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. (a)1 s
B. (b)3 s
C. (c)5 s
D. (d)both 1 s and 3 s

Answer

To solve the problem, we need to find the time(s) at which the body momentarily comes to rest. This occurs when the velocity of the body is zero.

### Step-by-Step Solution:

1. **Given Position Function**: The position of the body is given by the equation:
\[ x(t) = t^3 - 6t^2 + 9t \]

2. **Find Velocity**: The velocity \( v(t) \) is the derivative of the position function with respect to time \( t \). We differentiate \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t) \]
Using the power rule for differentiation:
\[ v(t) = 3t^2 - 12t + 9 \]

3. **Set Velocity to Zero**: To find when the body comes to rest, we set the velocity equation to zero:
\[ 3t^2 - 12t + 9 = 0 \]

4. **Simplify the Equation**: We can simplify this equation by dividing all terms by 3:
\[ t^2 - 4t + 3 = 0 \]

5. **Factor the Quadratic**: Now we factor the quadratic equation:
\[ (t - 1)(t - 3) = 0 \]

6. **Find the Roots**: Setting each factor to zero gives us the possible values for \( t \):
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]
\[ t - 3 = 0 \quad \Rightarrow \quad t = 3 \]

7. **Conclusion**: The body momentarily comes to rest at \( t = 1 \) second and \( t = 3 \) seconds.

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