Question

A particle moves along the positive branch of the curve $y=2x$ where $x=t^2,x$ and $y$ are measured in metre and $t$ in second. At $t=2s$, the velocity (in $m/s$) of the particle is:

• A. $\hat{i}+2\hat{j}$
• B. $2\hat{i}+4\hat{j}$
• C. $4\hat{i}+2\hat{j}$
• D. $4\hat{i}+8\hat{j}$

Answer 1

The correct option is D $4\hat{i}+8\hat{j}$

Given, $x=t^2$ and $y=2x$
$\Rightarrow y=2t^2(\because x=t^2)$.
To find velocity $\overrightarrow{v}$ at $t=2\text{sec}$. Position of particle is given by
$\overrightarrow{r}=x\hat{i}+y\hat{j}$
Put $x=t^2,y=2t^2$
$\therefore \overrightarrow{r}=t^2\hat{i}+2t^2\hat{j}$
Velocity $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=2t\hat{i}+4t\hat{j}$
$Att=2\text{sec}$
$\overrightarrow{v}=(2\times 2)\hat{i}+(4\times 2)\hat{j}=4\hat{i}+8\hat{j}$
Final Answer: (D)