Question

A particle $A$ starts from rest and moves with an acceleration of $5{\text{m/s}}^2$ due East while another particle $B$ moves with a constant velocity of $5\text{m/s}$ due North from the same position. Then the magnitude of relative position of $A$ w.r.t $B$ at 2 second is

• A. $10\text{m}$
• B. $10\sqrt{2}\text{m}$
• C. $200\text{m}$
• D. $\frac{10}{\sqrt{2}}\text{m}$

Answer 1

The correct option is B $10\sqrt{2}\text{m}$

$\text{For}A:x_A=ut+\frac{1}{2}a{t}^2{x}_A=(0\times 2)+\frac{1}{2}\times 5\times 2^2=10\text{m}$
$\text{For}B:x_B=5\times 2=10\text{m}$
Relative position of $A$ w.r.t $B$, ${\overrightarrow{x}}_{AB}=x_A-x_B=10\hat{i}-10\hat{j}\text{m}$
$\left|{\overrightarrow{x}}_{AB}\right|=\sqrt{{10}^2+{10}^2}=10\sqrt{2}\text{m}$