A particle starts from the origin at $t=0$ with an initial velocity of $3.0 \hat{i} \mathrm{~m} / \mathrm{s}$ and moves in the $x-y$ plane with a constant acceleration $(6.0 \hat{i}+4.0 \hat{j}) \mathrm{m} / \mathrm{s}^{2}$. The $x$-coordinate of the particle at the instant when its $y$-coordinate is $32 \mathrm{~m}$ is $\mathrm{D}$ meters. The value of $\mathrm{D}$ is:
Correct Option: , 3
(3) Using $S=u t+\frac{1}{2} a t^{2}$
$y=u_{y} t+\frac{1}{2} a_{y} t^{2}$ (along $y$ Axis)
$\Rightarrow \quad 32=0 \times t+\frac{1}{2}(4) t^{2}$
$\Rightarrow \frac{1}{2} \times 4 \times t^{2}=32$
$\Rightarrow t=4 \mathrm{~s}$
$S_{x}=u_{x} t+\frac{1}{2} a_{x} t^{2}$ (Along $x$ Axis)
$\Rightarrow x=3 \times 4+\frac{1}{2} \times 6 \times 4^{2}=60$
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