The displacement of a particle moving along the x-axis is given by equation `x=2t^(3)-21"t"^(2)+60t+6`.The possible acceleration of the particle when its velocity is zero is

Options

A. A) `-18 m//s^(2)`
B. B) `30 m//s^(2)`
C. C) `9 m//s^(2)`
D. D) `-9 m//s^(2)`

Answer

To solve the problem, we need to find the acceleration of a particle when its velocity is zero, given the displacement equation \( x = 2t^3 - 21t^2 + 60t + 6 \).

### Step-by-Step Solution:

1. **Differentiate the displacement equation to find velocity:**
The velocity \( v \) is the first derivative of displacement \( x \) with respect to time \( t \).
\[ v = \frac{dx}{dt} = \frac{d}{dt}(2t^3 - 21t^2 + 60t + 6) \]
Differentiating term by term:
\[ v = 6t^2 - 42t + 60 \]

2. **Set the velocity equation to zero:**
To find the time(s) when the velocity is zero, we set the velocity equation to zero:
\[ 6t^2 - 42t + 60 = 0 \]

3. **Simplify the quadratic equation:**
We can simplify the equation by dividing all terms by 6:
\[ t^2 - 7t + 10 = 0 \]

4. **Factor the quadratic equation:**
The quadratic can be factored as:
\[ (t - 5)(t - 2) = 0 \]
Thus, the solutions are:
\[ t = 5 \quad \text{and} \quad t = 2 \]

5. **Differentiate the velocity equation to find acceleration:**
The acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 42t + 60) \]
Differentiating term by term:
\[ a = 12t - 42 \]

6. **Calculate acceleration at \( t = 5 \):**
Substitute \( t = 5 \) into the acceleration equation:
\[ a = 12(5) - 42 = 60 - 42 = 18 \, \text{m/s}^2 \]

7. **Calculate acceleration at \( t = 2 \):**
Substitute \( t = 2 \) into the acceleration equation:
\[ a = 12(2) - 42 = 24 - 42 = -18 \, \text{m/s}^2 \]

8. **Conclusion:**
The possible accelerations of the particle when its velocity is zero are \( 18 \, \text{m/s}^2 \) and \( -18 \, \text{m/s}^2 \). Since the question asks for the possible acceleration when the velocity is zero, we can conclude that the answer is:
\[ \text{Possible acceleration when velocity is zero: } -18 \, \text{m/s}^2 \]