A particle starts from the origin at $$t = 0$$ with an initial velocity of $$3.0\hat{i}$$ m/s and moves in the $$x - y$$ plane with a constant acceleration $$(6.0\hat{i} + 4.0\hat{j})$$ m/s$$^2$$. The $$x-$$ coordinate of the particle at the instant when its $$y-$$ coordinate is $$32m$$ is $$D$$ meters. The value of $$D$$ is:

At the initial instant $$t = 0$$ the position vector of the particle is at the origin, so $$x_0 = 0$$ and $$y_0 = 0$$. The given initial velocity is $$\vec v_0 = 3.0\,\hat i$$ m s−1. This means

$$v_{0x} = 3.0\text{ m s}^{-1}, \qquad v_{0y} = 0\text{ m s}^{-1}.$$

The constant acceleration is $$\vec a = 6.0\,\hat i + 4.0\,\hat j$$ m s−2, so

$$a_x = 6.0\text{ m s}^{-2}, \qquad a_y = 4.0\text{ m s}^{-2}.$$

The kinematic equation for position under constant acceleration is first stated:

$$s = s_0 + v_0 t + \tfrac12 a t^2.$$

We will apply this formula separately to the y- and x-components.

y-component

Using $$y = y_0 + v_{0y} t + \frac12 a_y t^2$$ and substituting the known values, we obtain

$$y = 0 + 0\cdot t + \frac12 \,(4.0)\,t^2 = 2\,t^2.$$

The instant of interest is when $$y = 32\text{ m}$$, therefore

$$2\,t^2 = 32.$$

Dividing both sides by 2,

$$t^2 = 16.$$

Taking the positive square root (time cannot be negative),

$$t = 4\text{ s}.$$

x-component

Again using the same kinematic formula, now for the x-direction, we write

$$x = x_0 + v_{0x} t + \tfrac12 a_x t^2.$$

Substituting $$x_0 = 0,\; v_{0x} = 3.0\text{ m s}^{-1},\; a_x = 6.0\text{ m s}^{-2}$$ and the previously found $$t = 4\text{ s}$$ gives

$$x = 0 + 3.0\,(4) + \tfrac12\,(6.0)\,(4)^2.$$

Simplifying step by step, first the term with initial velocity:

$$3.0\,(4) = 12.$$

Next the term with acceleration:

$$\tfrac12\,(6.0)\,(4)^2 = 3.0\,(16) = 48.$$

Adding the two contributions to the displacement,

$$x = 12 + 48 = 60\text{ m}.$$

So the $$x$$-coordinate of the particle when its $$y$$-coordinate is 32 m is

$$D = 60\text{ m}.$$

Hence, the correct answer is Option C.