Relation between displacement x and time t is `x=2-5t+6t^(2)`, the initial acceleration will be :-

Options

A. `-3 ms^(-2)`
B. `12 ms^(-2)`
C. `2 ms^(-2)`
D. `-5 ms^(-2)`

Answer

To find the initial acceleration from the given displacement-time relation \( x = 2 - 5t + 6t^2 \), we will follow these steps:

### Step 1: Differentiate the displacement equation to find velocity
The first step is to differentiate the displacement \( x \) with respect to time \( t \) to find the velocity \( v \).

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

### Step 2: Calculate the derivative
Now we calculate the derivative term by term:
- The derivative of \( 2 \) is \( 0 \).
- The derivative of \( -5t \) is \( -5 \).
- The derivative of \( 6t^2 \) is \( 12t \).

Putting it all together, we have:

\[ v = 0 - 5 + 12t = 12t - 5 \]

### Step 3: Differentiate the velocity equation to find acceleration
Next, we differentiate the velocity \( v \) with respect to time \( t \) to find the acceleration \( a \).

\[ a = \frac{dv}{dt} = \frac{d}{dt}(12t - 5) \]

### Step 4: Calculate the derivative of velocity
Now we calculate the derivative:
- The derivative of \( 12t \) is \( 12 \).
- The derivative of \( -5 \) is \( 0 \).

Thus, we find:

\[ a = 12 \]

### Step 5: Identify the initial acceleration
Since the acceleration \( a \) is constant and does not depend on time, the initial acceleration is also \( 12 \, \text{m/s}^2 \).

### Final Answer
The initial acceleration is:

\[ \boxed{12 \, \text{m/s}^2} \]

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