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(D) Let the starting point be the origin $(0, 0)$.After time $t$,the position of the first particle is $(v_1 t, 0)$ and the second particle is $(0, v_

Vedclass · Accepted Answer

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

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(D) Let the starting point be the origin $(0, 0)$.After time $t$,the position of the first particle is $(v_1 t, 0)$ and the second particle is $(0, v_2 t)$.The equation of the line passing through these two points is $\frac{x}{v_1 t} + \frac{y}{v_2 t} = 1$,which simplifies to $v_2 x + v_1 y = v_1 v_2 t$.The third particle moves towards the north-east,meaning its velocity vector makes an angle of $45^\circ$ with the x-axis. Its position at time $t$ is $(v_3 t \cos 45^\circ, v_3 t \sin 45^\circ) = (\frac{v_3 t}{\sqrt{2}}, \frac{v_3 t}{\sqrt{2}})$.Since the third particle must lie on the same line,substitute its coordinates into the line equation:$v_2 (\frac{v_3 t}{\sqrt{2}}) + v_1 (\frac{v_3 t}{\sqrt{2}}) = v_1 v_2 t$.Dividing by $t$ and solving for $v_3$:$\frac{v_3}{\sqrt{2}} (v_1 + v_2) = v_1 v_2$.$v_3 = \frac{\sqrt{2} v_1 v_2}{v_1 + v_2}$.

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