At a particular instant velocity and acceleration of a particle are `(-hat(i)+hat(j)+2hat(k))m//s` and `(3hat(i)-hat(j)+hat(k))m//s^(2)` respectively at the given instant particle's speed is `:`

Options

A. `hat(j) + hat(k)`
B. `hat(i) + 2hat(j) - 3hat(k)`
C. `-hat(j)-hat(k)`
D. None of these

Answer

To find the speed of the particle, we need to calculate the magnitude of the velocity vector. The velocity vector is given as:

\[ \vec{v} = -\hat{i} + \hat{j} + 2\hat{k} \, \text{m/s} \]

### Step 1: Write down the components of the velocity vector
The components of the velocity vector \(\vec{v}\) are:
- \(v_x = -1\) (coefficient of \(\hat{i}\))
- \(v_y = 1\) (coefficient of \(\hat{j}\))
- \(v_z = 2\) (coefficient of \(\hat{k}\))

### Step 2: Use the formula for the magnitude of a vector
The magnitude of a vector \(\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}\) is given by the formula:

\[ |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

### Step 3: Substitute the components into the formula
Substituting the components we have:

\[ |\vec{v}| = \sqrt{(-1)^2 + (1)^2 + (2)^2} \]

### Step 4: Calculate the squares of the components
Calculating the squares:

\[ (-1)^2 = 1, \quad (1)^2 = 1, \quad (2)^2 = 4 \]

### Step 5: Add the squares
Now, add these values:

\[ |\vec{v}| = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 6: Conclusion
Thus, the speed of the particle is:

\[ \text{Speed} = \sqrt{6} \, \text{m/s} \]

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