`

Explanation

To find the speed of the particle at \( t = 2 \) seconds, we can follow these steps:

### Step 1: Calculate the Acceleration
Given the force \( \vec{F} = (2\hat{i} + \hat{j} + 3\hat{k}) \) N and the mass \( m = 1 \) kg, we can calculate the acceleration \( \vec{a} \) using Newton's second law:

\[ \vec{a} = \frac{\vec{F}}{m} \]

Substituting the values:

\[ \vec{a} = \frac{(2\hat{i} + \hat{j} + 3\hat{k})}{1} = 2\hat{i} + \hat{j} + 3\hat{k} \, \text{m/s}^2 \]

### Step 2: Use the Equation of Motion
We can use the equation of motion to find the final velocity \( \vec{v} \) after \( t = 2 \) seconds. The equation is:

\[ \vec{v} = \vec{u} + t \cdot \vec{a} \]

Where \( \vec{u} = (2\hat{i} + \hat{j}) \) m/s is the initial velocity, and \( t = 2 \) s. Now substituting the values:

\[ \vec{v} = (2\hat{i} + \hat{j}) + 2 \cdot (2\hat{i} + \hat{j} + 3\hat{k}) \]

Calculating \( 2 \cdot (2\hat{i} + \hat{j} + 3\hat{k}) \):

\[ = 4\hat{i} + 2\hat{j} + 6\hat{k} \]

Now substituting this back into the equation for \( \vec{v} \):

\[ \vec{v} = (2\hat{i} + \hat{j}) + (4\hat{i} + 2\hat{j} + 6\hat{k}) = (2 + 4)\hat{i} + (1 + 2)\hat{j} + 6\hat{k} \]

Thus,

\[ \vec{v} = 6\hat{i} + 3\hat{j} + 6\hat{k} \, \text{m/s} \]

### Step 3: Calculate the Speed
The speed is the magnitude of the velocity vector \( \vec{v} \):

\[ |\vec{v}| = \sqrt{(6)^2 + (3)^2 + (6)^2} \]

Calculating the squares:

\[ = \sqrt{36 + 9 + 36} = \sqrt{81} = 9 \, \text{m/s} \]

### Final Answer
The speed of the particle at \( t = 2 \) seconds is \( 9 \, \text{m/s} \).

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