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# Two trains each having a speed of $30{km}/{hr}\;$ are headed at each other on the same track. A bird that can fly at $60{km}/{hr}\;$flies off one train, when they are 60 km apart and heads directly for the other train. On reaching the other train it files directly back to the first train and so forth. The number of trips which the bird can make from one train to the other before they meet isA. 6B. 8C. 10D. 12E. infinite number

Last updated date: 07th Aug 2024
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Hint: This question is a direct one. The total distance covered by the bird and the trains are calculated for the subsequent trips. If the result of the distance in-between trains after the subsequent trips does not approach the value equal to zero, it means that the bird makes an infinite number of trips.
Formula used:
$s=\dfrac{d}{t}$

From given, we have the data,
The speed of the bird is $60{km}/{hr}\;$
$=60\times \dfrac{5}{18}{m}/{s}\;$
Therefore, the speed of the bird is, $\dfrac{50}{3}{m}/{s}\;$
The speed of the trains is $30{km}/{hr}\;$
$=30\times \dfrac{5}{18}{m}/{s}\;$
Therefore, the speed of the bird is, $\dfrac{25}{3}{m}/{s}\;$
Now we will compute the distance between the trains after the first trip, that is, trip 1
Let the time taken by the bird to fly from train A to train B be denoted by ‘t’.
The total distance travelled by bird and train
$\dfrac{50}{3}\times t+\dfrac{25}{3}\times t=60$
Upon solving for ‘t’ we get,
$t=\dfrac{12}{5}s$
The distance travelled by the bird is,
$\dfrac{50}{3}\times \dfrac{12}{5}=40m$
The distance travelled by both train is,
$\dfrac{25}{3}\times \dfrac{12}{5}\times 2=40m$
Therefore, the distance in-between trains after the first trip is, $60-40=20m$
Similarly, we will compute the distance between the trains after the first trip, that is, trip 2
Let the time taken by the bird to fly from train A to train B be denoted by ‘t’.
The total distance travelled by bird and train
$\dfrac{50}{3}\times t+\dfrac{25}{3}\times t=40$
Upon solving for ‘t’ we get,
$t=\dfrac{4}{5}s$
The distance travelled by the bird is,
$\dfrac{50}{3}\times \dfrac{4}{5}=\dfrac{40}{3}m$
The distance travelled by both train is,
$\dfrac{25}{3}\times \dfrac{4}{5}\times 2=\dfrac{40}{3}m$
Therefore, the distance in-between trains after the first trip is, $20-\dfrac{40}{3}=\dfrac{20}{3}m$
Similarly, we can compute the subsequent trips values, by finding the values of the distance travelled by bird and distance in between trains.
Therefore, the distance in-between trains after completing the n trips can be expressed in terms of a series, given as follows.
$\left( \dfrac{20}{3(n-1)} \right)m$
This above equation equals to zero, after the infinite number of trips. Thus, the bird has to fly an infinite number of trips.
As the number of trips which the bird can make from one train to the other before they meet is infinite.

So, the correct answer is “Option E”.

Note:
This question can be solved by the direct method, that is, simply substituting the values in the formula that relates the distance, speed and time. The units of the parameters should be taken care of. All the units must be mentioned using the SI system. As, in this case, we have converted the unit from km per hr to m per sec.