Question

Three particles start simultaneously from a point on a horizontal smooth plane. First particle moves with speed $v_1$ towards east, second particle moves towards north with speed $v_2$ and third-one moves towards north east. The velocity of the third particle, so that the three always lie on a line, is

• A. $\frac{v_1+v_2}{\sqrt{2}}$
• B. $\sqrt{v_1{v}_2}$
• C. $\frac{v_1{v}_2}{v_1+v_2}$
• D. $\sqrt{2}\frac{v_1{v}_2}{v_1+v_2}$

Answer 1

The correct option is D $\sqrt{2}\frac{v_1{v}_2}{v_1+v_2}$

Equation of line RS is $y=-mx+C$


$y=-\left(\frac{v_2}{v_1}\right)x+v_{2}t$
$v_{1}y=-v_{2}x+v_1{v}_{2}t$
Equation of line OP is
$y=x$

$\Rightarrow v_{1}x=-v_{2}x+v_1{v}_{2}t$
Point P is the point of intersection we get,
$x_p=y_p=\frac{v_1{v}_{2}t}{v_1+v_2}$
$\therefore OP=\sqrt{x_p^2+y_p^2}=\frac{\sqrt{2}{v}_1{v}_{2}t}{v_1+v_2}$
Hence, velocity of third particle is
$v_3=\frac{OP}{t}=\sqrt{2}\frac{v_1{v}_2}{v_1+v_2}$