# [Math] How to find direction of velocity V2 to reach an object travelling at velocity V1, such that it takes least time

calculus-of-variationsphysics

If an object A is currently at point P1 moving with constant velocity V1,
and there is another object, object B which currently at point P2 which can move with velocity v2, then what should be the direction of velocity vector V2, so that it reaches in minimum time possible?

### glossary

• P1,P2 – position vector of object A & object B respectively
• V1,V2 – velocity vector of object A & object B respectively
• v1,v2 – magnitude of V1 & V2 respectively

The question is with respect to 2D motion, from top view, no gravity or other force.

### Analogy

Consider a ball moving on ground with constant speed and I too have a ball which I can throw at a fixed speed. What should be the direction to reach the ball?

The notation in the problem statement seems misleading to me. We are given symbols $P_1,$ $P_2,$ $V_1,$ $V_2,$ $v_1,$ and $v_2,$ seemingly describing objects $1$ and $2$ in a completely symmetric fashion, whereas the problem statement actually is not symmetric.

The problem statement implies that $P_1,$ $P_2,$ $V_1,$ $v_1,$ and $v_2$ are all known, with $v_1 = \lVert V_1\rVert.$ The unknown entity is $V_2$; we have the constraint that $\lVert V_2\rVert = v_2,$ but as long as that constraint is satisfied, we can choose the direction of $V_2$ freely.

The mathematics of finding a suitable direction for $V_2$ were thoroughly hashed out in the answers to Intersection of two moving objects. There is an algebraic treatment which concludes with an equation that the angle $\phi$ (the direction of vector $V_2$) must meet in order to make Object $2$ meet Object $1.$ There are at most two solutions of the equation, but in some cases a solution of the equation gives a value of $\phi$ that causes Object $2$ to move away from Object $1$ (so they will never meet), while in other cases there are two values of $\phi$ that would cause the objects to meet, although only one of those values can be the one that makes them meet in the minimum possible time.

I find a geometric approach gives me more insight on the problem; in particular, once the geometry of the problem has been set up in this way (see this answer or this answer) it immediately gives us the equation $$v_2 \cos \phi = -v_1 \cos\theta,$$ where $\phi$ is the angle between $V_2$ and the line $P_2P_1$ and $\theta$ is the angle between $V_1$ and the line $P_2P_1,$ measuring both angles counterclockwise. (The algebraic treatment comes to an equivalent conclusion.) Another way to say this is:

In order for the two objects to meet, the component of $V_2$ perpendicular to the line between the objects must exactly cancel the component of $V_1$ perpendicular to the line between the objects.

This condition is necessary but not sufficient to solve the problem. A condition that is both necessary and sufficient is:

The vector $V_2 - V_1$ must point in the direction from $P_2$ to $P_1.$

Geometrically, we can construct representations of all possible vectors of the form $V_2 - V_1$ (given that $V_1$ is fixed and the magnitude but not the direction of $V_2$ is fixed) by drawing the vector $-V_1$ and then drawing a circle of radius $v_2$ around the tip of $V_1.$ We can then represent $V_2 - V_1$ by placing its tail at the tail of $-V_1$ and its head at any point on the circle; the direction from the center of the circle to the chosen point is the direction of $V_2.$

The diagram in this answer shows the case where $v_1 > v_2$ but where it is still possible to make the objects meet. In this case the circle intersects the line $P_1P_2$ in two places, giving two possible choices of a vector $V_2$ that will cause the objects to meet. The choice that causes the objects to meet in the minimum possible time is the choice for which the magnitude of $V_2 - V_1$ is larger. (In the diagram in the linked answer, that vector is $CG.$)

When $v_1 > v_2,$ it can also happen that there is no solution to the problem; object $2$ can only "catch" object $1$ if object $1$ is initially reducing its distance to object $2$ quickly enough.

When $v_2 \geq v_1,$ there is always a solution but only one solution for $V_2$ that allows object $2$ to "catch" object $1,$ so of course that is also the minimum-time solution. The solution is shown graphically in this answer.