# Given initial positions and velocities of two boats, do they collide?

This is a homework question from a precalculus class that I'm a TA for.

Boat $$A$$ is initially at position $$(1,4)$$ and moves at a constant velocity $$\langle 3,5 \rangle$$. Boat $$B$$ is at position $$(7,2)$$ and moves at a constant velocity of $$\langle 1,10 \rangle$$. Do the paths of the boats ever cross? If so where? Will the boats collide?

I wanted to write up a thorough solution to this exercise for my class, and figured I'd post it online to help anyone else who may wander across it.

• What is it that you want your students to practice? The simplest solution (I think) is to move to a frame in which one of the boats is at rest at the origin and see if the resulting forward ray from the other boat reaches the origin, i.e., that the difference in positions is a negative multiple of the difference in velocities. – amd Oct 25 '18 at 1:20
• @amd That's probably the simplest solution computationally, but I don't think it's very intuitive. In the situation both boats are moving, so within the calculations both boats should be moving too. The students are struggling to become comfortable thinking in terms of vectors anyways, and I don't think they are ready to make the mental leap of shifting the frame of reference from the origin to one of the boats. I'll probably share this thought with any of the students who have a good grasp on working with vectors though, because it is a very useful thought. – Mike Pierce Oct 25 '18 at 14:51

Let $$P_A(a)$$ denote the position of the boat $$A$$ at time $$a$$, and let $$P_B(b)$$ denote the position of boat $$B$$ at time $$b$$. From the initial positions and velocities given, we have: \begin{align} P_A(a) &= (1,4) + a\langle 3,5 \rangle &\qquad P_B(b) &= (7,2) + b\langle 1,10 \rangle \\ &= (3a+1, 5a+4) &\qquad &= (b+7,10b+2) \end{align} Now these equations give the paths of the boats starting at time $$a=b=0$$. The paths of the boats cross only if at some time $$a$$ and some time $$b$$ after each starts moving they have the same position. In terms of those equations, the paths of the boats will cross if there are positive times $$a$$ and $$b$$ such that $$P_A(a) = P_B(b)$$. Now the boats collide if not only is there a location where their paths cross, but if they are at that location at the same time. So the boats collide if $$P_A(a) = P_B(b)$$ for some positive $$a$$ equal to $$b$$. So we can proceed by setting $$P_A = P_B$$: $$(3a+1, 5a+4) = (b+7,10b+2) \implies \begin{cases} 3a+1=5a+4 \\ b+7=10b+2 \end{cases}\ \implies \begin{cases} 3a-b=6 \\ 5a-10b=-2 \end{cases}\,,$$ This, being a system of linear equations, has at most a single solution, which we can calculate to be $$a = \frac{62}{25}$$ and $$b = \frac{36}{25}$$. These are both positive times, so the paths of the boats do cross, but since this is the only solution and $$a \neq b$$, the boats do not collide. To find the actual coordinates where they do cross will be the location of boat $$A$$ at time $$a=\frac{62}{25}$$ (which should equal the location of $$B$$ at time $$b=\frac{36}{25}$$ if we've done our calculations correctly), which we can calculate: $$P_A\left(\frac{62}{25}\right) = \left(3\cdot\frac{62}{25}+1, 5\cdot\frac{62}{25}+4\right) = \left(\frac{211}{25} , \frac{410}{25} \right)\,.$$
• I would probably write the functions $P_A$ and $P_B$ originally with $t$ as the argument to both, then switch to $t_1$ and $t_2$ when looking for whether they cross. – Toby Bartels Nov 6 '18 at 22:23