The position of a particle is given by `vecr = 2t^(2) hati + 3t hatj + 4hatk` where t is in second and the coefficients have proper units for `vecr` to be in metre. The `veca(t)` of the particle at t = 1 s is 

Options

A. `4 m s^(-2)` along y-direction
B. `3 m s^(-2)` along x-direction
C. `4 m s^(-2)` along x-direction
D. `2 m s^(-2)` along z-direction

Explanation

To find the acceleration of the particle at \( t = 1 \, \text{s} \), we will follow these steps:

### Step 1: Write down the position vector
The position vector of the particle is given by:
\[ \vec{r}(t) = 2t^2 \hat{i} + 3t \hat{j} + 4 \hat{k} \]

### Step 2: Differentiate the position vector to find the velocity vector
To find the velocity vector \( \vec{v}(t) \), we differentiate the position vector \( \vec{r}(t) \) with respect to time \( t \):
\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(2t^2 \hat{i} + 3t \hat{j} + 4 \hat{k}) \]
Calculating the derivatives:
- The derivative of \( 2t^2 \) with respect to \( t \) is \( 4t \).
- The derivative of \( 3t \) with respect to \( t \) is \( 3 \).
- The derivative of the constant \( 4 \) is \( 0 \).

Thus, we have:
\[ \vec{v}(t) = 4t \hat{i} + 3 \hat{j} + 0 \hat{k} = 4t \hat{i} + 3 \hat{j} \]

### Step 3: Differentiate the velocity vector to find the acceleration vector
Next, we differentiate the velocity vector \( \vec{v}(t) \) with respect to time \( t \) to find the acceleration vector \( \vec{a}(t) \):
\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(4t \hat{i} + 3 \hat{j}) \]
Calculating the derivatives:
- The derivative of \( 4t \) with respect to \( t \) is \( 4 \).
- The derivative of \( 3 \) (a constant) is \( 0 \).

Thus, we have:
\[ \vec{a}(t) = 4 \hat{i} + 0 \hat{j} + 0 \hat{k} = 4 \hat{i} \]

### Step 4: Evaluate the acceleration at \( t = 1 \, \text{s} \)
At \( t = 1 \, \text{s} \):
\[ \vec{a}(1) = 4 \hat{i} \]

### Final Answer
The acceleration of the particle at \( t = 1 \, \text{s} \) is:
\[ \vec{a}(1) = 4 \hat{i} \, \text{m/s}^2 \]

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