IIT JEE , AIEEE Forum - Physics , Mechanics

kane

a taxi leaves the station X for station Y every 10 min. simultaneously,a taxi also leaves the station Y for station X every 10 min. the taxies move at the same constant speed and go from X to Y or vice versa in 2 hrs. how many taxies coming from the other side will meet each taxi enroute from Y to X?

ans:23

plz don't see from book and try to answer.i have the solution with me in book but i couldn't understand it.so plz try to explain me in your own words

rooney

Let velocity of car A be V.
So, total distance of path is 120V (since time taken to reach Y is 120min).

Let us assume for our convenience that the car leaves X JUST AS a car arrives at X from Y.

Now, relative velocity of Car A and the next car coming from Y is 2v. And the distance between A and the next car is 10V (since each car from Y leaves in every 10 min). So, A meets the next car at t = 10V/2V = 5 min.

So, it keeps meeting a car at every 5 min. So in 120 min, it meets 120/5 = 24 cars.

The last car is the one which is just leaving as A reaches Y. So, we dont count that (just like we didnt count the car reaching X when A was leaving X).

So , final answer is 24 - 1 = 23.

coolremo

Hey dude this one is really easy.it is like this question.a bus starts from A to B every ten minutes .AB=60km.a car starrts from B ,wit vel=60km/hr which is de vel of bus from A.so the dude in the car sees 11 buses in de road as there were 5 buses alredy and 6 buses which started wit him.just rack ur brains out dude!

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