If `x = (t^(3))/(3) - (5)/(2)t^(2) + 6t + 1`, then the value `(d^(2)x)/(d t^(2))`, when `(dx)/(d t)` is zero is :

Options

A. `+- 1`
B. `+-2`
C. `+- 3`
D. `+- 4`

Answer

To solve the problem step by step, we start with the given function:

\[ x = \frac{t^3}{3} - \frac{5}{2}t^2 + 6t + 1 \]

### Step 1: Differentiate \( x \) with respect to \( t \)

To find \( \frac{dx}{dt} \), we differentiate \( x \):

\[ \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) - \frac{d}{dt}\left(\frac{5}{2}t^2\right) + \frac{d}{dt}(6t) + \frac{d}{dt}(1) \]

Calculating each term:

- The derivative of \( \frac{t^3}{3} \) is \( t^2 \).
- The derivative of \( -\frac{5}{2}t^2 \) is \( -5t \).
- The derivative of \( 6t \) is \( 6 \).
- The derivative of \( 1 \) is \( 0 \).

So, we have:

\[ \frac{dx}{dt} = t^2 - 5t + 6 \]

### Step 2: Set \( \frac{dx}{dt} = 0 \)

We need to find the values of \( t \) for which \( \frac{dx}{dt} = 0 \):

\[ t^2 - 5t + 6 = 0 \]

### Step 3: Factor the quadratic equation

We can factor the quadratic:

\[ (t - 2)(t - 3) = 0 \]

Thus, the solutions are:

\[ t = 2 \quad \text{and} \quad t = 3 \]

### Step 4: Differentiate \( \frac{dx}{dt} \) to find \( \frac{d^2x}{dt^2} \)

Next, we differentiate \( \frac{dx}{dt} \) to find \( \frac{d^2x}{dt^2} \):

\[ \frac{d^2x}{dt^2} = \frac{d}{dt}(t^2 - 5t + 6) \]

Calculating the derivative:

- The derivative of \( t^2 \) is \( 2t \).
- The derivative of \( -5t \) is \( -5 \).
- The derivative of \( 6 \) is \( 0 \).

So, we have:

\[ \frac{d^2x}{dt^2} = 2t - 5 \]

### Step 5: Evaluate \( \frac{d^2x}{dt^2} \) at \( t = 2 \) and \( t = 3 \)

Now we will evaluate \( \frac{d^2x}{dt^2} \) at the points \( t = 2 \) and \( t = 3 \):

1. For \( t = 2 \):

\[ \frac{d^2x}{dt^2} = 2(2) - 5 = 4 - 5 = -1 \]

2. For \( t = 3 \):

\[ \frac{d^2x}{dt^2} = 2(3) - 5 = 6 - 5 = 1 \]

### Final Result

Thus, the values of \( \frac{d^2x}{dt^2} \) when \( \frac{dx}{dt} = 0 \) are:

- At \( t = 2 \): \( -1 \)
- At \( t = 3 \): \( 1 \)