Question

Two particles, 1 and 2, move with constant velocities $v_1$ and $v_2$ along two mutually perpendicular straight lines toward the intersection point $O$. At the moment $t = 0$ the particles were located at the distances $l_1$ and $l_2$ from the point $O$. How soon will the distance between the particles become the smallest? What is it equal to?

Answer 1

• Time when distance is minimum: $$t_{\min}=\frac{v_1l_1+v_2l_2}{v_1^2+v_2^2}$$
• Minimum distance between the particles: $$d_{\min}=\frac{|l_1v_2-v_1l_2|}{\sqrt{v_1^2+v_2^2}}$$

Answer 2

Place the intersection point $O$ at the origin. Assume particle 1 moves along the $x$-axis and particle 2 along the $y$-axis toward $O$. At $t=0$, their positions are:

$$P_1(0)=(l_1,0),\quad P_2(0)=(0,l_2)$$

Since the particles are moving toward $O$ with speeds $v_1$ and $v_2$ respectively, their positions at time $t$ are:

$$P_1(t)=(l_1-v_1t,\,0),\quad P_2(t)=(0,\,l_2-v_2t)$$

The distance between them is:

$$d(t)=\sqrt{(l_1-v_1t)^2+(l_2-v_2t)^2}$$

To find the time when $d(t)$ is minimum, minimize the square of the distance:

$$D(t)= (l_1-v_1t)^2+(l_2-v_2t)^2$$

Differentiate $D(t)$ with respect to $t$ and set to zero:

$$\frac{dD}{dt}=-2v_1(l_1-v_1t)-2v_2(l_2-v_2t)=0$$ $$\Rightarrow v_1(l_1-v_1t)+v_2(l_2-v_2t)=0$$ $$\Rightarrow v_1l_1+v_2l_2=t(v_1^2+v_2^2)$$

Thus, the time of minimum distance is:

$$t_{\min}=\frac{v_1l_1+v_2l_2}{v_1^2+v_2^2}$$

Now, substitute $t_{\min}$ into the expressions for the positions:

$$l_1-v_1t_{\min} = l_1-\frac{v_1(v_1l_1+v_2l_2)}{v_1^2+v_2^2} = \frac{l_1(v_1^2+v_2^2)-v_1^2l_1-v_1v_2l_2}{v_1^2+v_2^2} = \frac{v_2(l_1v_2-v_1l_2)}{v_1^2+v_2^2}$$ $$l_2-v_2t_{\min} = \frac{v_1(l_2v_1-v_2l_1)}{v_1^2+v_2^2}$$

Hence, the minimum distance is:

$$d_{\min}=\sqrt{\left(\frac{v_2(l_1v_2-v_1l_2)}{v_1^2+v_2^2}\right)^2+\left(\frac{v_1(l_2v_1-v_2l_1)}{v_1^2+v_2^2}\right)^2}$$

Notice that $(l_1v_2-v_1l_2)^2$ appears in both terms:

$$d_{\min} = \frac{|l_1v_2-v_1l_2|}{v_1^2+v_2^2}\sqrt{v_2^2+v_1^2} = \frac{|l_1v_2-v_1l_2|}{\sqrt{v_1^2+v_2^2}}$$