A body starts from origin and moves along x - axis so that its position at any instant is `x=4t^(2)-12t` where t is in second. What is the acceleration of particle?

Options

A. `4m//s^(2)`
B. `8m//s^(2)`
C. `24m//s^(2)`
D. `0m//s^(2)`

Answer

To find the acceleration of the particle given its position function, we can follow these steps:

### Step 1: Write down the position function
The position of the particle is given by:
\[ x(t) = 4t^2 - 12t \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^2 - 12t) \]

### Step 3: Differentiate the position function
Now, we differentiate \( 4t^2 - 12t \):
- The derivative of \( 4t^2 \) is \( 8t \).
- The derivative of \( -12t \) is \( -12 \).

So, the velocity function becomes:
\[ v(t) = 8t - 12 \]

### Step 4: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(8t - 12) \]

### Step 5: Differentiate the velocity function
Now, we differentiate \( 8t - 12 \):
- The derivative of \( 8t \) is \( 8 \).
- The derivative of \( -12 \) (a constant) is \( 0 \).

So, the acceleration function becomes:
\[ a(t) = 8 \]

### Final Answer
The acceleration of the particle is:
\[ a = 8 \, \text{m/s}^2 \]

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