Question

Two trains, each having a speed of 30km/h, are headed at each other on the same track. A bird flies off one train to another with a constant speed of 60km/h when they are 60km apart till before they crash. Find the distance covered by the bird and how many trips the bird can make from one train to other before they crash? (60km infinity,)

Answer 1

velocity of one train relative to other,
$V_{TT}=V_T+V_T=30+30= 60km/h$
it means, if the distance between trains is 60km then two trains will crash after 60/60 h = 1hr .
now velocity of bird with respect to train towards which it is moving will be 60km/h + 30km/h = 90 km/h,
so, time taken by bird for the 1st trip is
$t_1=\frac{60km}{90km/h}=\frac{2}{3}hr$

in this time train moves towards each other $\frac{2}{3}\times60=40km$
so, the remaining distance = 60 - 40 = 20km

similarly time taken by bird in 2nd trip = 20/90 hr = 2/9 hr = 2/3² hr
so, proceeding in this way time taken by bird in nth trip = 2/3ⁿ hr

now if the bird makes n trips till trains crash.
$1hr=t_1+t_2+t_3+......+t_n\\e.g.,\:=\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+......+\frac{2}{3^n}\\=\frac{2}{3}\left[\begin{array}{c}\frac{1-(1/3)^n}{1-(1/3)}\end{array}\right]\\1=1-\frac{1}{3^n}\\0=\frac{1}{3^n}\implies n=\infty$

hence, number of trips = infinity.

now distance covered by bird in 1st trip = 2/3 × 60 = 40 km
distance travelled by bird in 2nd trip = 2/9 × 60 = 40/3km...