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 step by step, we will first find the velocity of the particle at \( t = 1 \) second and then find the coordinates of the particle at the same time.

### Step 1: Understand the given information
We have the acceleration of the particle given as:
\[ \mathbf{a} = (2t \hat{i} + 3t^2 \hat{j}) \, \text{m/s}^2 \]
The initial velocity at \( t = 0 \) is:
\[ \mathbf{u} = 2 \hat{i} \, \text{m/s} \]
The particle starts from the origin, so the initial position is:
\[ \mathbf{r}_0 = 0 \hat{i} + 0 \hat{j} \]

### Step 2: Find the velocity as a function of time
Acceleration is the derivative of velocity with respect to time:
\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} \]
Thus, we can write:
\[ d\mathbf{v} = (2t \hat{i} + 3t^2 \hat{j}) dt \]
Integrating both sides from \( t = 0 \) to \( t = 1 \):
\[ \int_{\mathbf{u}}^{\mathbf{v}} d\mathbf{v} = \int_{0}^{1} (2t \hat{i} + 3t^2 \hat{j}) dt \]
This gives:
\[ \mathbf{v} - \mathbf{u} = \left[ \int_{0}^{1} 2t \, dt \hat{i} + \int_{0}^{1} 3t^2 \, dt \hat{j} \right] \]

Calculating the integrals:
\[ \int_{0}^{1} 2t \, dt = \left[ t^2 \right]_{0}^{1} = 1 \]
\[ \int_{0}^{1} 3t^2 \, dt = \left[ t^3 \right]_{0}^{1} = 1 \]

Thus, we have:
\[ \mathbf{v} - \mathbf{u} = (1 \hat{i} + 1 \hat{j}) \]
Substituting \( \mathbf{u} = 2 \hat{i} \):
\[ \mathbf{v} - 2 \hat{i} = 1 \hat{i} + 1 \hat{j} \]
\[ \mathbf{v} = (2 + 1) \hat{i} + 1 \hat{j} = 3 \hat{i} + 1 \hat{j} \]

### Step 3: Find the coordinates of the particle at \( t = 1 \)
The velocity is also the derivative of the position vector:
\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} \]
Thus:
\[ d\mathbf{r} = (2t \hat{i} + 3t^2 \hat{j} + 2 \hat{i}) dt \]
Integrating from \( t = 0 \) to \( t = 1 \):
\[ \int_{\mathbf{r}_0}^{\mathbf{r}} d\mathbf{r} = \int_{0}^{1} (2t + 2) \hat{i} + 3t^2 \hat{j} \, dt \]

Calculating the integrals:
\[ \int_{0}^{1} (2t + 2) \, dt = \left[ t^2 + 2t \right]_{0}^{1} = 1 + 2 = 3 \]
\[ \int_{0}^{1} 3t^2 \, dt = \left[ t^3 \right]_{0}^{1} = 1 \]

Thus:
\[ \mathbf{r} - \mathbf{r}_0 = (3 \hat{i} + 1 \hat{j}) \]
Since \( \mathbf{r}_0 = 0 \):
\[ \mathbf{r} = 3 \hat{i} + 1 \hat{j} \]

### Final Results
At \( t = 1 \) second, the velocity of the particle is:
\[ \mathbf{v} = 3 \hat{i} + 1 \hat{j} \, \text{m/s} \]
And the coordinates of the particle are:
\[ \mathbf{r} = (3, 1) \]