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 step by step, we will calculate the velocity and acceleration of the particle whose position is given by the equation \( x = 2 + 5t + 7t^2 \).

### Step 1: Find the Velocity

The velocity of the particle is the first derivative of the position with respect to time \( t \).

Given:
\[ x = 2 + 5t + 7t^2 \]

To find the velocity \( v(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(2 + 5t + 7t^2) \]

Calculating the derivative:
- The derivative of \( 2 \) is \( 0 \).
- The derivative of \( 5t \) is \( 5 \).
- The derivative of \( 7t^2 \) is \( 14t \).

Thus, the velocity function is:
\[ v(t) = 0 + 5 + 14t = 5 + 14t \]

### Step 2: Calculate Initial Velocity

The initial velocity is the velocity at \( t = 0 \):
\[ v(0) = 5 + 14(0) = 5 \, \text{m/s} \]

### Step 3: Calculate Velocity at \( t = 2 \) seconds

Now, we will find the velocity at \( t = 2 \) seconds:
\[ v(2) = 5 + 14(2) = 5 + 28 = 33 \, \text{m/s} \]

### Step 4: Find the Acceleration

The acceleration of the particle is the second derivative of the position with respect to time \( t \), or the first derivative of the velocity.

To find the acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(5 + 14t) \]

Calculating the derivative:
- The derivative of \( 5 \) is \( 0 \).
- The derivative of \( 14t \) is \( 14 \).

Thus, the acceleration function is:
\[ a(t) = 0 + 14 = 14 \, \text{m/s}^2 \]

### Summary of Results
- Initial Velocity \( v(0) = 5 \, \text{m/s} \)
- Velocity at \( t = 2 \) seconds \( v(2) = 33 \, \text{m/s} \)
- Acceleration \( a = 14 \, \text{m/s}^2 \)

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