The distance travelled by an object along the axes are given by ` x= 2 t^2 , y=t^2-4 t, z=3 t -5`. The initial velocity of the particle is .

Options

A. 10 units
B. 12 units
C. 5 units
D. 2 units

Explanation

To find the initial velocity of the particle given the equations of motion along the x, y, and z axes, we can follow these steps:

### Step 1: Write down the equations of motion
The equations of motion for the particle are given as:
- \( x = 2t^2 \)
- \( y = t^2 - 4t \)
- \( z = 3t - 5 \)

### Step 2: Find the velocity components
To find the velocity components, we need to differentiate each position function with respect to time \( t \).

1. **Velocity in the x-direction**:
\[ v_x = \frac{dx}{dt} = \frac{d(2t^2)}{dt} = 4t \]

2. **Velocity in the y-direction**:
\[ v_y = \frac{dy}{dt} = \frac{d(t^2 - 4t)}{dt} = 2t - 4 \]

3. **Velocity in the z-direction**:
\[ v_z = \frac{dz}{dt} = \frac{d(3t - 5)}{dt} = 3 \]

### Step 3: Write the velocity vector
The velocity vector \( \mathbf{v} \) can be expressed as:
\[ \mathbf{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} = 4t \hat{i} + (2t - 4) \hat{j} + 3 \hat{k} \]

### Step 4: Evaluate the velocity at \( t = 0 \)
To find the initial velocity, we evaluate the velocity vector at \( t = 0 \):
\[ v_x(0) = 4(0) = 0 \]
\[ v_y(0) = 2(0) - 4 = -4 \]
\[ v_z(0) = 3 \]

Thus, the initial velocity vector is:
\[ \mathbf{v}(0) = 0 \hat{i} - 4 \hat{j} + 3 \hat{k} \]

### Step 5: Calculate the magnitude of the initial velocity
The magnitude of the velocity vector is given by:
\[ |\mathbf{v}(0)| = \sqrt{(v_x(0))^2 + (v_y(0))^2 + (v_z(0))^2} \]
Substituting the values:
\[ |\mathbf{v}(0)| = \sqrt{(0)^2 + (-4)^2 + (3)^2} = \sqrt{0 + 16 + 9} = \sqrt{25} = 5 \]

### Final Answer
The initial velocity of the particle is \( 5 \) units.

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