The displacement (in metre) of a particle moving along x-axis is given by `x=18t +5t^(2)` Calculate (i) instantaneous acceleration.

Options

A. `18 ms^(-2)`
B. `10 ms^(-2)`
C. `5 ms^(-2)`
D. `1 ms^(-2)`

Explanation

To solve the problem of finding the instantaneous acceleration of a particle moving along the x-axis with the given displacement equation \( x = 18t + 5t^2 \), we will follow these steps:

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

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

So, the velocity equation becomes:
\[ v(t) = 18 + 10t \]

### Step 3: Differentiate the velocity to find acceleration
The acceleration \( a(t) \) is the derivative of the velocity \( v(t) \) with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(18 + 10t) \]
Again, applying the rules of differentiation:
- The derivative of the constant \( 18 \) is \( 0 \).
- The derivative of \( 10t \) is \( 10 \).

Thus, the instantaneous acceleration is:
\[ a(t) = 0 + 10 = 10 \, \text{m/s}^2 \]

### Conclusion
The instantaneous acceleration of the particle is:
\[ \boxed{10 \, \text{m/s}^2} \]

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