Three particles start simultaneously from a point on a horizontal… - Vidyadip

MCQ

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
$\frac{v_1 v_2}{v_1+v_2}$
✓
$\sqrt{2} \frac{v_1 v_2}{v_1+v_2}$

✓

Answer

Correct option: D.

$\sqrt{2} \frac{v_1 v_2}{v_1+v_2}$

d
(d)

Equation of line $R S$ is $y=-m x+C$

Or $y=-\left(\frac{v_2}{v_1}\right) x+v_2 t$

or $\quad v_1 y=-v_2 x+v_1 v_2 t$

Equation of line $O P$ is

$y=x$

Point $P$ is the point of intersection, we get

$x_P=y_P=\frac{v_1 v_2 t}{v_1+v_2}$

$O P=\sqrt{x_P^2+y_P^2}$

$=\frac{\sqrt{2} v_1 v_2 t}{v_1+v_2}$

$v_3=\frac{O P}{t}=\frac{\sqrt{2} v_1 v_2}{v_1+v_2}$

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