Whether a square can be traversed in finite time

Another answer: To get from height $y$ to height $\frac y2$, you need to travel a distance of at least $\frac y2$ at speeds less than $y$. This takes at least half a second. So no matter how close you are to 0, you still have more than half a second left.


You’ve already found that the existence question can be reduced to going straight from $B$ to $A$. On that leg, the $y$ coordinate $y(t)$ follows the ordinary first-order linear differentional equation $y'=y$, with solution $y=c\,\mathrm e^{-t}$. If you start at $t=0$, that determines the constant to be $c=1$, so your $y$ coordinate is $y=\mathrm e^{-t}$. This decays to $0$ exponentially and doesn’t reach $0$ in finite time.


Even easier...

When you are $y$ units away from the $x$-axis, you are one second away at current speed, and this speed can only decrease. No matter how much you approach, you are still always more than one second away.

Tags:

Optimization

Sequences And Series

Ordinary Differential Equations

Definite Integrals

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