The position of a particle moving in x-y plane changes with time `t` given by `vecx= 3t^2hati + 9thatj`. The acceleration of particle would be

Options

A. 2 m/s^2
B. 3 m/s^2
C. 4 m/s^2
D. 6 m/s^2

Answer

To find the acceleration of the particle given its position vector in the x-y plane, we will follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given as:
\[ \vec{x}(t) = 3t^2 \hat{i} + 9t \hat{j} \]

### Step 2: Differentiate the position vector to find the velocity vector
To find the velocity vector \(\vec{v}(t)\), we differentiate the position vector \(\vec{x}(t)\) with respect to time \(t\):
\[ \vec{v}(t) = \frac{d\vec{x}}{dt} = \frac{d}{dt}(3t^2 \hat{i} + 9t \hat{j}) \]
Calculating the derivatives:
\[ \vec{v}(t) = 6t \hat{i} + 9 \hat{j} \]

### Step 3: Differentiate the velocity vector to find the acceleration vector
Now, we differentiate the velocity vector \(\vec{v}(t)\) with respect to time \(t\) to find the acceleration vector \(\vec{a}(t)\):
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(6t \hat{i} + 9 \hat{j}) \]
Calculating the derivatives:
\[ \vec{a}(t) = 6 \hat{i} + 0 \hat{j} = 6 \hat{i} \]

### Step 4: Write the final answer
The acceleration of the particle is:
\[ \vec{a} = 6 \hat{i} \text{ m/s}^2 \]
The magnitude of the acceleration is \(6 \text{ m/s}^2\).

### Summary
The acceleration of the particle is \(6 \text{ m/s}^2\) in the direction of the x-axis.

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