1. Given data

• Initial position: origin, so $x_0=0$, $y_0=0$

• Initial velocity: $\vec u = 3\hat i$ m/s

  Hence, $u_x=3\text{ m/s},\qquad u_y=0$

• Constant acceleration: $\vec a = 6\hat i + 4\hat j\text{ m/s}^2$

  Hence, $a_x=6\text{ m/s}^2,\qquad a_y=4\text{ m/s}^2$

2. Write the equation of motion in the y-direction

Using $y=u_y t + \frac{1}{2}a_y t^2$ we get $y=0\cdot t + \frac{1}{2}(4)t^2=2t^2$

Given that $y=32$ m, $2t^2=32$ $t^2=16$ $t=4\text{ s}$

(We take positive time since motion starts at $t=0$.)

3. Now find the x-coordinate at $t=4$ s

Using $x=u_x t + \frac{1}{2}a_x t^2$ we get $x=3t+\frac{1}{2}(6)t^2$ $x=3t+3t^2$

Substitute $t=4$ s: $x=3(4)+3(4^2)$ $x=12+3(16)$ $x=12+48=60$

So, $D=60\text{ m}$

4. Check options

• A: 40
• B: 32
• C: 50
• D: 60

Therefore, the correct option is D.