Position of a particle moving along a straight line is given by `x=2t^(2)+t` . Find the velocity at t = 2 sec.

Options

A. 3
B. 6
C. 4
D. 9

Answer

To find the velocity of the particle at \( t = 2 \) seconds, we start with the position function given by:

\[ x(t) = 2t^2 + t \]

### Step 1: Differentiate the position function
The velocity \( v(t) \) is defined as the derivative of the position function with respect to time \( t \). We need to calculate \( \frac{dx}{dt} \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^2 + t) \]

### Step 2: Apply the power rule of differentiation
Using the power rule, we differentiate each term:

- The derivative of \( 2t^2 \) is \( 4t \).
- The derivative of \( t \) is \( 1 \).

Thus, we have:

\[ v(t) = 4t + 1 \]

### Step 3: Substitute \( t = 2 \) seconds into the velocity function
Now, we will substitute \( t = 2 \) seconds into the velocity function to find the velocity at that specific time:

\[ v(2) = 4(2) + 1 \]

### Step 4: Calculate the value
Calculating the above expression:

\[ v(2) = 8 + 1 = 9 \, \text{m/s} \]

### Final Answer
The velocity of the particle at \( t = 2 \) seconds is:

\[ \boxed{9 \, \text{m/s}} \]