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

Options

A. ` 10 unit`
B. ` 12 unit`
C. ` 5 unit `
D. ` 2 unit`

Explanation

To find the initial velocity of the particle, we need to follow these steps:

### Step 1: Write down the equations of motion
The position of the particle along the x, y, and z axes is given by:
- \( x = 2t^2 \)
- \( y = t^2 - 4t \)
- \( z = 3t - 5 \)

### Step 2: Differentiate the position functions to find the velocity components
The velocity components in the x, y, and z directions can be found by differentiating the position functions with respect to time \( t \).

- For the x-component of velocity \( v_x \):
\[ v_x = \frac{dx}{dt} = \frac{d(2t^2)}{dt} = 4t \]

- For the y-component of velocity \( v_y \):
\[ v_y = \frac{dy}{dt} = \frac{d(t^2 - 4t)}{dt} = 2t - 4 \]

- For the z-component of velocity \( v_z \):
\[ v_z = \frac{dz}{dt} = \frac{d(3t - 5)}{dt} = 3 \]

### Step 3: Evaluate the velocity components at \( t = 0 \)
To find the initial velocity, we substitute \( t = 0 \) into the velocity equations:

- For \( v_x \):
\[ v_x(0) = 4(0) = 0 \]

- For \( v_y \):
\[ v_y(0) = 2(0) - 4 = -4 \]

- For \( v_z \):
\[ v_z(0) = 3 \]

### Step 4: Combine the velocity components to find the initial velocity vector
The initial velocity vector \( \vec{v} \) can be expressed as:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} = 0 \hat{i} - 4 \hat{j} + 3 \hat{k} \]

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

### Final Answer
The initial velocity of the particle is \( 5 \text{ m/s} \).

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