Don't Go Too Far

After time $t$ , $A(0,3t)$ and $B(0,40-4t)$ .

$\therefore AB^2=(3t)^2+(40-4t)^2$

Diffrentiating and equating to zero,

$2\times 3t\times3+2\times(40-4t)(-4)=0 \implies t=6.4$

$\displaystyle AB=\sqrt{(3\times6.4)^2+(40-4\times6.4)^2 } = \large \boxed{\color{#3D99F6}{24}}$

There is a way to solve this problem without using differentiation.

Take particle $A$ as static. Relatively, particle $B$ has a downward partial speed of $3\text{ m/s}$ . By similar triangles, $B$ will cross the $y$ -axis at $-30$ , by a displacement of $50$ . The shortest distance is when the line connecting $A$ to the equation of motion of $B$ is perpendicular to each other.

image

Let $x$ be the shortest distance. By the area of the triangle, we have

$\begin{aligned} \frac{1}{2}\times 30 \times 40 &= \frac{1}{2}\times 50 \times x \\ x&= 24\end{aligned}$