Question

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

Answer 1

Given, $\overrightarrow{u}=(3\hat{i}+0\hat{j})\text{m/s}$ and

$\overrightarrow{a}=(6\hat{i}+4\hat{j}){\text{m/s}}^2$


Using, $s=ut+\frac{1}{2}a{t}^2$, for $y-\text{axis}$

$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow 32=0\times t+\frac{1}{2}\left(4\right)t^2$

$\Rightarrow \frac{1}{2}\times 4\times t^2=32$

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

For $\text{x-axis}$,

$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$\Rightarrow x=3\times 4+\frac{1}{2}\times 6\times 4^2=60\text{m}$

$\therefore D=60$

Hence, $\left(C\right)$ is the correct answer.