Acceleration of a particle in x-y plane varies with time as `a=(2t hati+3t^2 hatj) m//s^2` At time `t =0,` velocity of particle is `2 m//s` along positive x direction and particle starts from origin. Find velocity and coordinates of particle at `t=1s.`

Explanation

To solve the problem, we need to find the velocity and coordinates of the particle at \( t = 1 \) second given the acceleration function, initial velocity, and initial position.

### Step 1: Write down the given information.
- Acceleration:
\[ \vec{a} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{m/s}^2 \]
- Initial velocity at \( t = 0 \):
\[ \vec{u} = 2 \hat{i} + 0 \hat{j} \, \text{m/s} \]
- Initial position at \( t = 0 \):
\[ \vec{r_0} = 0 \hat{i} + 0 \hat{j} \, \text{m} \]

### Step 2: Find the velocity as a function of time.
To find the velocity, we integrate the acceleration function with respect to time:
\[ \vec{v}(t) = \vec{u} + \int \vec{a} \, dt \]
Calculating the integral:
\[ \int \vec{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = \left( t^2 \hat{i} + t^3 \hat{j} \right) + \vec{C} \]
where \( \vec{C} \) is the constant of integration.

At \( t = 0 \):
\[ \vec{v}(0) = 2 \hat{i} + 0 \hat{j} = \vec{u} \]
Thus, the constant of integration \( \vec{C} = 2 \hat{i} \).

So, the velocity function becomes:
\[ \vec{v}(t) = (2 \hat{i} + t^2 \hat{i} + t^3 \hat{j}) = (2 + t^2) \hat{i} + t^3 \hat{j} \]

### Step 3: Calculate the velocity at \( t = 1 \) second.
Substituting \( t = 1 \) into the velocity function:
\[ \vec{v}(1) = (2 + 1^2) \hat{i} + (1^3) \hat{j} = (2 + 1) \hat{i} + 1 \hat{j} = 3 \hat{i} + 1 \hat{j} \, \text{m/s} \]

### Step 4: Find the position as a function of time.
To find the position, we integrate the velocity function:
\[ \vec{r}(t) = \vec{r_0} + \int \vec{v}(t) \, dt \]
Calculating the integral:
\[ \int \vec{v}(t) \, dt = \int ((2 + t^2) \hat{i} + t^3 \hat{j}) \, dt = \left( 2t + \frac{t^3}{3} \right) \hat{i} + \left( \frac{t^4}{4} \right) \hat{j} + \vec{C'} \]
where \( \vec{C'} \) is the constant of integration.

At \( t = 0 \):
\[ \vec{r}(0) = 0 \hat{i} + 0 \hat{j} = \vec{r_0} \]
Thus, \( \vec{C'} = 0 \).

So, the position function becomes:
\[ \vec{r}(t) = \left( 2t + \frac{t^3}{3} \right) \hat{i} + \left( \frac{t^4}{4} \right) \hat{j} \]

### Step 5: Calculate the position at \( t = 1 \) second.
Substituting \( t = 1 \) into the position function:
\[ \vec{r}(1) = \left( 2(1) + \frac{1^3}{3} \right) \hat{i} + \left( \frac{1^4}{4} \right) \hat{j} = \left( 2 + \frac{1}{3} \right) \hat{i} + \left( \frac{1}{4} \right) \hat{j} \]
\[ = \left( \frac{6}{3} + \frac{1}{3} \right) \hat{i} + \left( \frac{1}{4} \right) \hat{j} = \frac{7}{3} \hat{i} + \frac{1}{4} \hat{j} \, \text{m} \]

### Final Results:
- Velocity at \( t = 1 \) second:
\[ \vec{v}(1) = 3 \hat{i} + 1 \hat{j} \, \text{m/s} \]
- Position at \( t = 1 \) second:
\[ \vec{r}(1) = \frac{7}{3} \hat{i} + \frac{1}{4} \hat{j} \, \text{m} \]