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I was thinking about this when flying on the plane which was approaching and slowing down.

Assume an object is approaching its target which is at a certain initial distance d at time t0.

It starts at a speed that will allow it to reach the target in exactly one hour (e.g. d=100km, it starts at 100 km/h).

Incrementally, it will slow down so that at every point in time, it will be exactly one hour far from reaching the target (after it had travelled 40 km being 60 km away, it will be travelling at 60 km/h).

After reaching a predefined minimum speed (10 km/h), it will keep its velocity constant (and it will need exactly one additional hour to reach the target).

How long will it take to reach the target?

I somehow assume the answer should be 2 hours, independently of initial distance, but it does not fit (because then it would not matter what the original distance is (which it should not since we are travelling faster in the beginning), but any shorter path is included in the longer path and the resulting equation cannot possibly be true (or can it be?))

I think I am missing the mathematical apparatus that is needed to solve this (is it differential equations?)

Can you please advise how to solve this?

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up vote 6 down vote accepted

Until the final speed, if $x$ is the distance to the target, we have $\frac {dx}{dt}=-x$. This is solved by $x=x_0e^{-t}$ where $x_0$ is the distance at time $0$. Then we look how long it takes to get to $10$, so we solve $10=100e^{-t}, -t=\ln \frac 1{10}, t=\ln 10$. Then, as you say, it takes an hour to get there so the total time is $1+\ln 10 \approx 3.303$ hours

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Note this text in the question: "After reaching a predefined minimum speed (10 km/h), it will keep its velocity constant (and it will need exactly one additional hour to reach the target)." – gt6989b Nov 15 '12 at 17:17
@gt6989b: missed that. I will fix – Ross Millikan Nov 15 '12 at 17:21
Thank you for the answer! If I understand this correctly, in generic terms, can we say that for the slowdown part: t = ln (x0/vf) where vf is the final speed? – Marek Nov 16 '12 at 8:30
@Marek: Yes. It would be little better to say $t = \ln \frac {v_0}{v_f}$ as that makes the units match. We have hidden a constant that has numeric value 1, but does have units of hr$^{-1}$ – Ross Millikan Nov 16 '12 at 14:05

Split the problem into two parts: the time it takes to get from 100 to 10 km, and from 10 to 0 km. Second part is trivial, so I'll focus on the first.

For the first part, let $x(t) = d(t) - 10$, so that when you're at the 10 km mark, $x = 0$. Then, since $v(t)=-d(t)$, $v(t)=-x(t)-10$. As a differential equation: $\frac{dx}{dt}=-x-10$.

I don't remember much from my ODE course other than this is a first-order differential equation, but google lead me to this. After relabeling the variables, you get $P=1$, $Q=-10$, and an integrating factor of $M(t)=e^t$.

This produces $x(t)=\frac{Q M(t) + C}{M(t)}$. Solve for $C$ using initial condition $x(0)=d(0)-10=90$, which gives you $C=100$ and $x(t)=100e^{-t}-10$.

To solve for the time it takes to travel the first 90 km, solve for $t$ when $x(t)=0$, which gives you $\ln 10$.

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thank you for the answer including steps and also for the link, it helped me understand the steps involved in the solving a little better even though I know nothing about ODE. – Marek Nov 16 '12 at 8:32

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