a boat in a river problem..........

@lovejeet-etHdkD • Oct 17, 2024

Oct 17, 2024

1.5K

i came across a nice tricky question:-

let there are two rivers............ river A and river B. water is flowing in river A with some speed, and stagnated ( not flowing) in river B. there are two boats, one in each river..... they have to travel a same distance, say x, and then return back at the starting point......
so which boat will return first????

note:- speed of water is less than the speed of boat, and both boats travel with same speed...........

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eternalthinker

@eternalthinker-Kvsi5Y • Jan 25, 2011

Isn't it the boat in the flowing river that will travel to and fro in the shortest time?
Reya

@reya-SMihdC • Jan 25, 2011

I guess the boat in River B.Which is correct?
lovejeet

@lovejeet-etHdkD • Jan 25, 2011

any explanation?????
and ya, boat is not traveling from one bank to another..........its traveling in the direction of water and then returning back..........
the diagram would make it clear.............

starting point <-----------------x------------------>half way, as boat will return back to starting pt....
-----------------------------------------------------------------
|\
|/
-----------------------------------------------------------------
rajeshchahar

@rajeshchahar-m1KsOz • Jan 25, 2011

boat in river A will come first
eternalthinker

@eternalthinker-Kvsi5Y • Jan 25, 2011

lovejeet
any explanation?????
I just substituted dummy values for both. And the boat A was a wee bit faster 😉
lovejeet

@lovejeet-etHdkD • Jan 25, 2011

yup............... the answer is B. but any mathematical explanation?????
eternalthinker

@eternalthinker-Kvsi5Y • Jan 28, 2011

The time taken by boats A and B will be in the ratio:
1/v(v^2 - a^2) : 1

Clearly, boat A takes lesser time.
Hence the result 😉
shreyasm89

@shreyasm89-jGobVm • Jan 28, 2011

@eternalthinker- I'm not able to understand your formula. Could you plz explain
freak16

@freak16-XgiLj6 • Jan 28, 2011

suppose x=3m,speed of boat=5m/s and speed of river=2m/s!t1=time taken by boat in river A,t2=time taken by boat in river B!
then,t1=3/(5+2) + 3/(5-2)
= 3/7 + 3/3
= 1.42 sec.
t2=(2*3)/5
= 1.2 sec
so t1>t2..means boat in river b will reach faster.....please someone explain...how the answer is A then??
am i doing the solution wrong??
lovejeet

@lovejeet-etHdkD • Jan 29, 2011

sorry for the mistake.................... freak you are correct............correct answer is B. i just wrote option A mistakenly......
the mathematical explanation is:-
let speed of the boat be s, speed of water be w, and distance be x
we have the relation, time= dis/speed....
now, case A:-
time, Ta = x/(s+w) + x/ (s-w)
= 2xs/ (s*s - w*w)
case B:-
time, Tb = x/s+ x/s = 2x/s

now, tb/ ta= (s*s- w*w)/ (s*s) = 1- (w*w)/(s*s)

clearly, ta is greater than tb...............
therefore correct answer is B.
Reya

@reya-SMihdC • Jan 29, 2011

Tb=x/s+x/s=2xs??

how come??Tb=2x/s 😀

Ta=2xs/(s*s-w*w)??How come?

Isn't Ta=xs-xw+s^2+sw/s^2-w^2 if the value of Ta=x/(s+w)+s/(s-w) ?

Do tb and Tb are same??

If yes then how did you calculate tb/ta=1-(w*w)/(s*s)??

Pls explain me!
freak16

@freak16-XgiLj6 • Jan 29, 2011

See Praveena
Tb=(2*x*s)/s^2-w^2
Ta-(2*x)/s

now divide Ta By Tb
you must get it
Reya

@reya-SMihdC • Jan 29, 2011

I know the answer!

@Freak:Just look at the value of Tb and Ta posted by lovejeet!

Isn't it wrong??
lovejeet

@lovejeet-etHdkD • Jan 29, 2011

thanks a lot praveena for replying and pointing out the mistake.........
i m sorry for the mistakes...................... sorry for my casual writing..........
the value of Ta,will be:-
Ta= x/(s+w) + x/ (s-w)
= [ x(s-w) +x (s+w) ]/ (s^2-w^2)
= 2xs/ (s^2- w^2)
ISHAN TOPRE

@ishan-nohePN • Jan 29, 2011

Its damn simple
Both the boats assuming that they with pedal same amount of energy, will return at same time.
1)Boat in river A:It will be easy in downstream but will face opposition in upstream.
Let Speed of boat be 'x' and that of river water be 'y'.
downstream velocity=x+y; upstream velocity=x-y;
Hency average velocity=(x+y+x-y)/2=2x/2= x
2)Boat in river B:It will have same velocity as 'x' any way😁