Solutions

Let $y_a(t)$ and $x_a(t)$ denote the position of particle $A$ on the vertical axis and the horizontal axis given time $t$, respectively. Likewise for $y_b(t)$ and $x_b(t)$ for particle $B$. We have $y_a(t)=0$ and $x_a(t)=3t+5$. Let $d(\alpha,\beta)$ be the distance between points $\alpha$ and $\beta$. We have

$$\begin{align*} d((x_b(t),y_b(t)),(0,0))^2&=x_b(t)^2+y_b(t)^2 \\ &=x_b(t)^2+(-\sqrt{3}x_b(t))^2 \\ &=4x_b(t)^2. \end{align*}$$

Therefore

$$d((x_b(t),y_b(t)),(0,0))=-2x_b(t),$$

which yields

$$d'((x_b(t),y_b(t)),(0,0))=-2x_b'(t),$$$$x_b(0)=-\frac{3}{2},$$$$x_b'(4)=-2.$$

Now, let $h(t)=d((x_a(t),y_a(t)),(x_b(t),y_b(t)))$. Then

$$\begin{align*} h(t)^2&=(x_a(t)-x_b(t))^2+(y_a(t)-y_b(t))^2 \\ &=(x_a(t)-x_b(t))^2+3x_b(t)^2. \end{align*}$$

and

$$2h(t)h'(t)=2(x_a(t)-x_b(t))(x_a'(t)-x_b'(t))+6x_b(t)x_b'(t).$$

Thus,

$$h'(t)=\frac{(x_a(t)-x_b(t))(x_a'(t)-x_b'(t))+3x_b(t)x_b'(t)}{\sqrt{(x_a(t)-x_b(t))^2+3x_b(t)^2}}.$$