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So I have a website with some job interview type questions on it. Yesterday I pulled a question out the depths of my memory:

A train enters a station at 50m/s where it starts to decelerate uniformly, coming to a complete stop at the end of the platform. The distance from the entrance to the station to the platform end is 500m. At the exact same moment as the train enters the station a Bee takes off from the buffers at the end of the platform and heads towards the train, when it reaches the front of the train it turns around and heads back to where it started. It continues to do this, start of the platform, front of the train, start of the platform back and forth until the train stops whereupon it is crushed.

If the speed of the Bee is 40m/s how far does the Bee travel? You can expand this question to determine, for example, the time of the first impact or, is there such a thing as (and if so what is it), a minimum speed the Bee needs to travel at in order to be able to perform this feat?

I know the first part is easy enough, classic puzzle and all that. But then i got carried away. It's the bit in bold I'm struggling with, i've been thinking about it for half a day and don't have an effective way of modelling it.

My thinking was this. Can we be sure that when the Bee reaches the train and turns around, flying in the same direction as the train that the Bee will be able to fly faster than the train is currently travelling. I thought this might be interesting because, the faster the Bee is travelling the further down the track it will be when it meets the train and therefore the faster the train will be going. Conversely the slower the Bee flies the slower the train will be going at that point where it meets the train. So modelling this might be interesting.

The question is here if anyone is interested. If you can help and I can use your answer I'll of course give you a mention.

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The acceleration of the train is $a_{T} = -\frac{5}{2}\mathrm{ms^{-1}}$.
Using basic calculus we can derive that:

$v_{T} = 50-\frac{5}{2}t$
$x_{T} = 50t-\frac{5}{4}t^2$

Let the speed of the bee be $v_{B}$. Then:

$x_{B} = 500-v_{B}t$

When the train first touches the bee, $x_{B} = x_{T}$ and, if the speed of the bee is to be minimal, it has to exactly equal the speed of the train, eg. $v_{B} = v_{T}$. Therefore we have:

$v_{B} = 50-\frac{5}{2}t$
$50t - \frac{5}{4}t^2 = 500-v_{B}t$

And, by substitution,

$50t - \frac{5}{4}t^2 = 500-(50-\frac{5}{2}t)t$

Solving gives $t = \frac{20}{3}$ or $t=20$. If $t=20$, $v_{B} = 0$, which makes sense - if the bee does not move, the train will have zero speed when it touches the bee. However, the more interesting solution is $t=\frac{20}{3}$, which gives $v_{B} = \frac{100}{3}\mathrm{ms^{-1}}$. This is the minimum speed of the bee.

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  • $\begingroup$ Excellent, that's the badger. If i can give you a link, or a cite or whatever, let me know. $\endgroup$ – user2633613 Oct 7 '15 at 11:56
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The average speed of the train during its deceleration is 25 m/s, since it started at 50 m/s and with uniform deceleration slowed to 0 m/s. During this time, the bee traveled at 40 m/s, which is $1.6$ times as fast as the average speed of the train. Thus, the bee traveled $1.6$ times as far, for a total of 800m.

The train traveled 500m during its deceleration, at an average speed of 25 m/s, so the time it took to stop is $t = 20s$. The acceleration is $a = -v_0/t = -2.5 ms^2$.

Using a bit of calculus or basic physics, we derive the formula for distance traveled under constant deceleration is $D = v_0 t + 1/2 a t^2$ (recall that $a$ is negative, in this case).

It takes 4 seconds for the train to decelerate to 40 m/s, so the distance traveled in that time is $D = 180m$. During that time, the bee traveled 160m. Combined, that is 340m, which is less than the distance from where the bee started to the entrance to the station. Thus, the train was definitely traveling at less than 40 m/s when the bee first reached it.

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