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 \( t = 2 \, \text{s} \), we need to determine the components of the velocity in the x and y directions. Here’s a step-by-step solution:

### 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 as a function of time:
\[ x = \frac{t^2}{2} \]

### Step 2: Find the expression for y in terms of t
Substituting the expression for \( x \) into the equation for \( y \):
\[ y = \frac{1}{2} \left(\frac{t^2}{2}\right)^2 = \frac{1}{2} \cdot \frac{t^4}{4} = \frac{t^4}{8} \]

### Step 3: Calculate the velocity components
The velocity components in the x and y directions are given by the derivatives of x and y with respect to time t.

#### Velocity in the x-direction (\( v_x \)):
\[ v_x = \frac{dx}{dt} \]
Differentiating \( x = \frac{t^2}{2} \):
\[ v_x = \frac{d}{dt}\left(\frac{t^2}{2}\right) = t \]

#### Velocity in the y-direction (\( v_y \)):
\[ v_y = \frac{dy}{dt} \]
Differentiating \( y = \frac{t^4}{8} \):
\[ v_y = \frac{d}{dt}\left(\frac{t^4}{8}\right) = \frac{4t^3}{8} = \frac{t^3}{2} \]

### Step 4: Evaluate the velocity components at \( t = 2 \, \text{s} \)
Now, we substitute \( t = 2 \) into the expressions for \( v_x \) and \( v_y \):

For \( v_x \):
\[ v_x = 2 \, \text{m/s} \]

For \( v_y \):
\[ v_y = \frac{(2)^3}{2} = \frac{8}{2} = 4 \, \text{m/s} \]

### Step 5: Write 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
Thus, the velocity of the particle at \( t = 2 \, \text{s} \) is:
\[ \vec{v} = 2 \hat{i} + 4 \hat{j} \, \text{m/s} \]

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