A particle moves along a straight line in such a way that its distance from a fixed point on the line, at any time t from the start, is given by the equation `s=6-2t+3^(3)`. Its acceleration after 1 second of motion is

Options

A. 12
B. 16
C. 18
D. None

Answer

To find the acceleration of the particle after 1 second of motion, we need to follow these steps:

### Step 1: Write down the position function
The position of the particle as a function of time \( t \) is given by:
\[ s(t) = 6 - 2t + 3t^3 \]

### Step 2: Differentiate the position function to find velocity
To find the velocity \( v(t) \), we differentiate the position function \( s(t) \) with respect to time \( t \):
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(6 - 2t + 3t^3) \]
Calculating the derivative:
\[ v(t) = 0 - 2 + 9t^2 = 9t^2 - 2 \]

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

### Step 4: Calculate the acceleration at \( t = 1 \) second
Now, we substitute \( t = 1 \) into the acceleration function:
\[ a(1) = 18 \times 1 = 18 \]

### Final Answer
The acceleration of the particle after 1 second of motion is:
\[ \boxed{18 \, \text{m/s}^2} \]

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