The $x$ and $y$ co-ordinates of a particle at any time are $x=5t^2-3t+6$ and $y=10t^2+2t+3$ respectively, where $x$ and $y$ are in $\text{meters}$ and $t$ in $\text{seconds}$. Find the acceleration of the particle at $t=1\text{s}$.

The correct option is B $(10\hat{i}+20\hat{j}){\text{m/s}}^2$
Given,
$x$ co-ordinate of the particle: $x=5t^2-3t+6$
$y$ co-ordinate of the particle:$y=10t^2+2t+3$

To find:
acceleration $a$ at $t=1\text{s}$

We know that,
Velocity, $v=\frac{dr}{dt}[r\to \text{displacement}]$
Acceleration , $a=\frac{dv}{dt}$

Along $x$ - axis:
$v_x=\frac{dx}{dt}=\frac{d(5t^2-3t+6)}{dt}=(10t-3)\text{m/s}$

And $a_x=\frac{d{v}_x}{dt}=10{\text{m/s}}^2$

Similarly, along $y$ - axis :

$v_y=\frac{dy}{dt}=(20t+2)\text{m/s}$

And $a_y=\frac{d{v}_y}{dt}=20{\text{m/s}}^2$

$\therefore \overrightarrow{a}=a_x\hat{i}+a_y\hat{j}=(10\hat{i}+20\hat{j}){\text{m/s}}^2$

Since acceleration is uniform, so it will remain same at all the time.

Therefore, acceleration of the particle at $(t=1\text{s})=(10\hat{i}+20\hat{j}){\text{m/s}}^2$.

Hence, option $\left(\text{B}\right)$ is the right choice.