Question

Two particles A and B are moving in XY plane. Their positions vary with time $t$, according to relation
$x_A\left(t\right)=4t^2,x_B\left(t\right)=7$
$y_A\left(t\right)=3t,y_B\left(t\right)=3+4t^2$
Distance between these two particles at $t=1\text{s}$ is:-

• A. $5\text{m}$
• B. $3\text{m}$
• C. $4\text{m}$
• D. $\sqrt{12}\text{m}$

Answer 1

The correct option is A $5\text{m}$

Position of particle A is $\overrightarrow{r_A}=4t^2\hat{i}+3t\hat{j}$
Position of particle B is $\overrightarrow{r_B}=7\hat{i}+(3+4t^2)\hat{j}$
position of A wrt B is
$\overrightarrow{△r}=\overrightarrow{r_A}-\overrightarrow{r_B}$
$=(4t^2-7)\hat{i}+(3t-3-4t^2)\hat{j}$
At $t=1$
$\overrightarrow{|△r}{|}_{t=1}=-3\hat{i}-4\hat{j}$
$|\overrightarrow{△r}{|}_{t=1}=\sqrt{3^2+4^2}=5\text{m}$