The displacement x of a particle moving along x-axis at time t is given by `x^2 =2t^2 + 6t`. The velocity at any time t is

Options

A. `(4t + 6)/x`
B. `(2t - x)/x`
C. `4t + 6`
D. `(2t + 3)/x`

Answer

To find the velocity of the particle at any time \( t \), we start with the given equation for displacement:

\[ x^2 = 2t^2 + 6t \]

### Step 1: Differentiate both sides with respect to time \( t \)

We will differentiate the equation \( x^2 = 2t^2 + 6t \) with respect to \( t \).

Using the chain rule on the left side, we have:

\[ \frac{d}{dt}(x^2) = 2x \frac{dx}{dt} \]

For the right side, we differentiate \( 2t^2 + 6t \):

\[ \frac{d}{dt}(2t^2 + 6t) = 4t + 6 \]

### Step 2: Set the derivatives equal to each other

Now we set the derivatives from both sides equal to each other:

\[ 2x \frac{dx}{dt} = 4t + 6 \]

### Step 3: Solve for \( \frac{dx}{dt} \)

To isolate \( \frac{dx}{dt} \), we rearrange the equation:

\[ \frac{dx}{dt} = \frac{4t + 6}{2x} \]

### Step 4: Simplify the expression

We can simplify the right side:

\[ \frac{dx}{dt} = \frac{2t + 3}{x} \]

### Conclusion

Thus, the velocity \( v \) at any time \( t \) is given by:

\[ v = \frac{dx}{dt} = \frac{2t + 3}{x} \]