# Hint to Puzzle #53: The train, the Station and the Bee.

53. 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 where up on 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?

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?

This puzzle for the first part at least is a lot easier than it sounds. We know the speed of the Bee, so what is the one thing we need to work out the distance it travels? Seriously it's that easy.

For the second part you will need your equations of motion:

v = u + at

s = (u + v)t/2

s = ut + ½at

v

And this graph will give you an idea of what is going on.

Good Luck.

Where next?

Questions Answer

For the second part you will need your equations of motion:

v = u + at

s = (u + v)t/2

s = ut + ½at

^{2}v

^{2}= u^{2}+ 2asAnd this graph will give you an idea of what is going on.

Good Luck.

Where next?

Questions Answer

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