Motion of a particle is given by equation `S = (3t^(3) + 7 t^(2) + 14 + t + 8)`m, The value of acceleration of the particle at t = 1 sec, is:

Options

A. `10 m//s^(2)`
B. `32 m//s^(2)`
C. `23 m//s^(2)`
D. `16 m//s^(2)`

Explanation

To find the acceleration of the particle at \( t = 1 \) second given the position function \( S(t) = 3t^3 + 7t^2 + t + 14 + 8 \), we will follow these steps:

### Step 1: Differentiate the position function to find the velocity
The first step is to differentiate the position function \( S(t) \) with respect to time \( t \) to find the velocity \( v(t) \).

\[ v(t) = \frac{dS}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + t + 14 + 8) \]

### Step 2: Calculate the derivative
Now, we will calculate the derivative term by term:

- The derivative of \( 3t^3 \) is \( 9t^2 \).
- The derivative of \( 7t^2 \) is \( 14t \).
- The derivative of \( t \) is \( 1 \).
- The constants \( 14 \) and \( 8 \) have derivatives of \( 0 \).

Putting it all together, we have:

\[ v(t) = 9t^2 + 14t + 1 \]

### Step 3: Differentiate the velocity function to find the acceleration
Next, we differentiate the velocity function \( v(t) \) with respect to time \( t \) to find the acceleration \( a(t) \).

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2 + 14t + 1) \]

### Step 4: Calculate the derivative of the velocity
Now, we will calculate the derivative term by term:

- The derivative of \( 9t^2 \) is \( 18t \).
- The derivative of \( 14t \) is \( 14 \).
- The derivative of \( 1 \) is \( 0 \).

Putting it all together, we have:

\[ a(t) = 18t + 14 \]

### Step 5: Substitute \( t = 1 \) into the acceleration function
Now, we need to find the acceleration at \( t = 1 \):

\[ a(1) = 18(1) + 14 \]

Calculating this gives:

\[ a(1) = 18 + 14 = 32 \, \text{m/s}^2 \]

### Final Answer
The value of acceleration of the particle at \( t = 1 \) second is \( 32 \, \text{m/s}^2 \).

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