Position of particle moving along x-axis is given as `x=2+5t+7t^(2)` then calculate :

Options

A. Velocity `(i.e(dx)/(dt))` of particle
B. Initial velocity`(i.e(dx)/(dt)" att"=0)`
C. Velocity at `t=2sec`
D. Accleration`(i.e(d^(2)x)/(dt^(2)))` of particle

Answer

To solve the problem, we need to calculate the velocity, initial velocity, velocity at 2 seconds, and acceleration of a particle moving along the x-axis, given its position as \( x = 2 + 5t + 7t^2 \).

### Step-by-Step Solution:

1. **Calculate the Velocity**:
The velocity \( v \) of the particle is the first derivative of the position \( x \) with respect to time \( t \).
\[ v = \frac{dx}{dt} \]
Given \( x = 2 + 5t + 7t^2 \), we differentiate:
\[ v = \frac{d}{dt}(2 + 5t + 7t^2) = 0 + 5 + 14t = 14t + 5 \]

2. **Calculate the Initial Velocity**:
The initial velocity \( u \) is the velocity at \( t = 0 \).
\[ u = v(t=0) = 14(0) + 5 = 5 \, \text{m/s} \]

3. **Calculate the Velocity at 2 Seconds**:
To find the velocity at \( t = 2 \) seconds, substitute \( t = 2 \) into the velocity equation.
\[ v(t=2) = 14(2) + 5 = 28 + 5 = 33 \, \text{m/s} \]

4. **Calculate the Acceleration**:
The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \) or the second derivative of the position \( x \).
\[ a = \frac{dv}{dt} = \frac{d}{dt}(14t + 5) = 14 \]
Therefore, the acceleration is constant at \( 14 \, \text{m/s}^2 \).

### Summary of Results:
- Velocity: \( v = 14t + 5 \, \text{m/s} \)
- Initial Velocity: \( u = 5 \, \text{m/s} \)
- Velocity at 2 seconds: \( v(2) = 33 \, \text{m/s} \)
- Acceleration: \( a = 14 \, \text{m/s}^2 \)