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 solve the problem, we need to find the average acceleration of a particle moving along the X-axis, given its displacement equation \( x = 18t + 5t^2 \). We will follow these steps:

### Step 1: Find the expression for velocity
The velocity \( v \) of the particle is the first derivative of displacement \( x \) with respect to time \( t \).

\[ v = \frac{dx}{dt} = \frac{d}{dt}(18t + 5t^2) \]

### Step 2: Differentiate the displacement equation
Calculating the derivative:

\[ v = 18 + 10t \]

### Step 3: Calculate the initial velocity at \( t_1 = 2s \)
Substituting \( t_1 = 2 \) seconds into the velocity equation:

\[ v(t_1) = 18 + 10(2) = 18 + 20 = 38 \, \text{m/s} \]

### Step 4: Calculate the final velocity at \( t_2 = 4s \)
Substituting \( t_2 = 4 \) seconds into the velocity equation:

\[ v(t_2) = 18 + 10(4) = 18 + 40 = 58 \, \text{m/s} \]

### Step 5: Calculate the average acceleration
The average acceleration \( a_{avg} \) over the time interval from \( t_1 \) to \( t_2 \) is given by:

\[ a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1} \]

Substituting the values we found:

\[ a_{avg} = \frac{58 - 38}{4 - 2} = \frac{20}{2} = 10 \, \text{m/s}^2 \]

### Final Answer
The average acceleration during the interval \( t_1 = 2s \) and \( t_2 = 4s \) is \( 10 \, \text{m/s}^2 \).

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