One point following another moving in a straight line? There is a plane with two points on it, let's say A and B.
A starts at an arbitrary constant point, let's say $(0, 0)$, and $B$ at a point that needs to be tested, which we'll call $(c, d)$.
A moves upwards in a straight line at speed $V$. At any given point in time, $B$ will move at that same speed $V$ from wherever it is at that moment towards wherever $A$ is at the same time until $B$ contacts $A$, if that will ever happen at all. In what range of points must $B$ start in order to ever catch $A$?
I've determined (at least I think) that, for $B$, $$\frac{dy}{dx} = \frac{y(A) - y(B)}{x(A) - x(B)} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ and that if $B$ starts below $A$, or is ever at a point below $A$, it will never reach $A$. Also, for $A$, $\frac{dy}{dx} = \frac{V}{0}$ which is undefined, though $\frac{dy}{dt} = V$, and $\frac{dx}{dt} = 0$. Also, for $A, x = 0$ and $y = Vt$, where $t$ is time.
How can one determine if $B$ will reach $A$, given a starting point $(c, d)$? Can equations, likely using integrals, be formed to show the $x$ and $y$ values of $B$ as functions of $t$? If so, how?
 A: The points meet at some time $t \geq 0$ if and only if the initial position of $B$ is $(0,y_0)$ for some $y_0 \geq 0$. In other words, $B$ must start directly above $A$.
We may assume that $V=1$ since we can speed up or slow down time by a constant factor. (If rigor is desired here we can use what Jack Lee calls the "rescaling lemma" in Introduction to Smooth Manifolds (Lemma 9.3). That book is way overkill for this problem, but I don't know a more appropriate reference.)
Let $(x_B(t), y_B(t))$ be the position vector of $B$ at time $t \geq 0$. The corresponding velocity vector is
$$
(x_B', y_B')
= \left(\frac{-x_B}{\sqrt{x_B^2 + (y_B - t)^2}}, \frac{-(y_B - t)}{\sqrt{x_B^2 + (y_B - t)^2}}\right).
$$
My approach will be to make the change of coordinates $\tilde y = y - t$. This corresponds to transforming to a coordinate system that moves with $A$ so that $A$ remains fixed at the origin. The velocity of $B$ in this new coordinate system is
$$
(x_B', \tilde y_B')
= (x_B', y_B' - 1)
= \left(\frac{-x_B}{\sqrt{x_B^2 + \tilde y_B^2}},
        \frac{-\tilde y_B}{\sqrt{x_B^2 + \tilde y_B^2}} - 1 \right).
$$
As the velocity only depends explicitly on the position, the trajectory $(x_B(t), \tilde y_B(t))$ is an integral curve of the vector field
$$
F(x,\tilde y)
= \left(\frac{-x}{\sqrt{x^2 + \tilde y^2}},
        \frac{-\tilde y}{\sqrt{x^2 + \tilde y^2}} - 1 \right)
$$
on $\mathbb R^2 \setminus \{(0,0)\}$. Note that


*

*Each integral curve exists either (i) until it hits the origin or (ii) forever (this can be deduced from the fact that $F$ is bounded and continuous, using among other tools the Peano existence theorem); and

*For $x \neq 0$, the ratio $\displaystyle\frac{F_2}{F_1} = \frac{\tilde y + \sqrt{x^2 + \tilde y^2}}{x}$ of the components of $F$ (i.e. the slope of $F$) is constant along each radial ray of constant polar angle, and is positive iff $x > 0$ and negative iff $x < 0$.


Now, if $B$ starts directly above $A$ then they will obviously collide. And as you noted, if $B$ is ever side-by-side with or below $A$ it will never catch up.
The remaining case is the one where $x_B(0) \neq 0$ and $\tilde y_B(0) > 0$. Because $\tilde y'_B(t) < -1$ when $\tilde y_B(t) > 0$, the trajectory reaches the line $\tilde y = 0$ in finite time. Let $T > 0$ be the earliest such time.
Assume for the sake of contradiction that $(x_B(T), \tilde y_B(T)) = (0,0)$. We define sequences $(t_n)_{n \geq 0}$ of times and $(u_n)_{n \geq 0}$ of $\tilde y$-coordinates (plus, for convenience, $x_n = x_B(t_n)$ and $\tilde y_n = \tilde y_B(t_n)$) as follows.
The initial conditions will be $t_0 = 0$ and $u_0 = \tilde y_B(0)$.
Next, suppose $t_n < T$ and $u_n > 0$ are such that $(x_n, \tilde y_n)$ and $(x_0, u_n)$ are at the same polar angle. In particular we have $$\operatorname{sgn} x_n = \operatorname{sgn} x_0 \neq 0,$$ where $\operatorname{sgn}$ is the sign function. Choose a $t_{n+1} \in (t_n, T)$ such that
\begin{equation} \tag{*} \label{ratio}
\frac{\tilde y_n}{x_n}
= \frac{F_2(x_{n+1}, \tilde y_{n+1})}{F_1(x_{n+1}, \tilde y_{n+1})}
\equiv \frac{\tilde y_{n+1} + \sqrt{x_{n+1}^2 + \tilde y_{n+1}^2}}{x_{n+1}};
\end{equation}
such a time exists due to Cauchy's mean value theorem. Then
\begin{align*}
\operatorname{sgn} x_{n+1}
&= \operatorname{sgn}(\tilde y_n / x_n) && \text{(by item 2 of the earlier list since $x_{n+1} \neq 0$)} \\
&= \operatorname{sgn} x_n && \text{(since $t_n < T \implies \tilde y_n > 0$)} \\
&= \operatorname{sgn} x_0.
\end{align*}
Thus, there is a (unique) $u_{n+1} > 0$ such that $(x_0, u_{n+1})$ has the same polar angle as $(x_{n+1}, \tilde y_{n+1})$. This finishes the construction.
However, observe that by \eqref{ratio} and item 2,
$$
\frac{u_n}{x_0}
= \frac{u_{n+1} + \sqrt{x_0^2 + u_{n+1}^2}}{x_0},
$$
which implies $u_{n+1} < u_n - \lvert x_0 \rvert$. This means that $u_n < u_0 - n \lvert x_0 \rvert$ for all $n > 0$, contradicting the fact that the $u_n$ are all positive.
Thus even in this last case $B$ eventually falls side-by-side with $A$, after which $B$ is condemned to make a futile chase from below forever.
A: A starts from (0,0)and B starts from (c,d)and both A and B move at a speed of V. After time t,say, B catches up with A at a point, say, (a,b).Then (a,b) is at a distance Vt from the origin and from (c,d).
Therefore, B can start from any point on the circle centered (a,b) and passing through the origin in order that it can meet A at (a,b). 
(a,b) is at a distance of Vt both from the origin and from (c,d).
A: If B chases A with higher speed then eventually it can meet or overtake A. Else never.
This comes from time t = Distance to be covered/ relative speed.
