At time t, the position of a body moving along the x - axis is `x=t^3-6t^(2)+9tm` The deceleration of the body at 1 s is

Options

A. `6 m s ^(-2)`
B. `4 m s ^(-2)`
C. `8 m s ^(-2)`
D. None

Answer

To find the deceleration of the body at \( t = 1 \) second, we will follow these steps:

### Step 1: Write down the position function
The position of the body is given by:
\[ x(t) = t^3 - 6t^2 + 9t \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t) \]
Calculating the derivative:
\[ v(t) = 3t^2 - 12t + 9 \]

### Step 3: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) \]
Calculating the derivative:
\[ a(t) = 6t - 12 \]

### Step 4: Evaluate the acceleration at \( t = 1 \) second
Now, we substitute \( t = 1 \) into the acceleration function:
\[ a(1) = 6(1) - 12 = 6 - 12 = -6 \, \text{m/s}^2 \]

### Step 5: Interpret the result
The negative sign indicates that the body is decelerating. The magnitude of the deceleration is:
\[ \text{Deceleration} = 6 \, \text{m/s}^2 \]

### Final Answer
The deceleration of the body at \( t = 1 \) second is \( 6 \, \text{m/s}^2 \).

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