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, we will follow these steps:

### Step 1: Write down the position equation
The position of the particle is given by:
\[ S = 3t^3 + 7t^2 + 14 + t + 8 \]

### Step 2: Differentiate the position equation to find velocity
To find the velocity \( v \), we differentiate the position \( S \) with respect to time \( t \):
\[ v = \frac{dS}{dt} = \frac{d}{dt}(3t^3 + 7t^2 + 14 + t + 8) \]
Calculating the derivative:
\[ v = 9t^2 + 14t + 1 \]

### Step 3: Differentiate the velocity equation to find acceleration
Now, we differentiate the velocity \( v \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(9t^2 + 14t + 1) \]
Calculating the derivative:
\[ a = 18t + 14 \]

### Step 4: Substitute \( t = 1 \) to find acceleration at that instant
Now we substitute \( t = 1 \) second into the acceleration equation:
\[ a = 18(1) + 14 = 18 + 14 = 32 \, \text{m/s}^2 \]

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

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