A particle moves along the curve `y = x^2 /2.` Here x varies with time as `x = t^2 /2.` Where x and y are measured in metres and t in seconds. At `t = 2 s,` the velocity of the particle (in `ms^-1`) is

Options

A. `4 hat i + 6 hat j`
B. `2 hat i + 4 hat j`
C. `4 hat i + 2 hat j`
D. `4 hat i + 4 hat j`

Explanation

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

### Step 1: Determine the expressions for \( x \) and \( y \)
We are given:
- The equation of the curve:
\[ y = \frac{x^2}{2} \]
- The expression for \( x \) in terms of time \( t \):
\[ x = \frac{t^2}{2} \]

### Step 2: Substitute \( x \) into the equation for \( y \)
We can substitute the expression for \( x \) into the equation for \( y \):
\[ y = \frac{\left(\frac{t^2}{2}\right)^2}{2} = \frac{t^4}{8} \]

### Step 3: Differentiate \( x \) and \( y \) with respect to \( t \) to find velocities
Now we differentiate \( x \) and \( y \) with respect to \( t \):
- For \( x \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^2}{2}\right) = t \]
- For \( y \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}\left(\frac{t^4}{8}\right) = \frac{4t^3}{8} = \frac{t^3}{2} \]

### Step 4: Calculate \( v_x \) and \( v_y \) at \( t = 2 \) seconds
Now we will evaluate \( v_x \) and \( v_y \) at \( t = 2 \) seconds:
- For \( v_x \):
\[ v_x = 2 \]
- For \( v_y \):
\[ v_y = \frac{(2)^3}{2} = \frac{8}{2} = 4 \]

### Step 5: Write the velocity vector
The velocity vector \( \mathbf{v} \) can be expressed as:
\[ \mathbf{v} = v_x \hat{i} + v_y \hat{j} = 2 \hat{i} + 4 \hat{j} \]

### Step 6: Calculate the magnitude of the velocity
The magnitude of the velocity \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \, \text{ms}^{-1} \]

### Final Answer
The velocity of the particle at \( t = 2 \) seconds is:
\[ \mathbf{v} = 2 \hat{i} + 4 \hat{j} \quad \text{or} \quad 2\sqrt{5} \, \text{ms}^{-1} \text{ (magnitude)} \]