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The problem I've been given to solve is read as follows.

Two-thirds of the way across a narrow railroad bridge, Willy Gope hears a train coming towards him. He knows that the train's speed is 45 miles per hour and that he has just enough time to reach safety at either end of the bridge. How fast can Willy run?

My solution is as follows.


Let $L$ represent the length of the bridge, $D$ the distance of the train from the beginning of the bridge, and $v$ Willie's velocity. We know that Willie can run away from the train for a distance of $2L/3$ to reach the other end of the bridge, and the time it will take him to do this is the same as the time it takes for the train to reach the end of the bridge, which is $(L+D)/(45 \mathrm{mph})$, so we can say that: $$v =\frac{2L/3}{(L+D)/(45 \mathrm{mph})} = \frac{30L}{D+L} \mathrm{mph}$$

We also now that Willy can run towards the train, for a distance of $L/3$, and the time it will take him to do this is the same as the time it will take for the train to reach the beginning of the bridge, which is $D/(45 \mathrm{mph})$, so we can say that: $$v = \frac{L/3}{D/(45 \mathrm{mph})} = \frac{15L}{D} \mathrm{mph}$$

Setting these two expressions for Willie's velocity equal to eachother yields: $$\frac{30L}{D+L} \mathrm{mph} = \frac{15L}{D} \mathrm{mph}$$ $$\frac{2}{D+L} = \frac{1}{D}$$ $$2D = D+L$$ $$D = L$$

So therefore his velocity is given by:

$$v = \frac{15L}{D} \mathrm{mph} = \frac{15L}{L} \mathrm{mph} = 15 \mathrm{mph}$$


Does anyone see any errors in my solution, especially since this problem doesn't initially seem like it has enough information? Does anyone have an alternative way to solve this problem that would be simpler, faster, or simply provide a fresh perspective?

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1 Answer 1

Your solution is correct. Here is another way to look at it (just to confirm):

Time to reach the far end is $$ t_f=\frac{D+L}{v_t}=\frac{2L}{3v}$$

Time to reach the closer end is $$t_c=\frac{D}{v_t}=\frac{L}{3v}$$

The first equation can be rewritten as $$\frac{D}{v_t}+\frac{L}{v_t}=\frac{2L}{3v}$$

But, from the second we have $\frac{D}{v_t}=\frac{L}{3v}$, so that $$\frac{L}{3v}+\frac{L}{v_t}=\frac{2L}{3v}$$

Dividing by $L$, we get $$v=\frac{v_t}{3}=15 mph$$

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