A particle moves along the positive branch of the curve `y = (x^(2))/(2)` where `x = (t^(2))/(2),x` and y are measured in metres and t in second. At `t = 2s`, the velocity of the particle is

Options

A. `2 hati - 4 hatj ms^(-1)`
B. `4 hati +2 hatj ms^(-1)`
C. `2 hati + 4hatj ms^(-1)`
D. `4 hati - 2hatj ms^(-1)`

Explanation

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

### Step 1: Understand the equations given
The position of the particle is defined by:
- \( x = \frac{t^2}{2} \)
- \( y = \frac{x^2}{2} \)

### Step 2: Substitute \( x \) in the equation for \( y \)
First, we will substitute \( x \) into the equation for \( y \):
\[ y = \frac{x^2}{2} = \frac{1}{2} \left(\frac{t^2}{2}\right)^2 = \frac{1}{2} \cdot \frac{t^4}{4} = \frac{t^4}{8} \]

### Step 3: Differentiate \( x \) with respect to \( t \) to find the x-component of velocity
The x-component of velocity \( v_x \) is given by:
\[ v_x = \frac{dx}{dt} \]
Differentiating \( x = \frac{t^2}{2} \):
\[ \frac{dx}{dt} = \frac{1}{2} \cdot 2t = t \]
At \( t = 2 \) seconds:
\[ v_x = 2 \, \text{m/s} \]

### Step 4: Differentiate \( y \) with respect to \( t \) to find the y-component of velocity
The y-component of velocity \( v_y \) is given by:
\[ v_y = \frac{dy}{dt} \]
Differentiating \( y = \frac{t^4}{8} \):
\[ \frac{dy}{dt} = \frac{1}{8} \cdot 4t^3 = \frac{t^3}{2} \]
At \( t = 2 \) seconds:
\[ v_y = \frac{(2)^3}{2} = \frac{8}{2} = 4 \, \text{m/s} \]

### Step 5: Combine the components to get the velocity vector
The velocity vector \( \vec{v} \) can be expressed as:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = 2 \hat{i} + 4 \hat{j} \, \text{m/s} \]

### Final Answer
The velocity of the particle at \( t = 2 \) seconds is:
\[ \vec{v} = 2 \hat{i} + 4 \hat{j} \, \text{m/s} \]

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