Starting from the origin at time t = 0, with initial velocity $$5\hat{j}$$ ms$$^{-1}$$, a particle moves in the $$x$$-$$y$$ plane with a constant acceleration of $$\left(10\hat{i} + 4\hat{j}\right)$$ ms$$^{-2}$$. At time t, its coordinates are (20 m, $$y_0$$ m). The values of t and $$y_0$$ are, respectively:

We are told that at the initial instant $$t = 0$$ the particle is at the origin, so its position vector is $$\vec r_0 = 0\hat i + 0\hat j$$. The initial velocity is given as $$\vec u = 0\hat i + 5\hat j\;{\rm m\,s^{-1}}$$. The acceleration is constant and equal to $$\vec a = 10\hat i + 4\hat j\;{\rm m\,s^{-2}}$$.

For motion with constant acceleration we use the kinematic equation for the position vector:

$$\vec r = \vec r_0 + \vec u\,t + \tfrac12 \vec a\,t^2.$$

Because the motion is in the plane, we treat the x and y components separately.

x-component

The initial $$x$$-coordinate is zero, and the initial $$x$$-velocity is also zero, so for the $$x$$-coordinate we have

$$x = 0 + 0\cdot t + \tfrac12(10)\,t^2.$$

Simplifying the right-hand side step by step,

$$x = \tfrac12 \times 10 \times t^2 = 5t^2.$$

The problem states that at time $$t$$ the particle is at $$x = 20\,{\rm m}$$. Hence

$$5t^2 = 20.$$

Dividing both sides by 5,

$$t^2 = 4.$$

Taking the positive square root (because time is positive),

$$t = 2\;{\rm s}.$$

y-component

Now we find the corresponding $$y$$-coordinate. The kinematic equation in the $$y$$ direction is

$$y = 0 + (5)\,t + \tfrac12(4)\,t^2.$$

First substitute the value $$t = 2\,{\rm s}$$ just obtained:

$$y = 5\,(2) + \tfrac12 \times 4 \times (2)^2.$$

Calculate each term carefully. The first term is

$$5 \times 2 = 10.$$

The second term involves several steps:

$$\tfrac12 \times 4 = 2,$$

and

$$(2)^2 = 4,$$

so

$$2 \times 4 = 8.$$

Adding the two contributions gives

$$y = 10 + 8 = 18\;{\rm m}.$$

Therefore $$y_0 = 18\;{\rm m}$$ when the particle is at $$x = 20\;{\rm m}$$.

We have now found both required quantities: $$t = 2\,{\rm s}$$ and $$y_0 = 18\,{\rm m}$$.

These match the first option in the list.

Hence, the correct answer is Option A.