At time t , the position of a body moving along the x - axis is `x=(t^3-6t^2+9t)m` Then, it momentarily comes to rest at

Options

A. 1 s
B. 3 s
C. 5 s
D. both 1 s and 3 s

Answer

To determine when the body momentarily comes to rest, we need to find the time \( t \) when the velocity of the body is zero. The position of the body is given by the equation:

\[ x(t) = t^3 - 6t^2 + 9t \]

### Step 1: Find the Velocity
The velocity \( v(t) \) is the derivative of the position function \( x(t) \) with respect to time \( t \). We can find the velocity by differentiating \( x(t) \):

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t) \]

Using the power rule of differentiation, we differentiate each term:

\[ v(t) = 3t^2 - 12t + 9 \]

### Step 2: Set the Velocity to Zero
To find when the body comes to rest, we set the velocity \( v(t) \) equal to zero:

\[ 3t^2 - 12t + 9 = 0 \]

### Step 3: Simplify the Equation
We can simplify the equation by dividing all terms by 3:

\[ t^2 - 4t + 3 = 0 \]

### Step 4: Factor the Quadratic Equation
Next, we factor the quadratic equation:

\[ (t - 3)(t - 1) = 0 \]

### Step 5: Solve for \( t \)
Setting each factor equal to zero gives us the possible values for \( t \):

1. \( t - 3 = 0 \) → \( t = 3 \)
2. \( t - 1 = 0 \) → \( t = 1 \)

### Conclusion
The body momentarily comes to rest at \( t = 1 \) seconds and \( t = 3 \) seconds.

### Final Answer
The body comes to rest at \( t = 1 \) seconds and \( t = 3 \) seconds.

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