The displacement (in metre) of a particle moving along X-axis is given by `x=18t+5t^(2)`. The average acceleration during the interval `t_(1)=2s` and `t_(2)=4s` is

Options

A. `13 ms^(-2)`
B. `10 ms^(-2)`
C. `27 ms^(-2)`
D. `37 ms^(-2)`

Explanation

To find the average acceleration of the particle during the time interval from \( t_1 = 2 \, \text{s} \) to \( t_2 = 4 \, \text{s} \), we will follow these steps:

### Step 1: Write down the displacement equation
The displacement \( x \) of the particle is given by:
\[ x = 18t + 5t^2 \]

### Step 2: Differentiate the displacement to find the velocity
The velocity \( v \) is the derivative of the displacement with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(18t + 5t^2) \]
Using the power rule of differentiation:
\[ v = 18 + 10t \]

### Step 3: Calculate the initial velocity \( v_1 \) at \( t_1 = 2 \, \text{s} \)
Substituting \( t = 2 \) into the velocity equation:
\[ v_1 = 18 + 10(2) = 18 + 20 = 38 \, \text{m/s} \]

### Step 4: Calculate the final velocity \( v_2 \) at \( t_2 = 4 \, \text{s} \)
Substituting \( t = 4 \) into the velocity equation:
\[ v_2 = 18 + 10(4) = 18 + 40 = 58 \, \text{m/s} \]

### Step 5: Calculate the change in velocity
The change in velocity \( \Delta v \) is given by:
\[ \Delta v = v_2 - v_1 = 58 - 38 = 20 \, \text{m/s} \]

### Step 6: Calculate the time interval
The time interval \( \Delta t \) is:
\[ \Delta t = t_2 - t_1 = 4 - 2 = 2 \, \text{s} \]

### Step 7: Calculate the average acceleration
The average acceleration \( a_{avg} \) is given by:
\[ a_{avg} = \frac{\Delta v}{\Delta t} = \frac{20 \, \text{m/s}}{2 \, \text{s}} = 10 \, \text{m/s}^2 \]

### Final Answer
The average acceleration during the interval from \( t_1 = 2 \, \text{s} \) to \( t_2 = 4 \, \text{s} \) is:
\[ \boxed{10 \, \text{m/s}^2} \]