Train and horseman velocity

Question

This is a more or less easy exercise but there is one point I do not understand.

We have a train starting at $(0|0) $with velocity $v_t=20 m/s$ straight in the $y$-direction and a horseman at $H=(100|100)$ with velocity $v_h=15 m/s$

There are two questions now:

1) In which direction does the horseman need to ride to meet the train at the same time on the line of the train.

2) What is the minimum velocity for the horseman to meet the train? In which direction does he have to go now?

You can draw a picture and will immediately see two things (to solve 1):

$(y-100)^2+100^2=x^2$ where is y the lenght from $0$ to the meeting point $M=(0|y)$ on the $y$-axis and $x$ is the lenght fromthe horseman to the meeting point. We also see that $\frac{x}{y}=\frac{15}{20}$. Now we can solve the two equations and get $y$ and $x$, the direction is therefore $HM$.

For the second part I have $\frac{\sqrt{(y-100)^2+100^2}}{20}=\frac{x}{v_{min}}$. Now I want to get somehow an equation like $x_{1,2}=...$ where only $v_{min}$ appears. Setting the root inside to 0 will lead to $v_{min}$, but how?

Answer 1

For the second, the distance the train covers is $y$ and the time is $\frac y{20}$. The horse covers $\sqrt{(y-100)^2+100^2}$ in time $\frac {\sqrt{(y-100)^2+100^2}}{v_h}$. Setting these equal we have $\frac y{20}=\frac {\sqrt{(y-100)^2+100^2}}{v_h}$ or $v_h=\frac {20\sqrt{(y-100)^2+100^2}}{y}$. Now you can take the derivative with respect to $y$ and get home.

Answer 2

Let $P$ be the initial position of the horse, let $N$ be the nearest point to $P$ on the track, and let $M$ be the place where horse and train will meet. Let $\theta=\angle MPN$. Note that $PN=100$. Let $s$ be the speed of the horse.

The train will have travelled $100+100\tan\theta$, taking time $5+5\tan\theta$. In that time the horse travels through a distance $100\sec\theta$. It follows that $$(5+5\tan\theta)s=100\sec\theta.$$ Thus $$\frac{1}{s}=\frac{1}{20}\left(\sin\theta+\cos\theta\right).$$ We want to maximize the right-hand side, given that $0\le \theta\le \pi/2$.

Equivalently, we want to maximize $(\sin\theta+\cos\theta)^2$, which is $1+\sin 2\theta$, and reaches a maximum at $\theta=\frac{\pi}{4}$.