Question

Two trains each having a speed of $30\text{km/h}$ are headed at each other on the same straight track. A bird that can fly at $60\text{km/h}$ flies off from one train when they are $60\text{km}$ apart and heads directly for the other train. On reaching the other train, it flies directly back to the first and so forth. How many trips $\left(n\right)$ can the bird make from one train to the other before the trains crash. Find the total distance that the bird travels assuming bird to be a point object.

• A. $n=\infty ,60\text{km}$
• B. $n=\infty ,50\text{km}$
• C. $n=60,60\text{km}$
• D. $n=60,50\text{km}$

Answer 1

The correct option is A $n=\infty ,60\text{km}$

Relative speed of one train w.r.t. another train is $v_{rel}=30+30=60\text{km/hr}$
Relative distance covered by one train w.r.t. another train is $d_{rel}=60\text{km}$
Time taken for the train to meet
$t=\frac{60}{60}=1\text{hour}$
Relative velocity between bird and train on which bird is approaching is $(60\text{km/h}+30\text{km/h}=90\text{km/h})$
For the first trip time taken by bird
$t_1=\frac{60}{90}=\frac{2}{3}\text{h}$
Separation between the trains after first trip of bird
$s_1=d-v_{rel}{t}_1=60-60\times \frac{2}{3}=20\text{km}$
For the second trip, time taken
$t_2=\frac{20}{90}=\frac{2}{3^2}\text{h}$
Separation between the trains after second trip of bird is
$s_2=20-60\times \frac{2}{3^2}=\frac{20}{3}\text{km}$
We can see that for third trip time
$t_3=\frac{2}{3^3}\text{h}$
For $n^{th}$ trip $t_n=\frac{2}{3^n}\text{h}$
Hence total time is going in geometric series
$T=2(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.........+\frac{1}{3^n})$
$=\frac{\frac{2}{3}(1-\frac{1}{3^n})}{1-\frac{1}{3}}=1-\frac{1}{3^n}$
But we know that available total time is 1 hour
$1=1-\frac{1}{3^n}$
Hence, $n=\infty$ (bird will do infinite number of trips before collision)
Total distance travelled by the bird
$vt=60\times 1=60\text{km}$