2 cars and the fly problem - calculate the total distance covered by the fly

Question

2 cars approach each other, with 20 km between them. The speed of each car is 10 kmph. At 20 km apart from each, a fly starts traveling from one car towards another at 15 kmph. Once it reaches the other car, it turns back and starts towards the first car. It continues to do this until the cars meet/collide. How much distance does the fly cover in total?

The answer is fairly simple for this one. However, this can be solved by summing the infinite geometric series as well, by considering relative motion of the fly w.r.t. one or the other car and generating a geometric series of distances which can then be summed.

I am trying to solve this question by assuming the relative motion of one car and the fly w.r.t. only car, instead of switching between the cars. But I am unable to solve it this way. I don't understand the issue with this approach.

Would really appreciate it if somebody could help me understand what I'm doing wrong.

Edit: details of my approach

At the moment the fly starts from train A, i assume the frame of reference of train B which is moving towards A and the fly.

Relative velocity of fly = 25 kmph Relative velocity of train A w.r.t. B = 20 kmph

When the frame of reference of train B was assumed (to calculate relative velocity for ease of calculation), the fly had just started from A and the distance between them was 20 km.

Hence the distance covered by fly in first trip until it reaches train B = 20 km

Time taken by fly to cover the above distance= 20/25 hr = 4/5 hr

Distance covered by train A during this time = (4/5)*20 = 16 km

Distance left between train B (and fly which is now at train B) and train A = 4 km

Assuming the frame of reference of train B again as the fly starts moving towards train A -

Relative speed of fly w.r.t. train B = 5 kmph (15-10)

Relative speed of train A w.r.t B = 20 kmph

The distance between fly and train A = 4 km

Both meet at a point which divides this 4 km in 1:4 ratio since the ratio of relative speed of fly w.r.t. B and relative speed of A w.r.t. B is 1:4 (5 kmph : 20 kmph). Hence, fly covers 4/5 km (4/(1+4))

And so on and so forth. However through this method, i am unable to get to the correct answer which is 15 km. The fly's first journey itself, from A to B, is 20 km (> 15 km) in my approach. I am just not able to understand what I'm overlooking.

Kindly help with your analysis of my mistaken approach and please let me know where am I exactly messing it up.

Answer 1

The trouble with frame of reference to a car is that when the fly leaves a car and the distance between them is $D$, the fly won't fly the entire distance $D$ because the other car will have moved toward the fly in that time. But if we fix the car then how do we take that into account if we think of $B$ as still?

If the car $B$ is still then the ground is moving under the fly at $10$kmh toward car $B$ so although the fly did fly $20$km in $t$ but the ground moved $10t$ so the distance of the fly over the ground is $20 -10t$.

So to begin with. The fly is flying at $25$kmh so it takes $\frac {20}{25}=\frac 45$hr for the fly to reach the train. And the fly flew $20$km but the ground moved $10\times \frac 45= 8$km so the fly only flew $20-8=12$ over the ground. Meanwhile train $A$ and $20$kmh has got $20\times \frac 45 = 16$km and the trains are now $20-16=4$ km apart.

We need to have the fly fly back. The fly flies away from $B$ at $5$kmh while train $A$ comes toward and $20$kmh. So the time it takes for the fly to meet the train is $\frac {4}{20+5}=\frac 4{25}$hr. In this time the fly flew $5\times 4{25}= \frac 45$ km but the ground moved $10\times \frac 4{25}=\frac 85$ so the fly *over ground flew $\frac {12}5$km. As that to the $12$ earlie and the fly has flowe $12 + \frac {12}5$ km.

Meanwhile in $\frac 4{25}$hr train $A$ and move $20\times \frac 4{25}=\frac {16}5$ and the trains are now $4 - \frac {16}5= \frac {4}5$km apart.

Compared to the original $20$ km this is $\frac {\frac {4}5}{20}= \frac {1}{25}$ the original distance. Each leg the distance is $\frac {1}{25}$ of the one before.

So the fly over ground will fly $(12+\frac {12}5)\sum (\frac {1}{25})^k = \frac{72}5 \times \frac 1{\frac {24}{25}} =\frac {72\times 25}{5\times 24}=15$km.

....

But I wouldn't do it with a fixed $B$.

.....

If we do it with the mid point and crash as the center.

We have the speed of fly plus speed of train B is $25$kmp. So first leg is $\frac {20}{25}=\frac 45$. So the fly fly $15\times \frac 45 = 12$.

Trains are now $20 - (10+10)\times 45= 4$km apart. Now by symmetry everything is as before but we have only to do $\frac 15$ the distance.

So the answer if $12 \times \sum \frac 1{5}^2 = 12\times \frac 1{\frac 45}=15$km.

Much simpler.

Answer 2

To sum the series, here is how I would approach it...

The speed of the fly is $15 \frac kh$ the speed of each car is $10 \frac kh$ at each leg the fly covers $\frac 35$ of the distance between it and the oncoming car before they meet.

In the meantime, the trailing car closes the distance between the two cars. Each leg is $\frac {1}{5}$ the previous leg.

$(20)(\frac 35)\sum_\limits{n=0}^\infty (\frac 15)^n\\ (20)(35)\left(\frac {1}{1-\frac 15}\right)\\ (20)(\frac 35)(\frac 54) = 15$

Now, if we put this in a moving frame of reference such that one car is stationary, the speed of the fly is $25 \frac kh$ and the trailing car is $20\frac kh$

The fly covers the distance of the first leg in $\frac 45$ hour. The second leg is $\frac 15$ the first leg.

Total time is $\frac 45 \sum (\frac 15)^n = (\frac 54)(\frac 54) = 1$ hour.

Transition back to the original frame of reference, the fly travels $15$ km.