A body is moving according to the equation `x = at +bt^(2) - ct^(3)` where x = displacement and a,b and c are constants. The acceleration of the body is

Options

A. `a +2bt`
B. `2b +6ct`
C. `2b - 6ct`
D. `3b - 6ct^(2)`

Answer

To find the acceleration of the body moving according to the equation \( x = at + bt^2 - ct^3 \), we will follow these steps:

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

\[ v = \frac{dx}{dt} = \frac{d}{dt}(at + bt^2 - ct^3) \]

### Step 2: Apply the differentiation
Now, we will differentiate each term:

1. The derivative of \( at \) is \( a \).
2. The derivative of \( bt^2 \) is \( 2bt \).
3. The derivative of \( -ct^3 \) is \( -3ct^2 \).

Putting it all together, we have:

\[ v = a + 2bt - 3ct^2 \]

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

\[ a = \frac{dv}{dt} = \frac{d}{dt}(a + 2bt - 3ct^2) \]

### Step 4: Apply the differentiation
Now, we will differentiate each term again:

1. The derivative of \( a \) (a constant) is \( 0 \).
2. The derivative of \( 2bt \) is \( 2b \).
3. The derivative of \( -3ct^2 \) is \( -6ct \).

Putting it all together, we have:

\[ a = 0 + 2b - 6ct = 2b - 6ct \]

### Final Result
Thus, the acceleration of the body is given by:

\[ a = 2b - 6ct \]

---