Question

Two trains each having a speed of 30 km/h are headed at each other on the same straight track. A bird
that can fly at 60 km/h flies off from one train when they are 60 km apart and heads directly for the
other train. On reaching the other train it flies directly back to the first, and so forth.
(a) How many trips the bird can make from one train to the other before they crash ?
(6) What is the total distance the bird travels ?​

Answer 1

Answer:

No. of trips  =  ∞

Total distance = 60 km.

Explanation:

Given :

Two train of speed 30 km / hr .

Speed of bird is 60 km / hr .

Given , distance between them is 60 m .

First , finding relative velocity of train A w.r.t train B

$\displaystyle{V_{AB}=30-(-30) \ km/hr}\\\\\\\displaystyle{V_{AB}=60 \ km/hr}$

Now time taken = displacement / velocity

Time taken = 60 km / 60 km / hr

Time taken = 1 hr .

Now relative velocity w.r.t Bird and train B.

$\displaystyle{V_{BB^{'}}=30-(-60) \ km/hr}\\\\\\\displaystyle{V_{BB^{'}}=90 \ km/hr}$

First tips t₁  time taken

Time taken = 60 / 90    [ 60 km distance given in question ]

Time taken = 2 / 3  hr .

Distance covers in t₁ time :

Distance = 60 × 2 / 3 km

Distance = 40 km .

Remaining distance = 60 - 40 = 20 km.

Second trips : t₂

Distance = 20 / 90 m

Distance = 2 / 3² m

We can see it is in form of GP.

$\displasystyle{Total \ trip = t_1 + t_2 +t_3 \ ..... \ t_n}$

$\displaystyle{T=\dfrac{2}{3}+\dfrac{2}{3^2} +\dfrac{2}{3^3} \ .... \ \dfrac{2}{3^n}}\\\\\\\displaystyle{T=\dfrac{2}{3}\left[1+\dfrac{1}{3} +\dfrac{1}{3^2} \ .... \ \dfrac{1}{3^{n-1}}\right]}\\\\\\\displaystyle{T=\dfrac{2}{3}\left[\dfrac{(1-(1/3)^n}{1-1/3}\right]}\\\\\\\displaystyle{T=1-\left(\dfrac{1}{3}\right)^n}$

But we already find that total time is taken 1 hr .

So the second term should be zero .

$\displaystyle{\left(\dfrac{1}{3}\right)^n=0}\\\\\\\displaystyle{3^n= \infty}\\\\\displaystyle{n= \infty}$

Thus the number of trips is infinite .

Total distance of bird :

Distance = velocity × time

Total distance of bird = 60  × 1 = 60 km

Hence we get answer .

Answer 2

SOLUTION:-

Given:

Speed of each train= 30km/h &

distance between them=60km

As they are the moving in opposite direction and same speed, So they will crash at middle of the distance them, i.e.30km from each train.

CASE 1️⃣

We know that time:

$= > Time = \frac{Distance}{Speed}$

$Time = \frac{60}{60 + 30} \\ \\ = > Time = \frac{60}{90} \\ \\ = > Time = \frac{2}{3} h$

$Total \: time = \frac{2}{3} + \frac{2}{ {3}^{2} } + ........ + \frac{2}{ {3}^{n} } = 1 - ( \frac{1}{3} ) {}^{n} \\ \\ but \: 1 - ( { \frac{1}{3} )}^{n} = 1 \: when \: n = ∞ \\ where, \: n = number \: of \: trips$

CASE 2️⃣

Since, the speed of bird is 60km/hrs.

In this problem the bird fly at a constant speed for the entire time and the trains travel until colliding. The distance the bird travels is thus, d= vt.

We need to compute the time that the trains run before colliding. By symmetry, we can argue that the trains will collide in the middle, at 30km.

Time taken to travel 30km by each train

=) 30/30

=) 1hrs

Therefore,

Speed of bird= 60km & time= 1hrs

So, Distance travelled by bird in 1 hrs

=) Speed× time

=) 60× 1

=) 60km

As trains will come closer to each other, bird can make any number of trips because distance between trains will approch to 0.

So, time taken by bird will also tends to 0.

Therefore,

Bird can infinite number of trips.

Hope it helps ☺️