Question

Two trains each having a speed of $30\text{kmph}$ are headed at each other on the same straight track. A bird that can fly at $60\text{kmph}$ 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 can the bird make from one train to the other before the crash. Also find the total distance that the bird travels.

• A. $\text{trips}=\infty ,\text{distance}=60km$
• B. $\text{trips}=30,\text{distance}=30km$
• C. $\text{trips}=5,\text{distance}=60km$
• D. $\text{trips}=5,\text{distance}=30km$

Answer 1

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

$v_{rel}=30+30=60\text{kmph}$ (for trains)
$d_{rel}=60\text{km}$
Time taken for the trains to meet $=\frac{60}{60}=1\text{hr}$
Relative velocity between the bird and train
$=60+30=90\text{kmph}$
For $1^{st}$ trip $t=\frac{60}{90}=\frac{2}{3}\text{hrs}$
Separation between the trains after 1st trip $\left(d_1\right)=d-v_{rel}{t}_1=60-60\times \frac{2}{3}=20\text{km}$
For $2^{nd}$ trip $t_2=\frac{20}{90}=\frac{2}{3^2}\text{hrs}$
$d_2=20-\frac{2}{9}\times 60=\frac{20}{3}\text{km}$
$t_3=\frac{20}{3\left(90\right)}=\frac{2}{3^3}\text{hrs}$
$\therefore t_n=\frac{2}{3^n}$
Hence total time $T=t_1+t_2+\dots \dots +t_n$
$=2[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \dots +\frac{1}{3^n}]=\frac{2}{3}\frac{[1-{\left(\frac{1}{3}\right)}^n]}{[1-\frac{1}{3}]}$
$=1-\frac{1}{3^n}$
But $T=1\text{hrs}\Rightarrow 1=1-\frac{1}{3^n}\Rightarrow \frac{1}{3^n}=0$
$\Rightarrow n=\infty$
Total distance travelled by the bird $=vt=60\times 1=60\text{km}$