Henry Ernest Dudeney/Puzzles and Curious Problems/72 - The Fly and the Motor-Cars/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $72$

The Fly and the Motor-Cars
A road is $300$ miles long.
A motor-car, $A$, starts at noon from one end and goes throughout at $50$ miles an hour,
and at the same time another car, $B$, going uniformly at $100$ miles an hour, starts from the other end,
together with a fly travelling $150$ miles an hour.
When the fly meets the car $A$, it immediately turns and flies towards $B$.
$(1)$ When does the fly meet $B$?
The fly then turns towards $A$ and continues flying backwards and forwards between $A$ and $B$.
$(2)$ When will the fly be crushed between the two cars if they collide and it does not get out of the way?


Solution

$(1): \quad$ At $13:48$, at which point they are $120$ miles from $A$.
$(2): \quad$ At $14:00$, at which point they are $100$ miles from $A$.


Proof

Let $d_1$ miles from $A$ be the point at which the fly first meets $A$.

Let $d_2$ miles from $A$ be the point where $B$ is at that time.

Let $t_1$ hours after noon be the time at which this happens.


Let $d_3$ miles from $A$ be the point at which the fly then meets $B$.

Let $t_2$ hours after noon be the time at which this happens.


First we investigate where everybody is when the fly first meets $A$.

We have:

\(\text {(1)}: \quad\) \(\ds d_1\) \(=\) \(\ds 50 t_1\) A motor-car, $A$, starts at noon from one end and goes throughout at $50$ miles an hour,
\(\text {(2)}: \quad\) \(\ds 300 - d_2\) \(=\) \(\ds 100 t_1\) and at the same time another car, $B$, going uniformly at $100$ miles an hour, starts from the other end,
\(\text {(3)}: \quad\) \(\ds 300 - d_1\) \(=\) \(\ds 150 t_1\) together with a fly travelling $150$ miles an hour.
\(\ds \leadsto \ \ \) \(\ds 300\) \(=\) \(\ds \paren {50 + 150} t_1\) $(1) + (2)$
\(\ds \leadsto \ \ \) \(\ds t_1\) \(=\) \(\ds \dfrac {300} {200}\)
\(\ds \) \(=\) \(\ds 1 \tfrac 1 2\)
\(\ds \leadsto \ \ \) \(\ds d_1\) \(=\) \(\ds 75\) substituting for $t$ in $(1)$ and simplifying
\(\ds \leadsto \ \ \) \(\ds 300 - d_2\) \(=\) \(\ds 100 \times \dfrac 3 2\) substituting for $t$ in $d_2$
\(\ds \leadsto \ \ \) \(\ds d_2\) \(=\) \(\ds 300 - 150 = 150\)

So when the fly first meets $A$, they are $75$ miles from $A$, while $B$ is $150$ miles from $A$ (which also happens to be $150$ miles from $B$).

This happens at $1 \tfrac 1 2$ hours after noon


Now we investigate when the fly first meets $B$.

At this stage we are not interested in what happens to $A$, just $B$ and the fly.

We have:

\(\text {(4)}: \quad\) \(\ds 150 - d_3\) \(=\) \(\ds 100 \paren {t_2 - \dfrac 3 2}\) ... $B$, going uniformly at $100$ miles an hour ...
\(\text {(5)}: \quad\) \(\ds d_3 - 75\) \(=\) \(\ds 150 \paren {t_2 - \dfrac 3 2}\) ... a fly travelling $150$ miles an hour ...
\(\ds \leadsto \ \ \) \(\ds 150 - 75\) \(=\) \(\ds \paren {100 + 150} \paren {t_2 - \dfrac 3 2}\) $(4) + (5)$
\(\ds \leadsto \ \ \) \(\ds t_2\) \(=\) \(\ds \dfrac 3 2 + \dfrac {75} {250}\)
\(\ds \) \(=\) \(\ds 1 \tfrac 4 5\)
\(\ds \leadsto \ \ \) \(\ds d_3\) \(=\) \(\ds 75 + 45 = 120\) substituting for $t_2$ into $(5)$ and simplifying


So the fly meets $B$ at $120$ miles from $A$, which happens at $1 \tfrac 4 5$ hours after noon, or at $13:48$.

$\Box$


Let $d_4$ miles from $A$ be the point at which the cars collide.

Let this happen at $t_3$ hours after noon.

We have:

\(\text {(6)}: \quad\) \(\ds d_4\) \(=\) \(\ds 50 t_3\) A motor-car, $A$, starts at noon from one end and goes throughout at $50$ miles an hour,
\(\text {(6)}: \quad\) \(\ds 300 - d_4\) \(=\) \(\ds 100 t_3\) and at the same time another car, $B$, going uniformly at $100$ miles an hour, starts from the other end
\(\ds \leadsto \ \ \) \(\ds 300\) \(=\) \(\ds \paren {50 + 100} t_3\) $(6) + (7)$
\(\ds \leadsto \ \ \) \(\ds t_3\) \(=\) \(\ds 2\)
\(\ds \leadsto \ \ \) \(\ds d_4\) \(=\) \(\ds 100\)

Thus $A$ and $B$ collide at $2$ hours past noon, $100$ miles from $A$.

We do not need to calculate the path of the fly, despite the fact that the naïve reader may be tempted into trying.

$\blacksquare$


Also see


Sources

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