IIT JEE , AIEEE Forum - Physics , Mechanics

tushar

Q . Two boats A and B move away from buoy anchored at the middle of a river along mutually perpendicular straight lines, the boat A along the river and the boat B across the river . Having moved off an equal distance from the buoy the boat returned. Find the times of motion ta/tb if the velocity of each boat with respect to water is n times greater than stream velocity. ?

kirtana

hiii........
it'll help if u draw a diagram and solve....

ok...

Vb = Vr

FOR BOAT A
downstream velocity = 1.2 Vr + Vr
= 2.2 Vr
let the distance travelled be = d
therefore the time taken ( t1) = d/ 2.2 Vr

upstrean velocity = 1.2 Vr - Vr = 0.2 Vr
time ( t2) = d/ 0.2 Vr

therefore total time t1 + t2 = d/ 2.2 Vr + d/ 0.2 Vr = 6d / 1.1 Vr

FOR BOAT B
(u reqiure the diagram here)

Vb2 = Vbr2 + Vr2 ( Vbr is the velocity with respect to the stream )
Vbr =

Vb2 - Vr2

by solving.......
Vbr = 2 Vr

0.11

time is same to go and return..
so total time= 2 * ( d / 2 Vr ^[

0.11)

= d / Vr

0.11

ratio of the two values of time = (6 *^]

0.11) / 1.1 = 1.8 / 1 = 1.8

answer is 1.8 : 1

hope this is right.... cheers

kirtana

hii
Vb2 means Vb²
so the equation is
Vb² = Vbr² + Vr² ( Vbr is the velocity with respect to the stream )
Vbr =

Vb² - Vr²

sorry....

tushar

sorry , but the answer is given in terms of n.

cvramana

The time of travel of A can be found easily by

T_A = L / (V₁ + V₂) + L / (V₁ - V₂) = 2 n L V₂ / (n² -1) ----------(1)

For the time of travel of B we should know that the direction of velocity of B should be at angle q (with respect to the perpendicular to the flow of water) so that it will going perpendicular to A and is given by

Sin q = V₂ / V₁ --------------------- (2)

And the resultant velocity is given by

V₁ cos q ----------------------------(3)

Therefore

T_B = 2 L / Ö(V₁² - V₂²) = 2 L V₂ / Ö (n² - 1) ---------------(4)

Therefore

T_A / T_B = n / Ö (n² - 1)

Given n value we can find the ratio.

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