Question

A particle starts its $2D$ motion from origin such that its velocity in $x-$direction remains constant at $3\text{m/s}$ and in $y-$direction its acceleration varies with time as $a_y=(t^2+2){\text{m/s}}^2$. Find the position vector of particle at $t=2\text{sec}$

• A. $\overrightarrow{r}=(6\hat{i}+5.33\hat{j})\text{m}$
• B. $\overrightarrow{r}=(6\hat{i}+1.33\hat{j})\text{m}$
• C. $\overrightarrow{r}=(6\hat{i}-5.33\hat{j})\text{m}$
• D. $\overrightarrow{r}=(6\hat{i}-1.33\hat{j})\text{m}$

Answer 1

The correct option is A $\overrightarrow{r}=(6\hat{i}+5.33\hat{j})\text{m}$

Given:
$v_x=3\Rightarrow \frac{dx}{dt}=3\Rightarrow \int dx=\int 3dt\Rightarrow x=3t$
Therefore, at $t=2\text{sec},x=6\text{m}$
Similarly, $a_y=t^2+2\Rightarrow \frac{d{v}_y}{dy}=t^2+2\Rightarrow \int d{v}_y=\int (t^2+2)dt\Rightarrow v_y=\frac{t^3}{3}+2t\Rightarrow \frac{dy}{dt}=\frac{t^3}{3}+2t\Rightarrow \int dy=\int (\frac{t^3}{3}+2t)dt\Rightarrow y=\frac{t^4}{12}+t^2$
At $t=2\text{sec}$,
$y=\frac{16}{12}+4=5.33\text{m}$
So, the position vector of the particle is $\overrightarrow{r}=(6\hat{i}+5.33\hat{j})\text{m}$