### by Howard Williams

#### Published Sunday November 27 2022 (link)

On rugby international morning, I found myself, along with eight friends, in a pub 5.8 miles from the match ground. We were enjoying ourselves, and so wished to delay our departure for the ground until the last possible minute. The publican, wishing to keep our custom for as long as possible, offered to help us get there by carrying us, one at a time, as pillion passengers on his motorbike.

We could walk at 2.5mph and the bike would travel at 30mph. We all left the pub together, and arrived at the ground in time for kick-off.

Ignoring the time taken getting on and off the bike, what was our minimum travelling time in minutes?

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In order to minimise the total time taken for the friends to travel from the pub to the playing ground, it is important that the barman and all the friends start moving at the same time and then move continuously at their respective speeds until they all arrive simultaneously at the playing ground.

Let the distance from the pub to the ground be $$d$$, the the pedestrian walking speed of the $$n$$ friends be $$p$$ and the speed of the barman on the motorbike be $$b$$.  If the friends each walk a distance $$w$$ the time taken to complete their journey is $t=\frac{w}{p}+\frac{d-w}{b}$ The barman will travel backwards and forwards a total of $$(2n-1)$$ times and will hence travel $$(2n-1)$$ times the distance $$d$$ less the total distance travelled by all the friends ($$2nw$$).  This will hence take a time $$t$$ given by:$t=\frac{(2n-1)d-2nw}{b}$Equating these two times and solving for $$w$$ now gives:$w=\frac{2(n-1)dp}{(2n-1)p+b}=\frac{(k-1)dp}{kp+b}$ where $$k=2n-1$$. Finally we can substitute for $$w$$ in the first equation to find the minimum total time for transfer:$t_{min}=\left(\frac{kb+p}{kp+b}\right)\left(\frac{d}{b}\right)$