CBSE Class 12-science Answered
River speed varies as v = t/2 and a man wants to reach directly opposite bank of river .his velocity in still water is 5 and river width is 48.
a. Find directikn which msn should swim and time he takes to cross
B. Find trajectory
Ans is 37 degree and 53 degree and time is 12 ,16 s
Asked by cudio1212 | 30 Oct, 2020, 11:27: PM
Expert Answer
Figure shows a river in which water flow speed v = t2 m/s , where t is time
Let swimmer who swims 5 m/s in still water swimming in the direction that makes angle θ with horizontal bank as shown in figure.
It is given that swimmer reaches the opposite bank exactly the same horizontal location , then we have
........................... (1)where τ is time to cross the river , τ = 48 / ( 5 sinθ ) = 9.6 / sinθ ................(2)
From eqn.(1) , we get , 5 cosθ = 2.4 / sinθ ................(3)
from eqn.(2) and (3) , we get, sinθ cosθ = 0.48 or sin2θ = 0.96 ..................... (4)
Hence 2θ = sin-1 ( 0.96 ) = 74o or θ = 37o
River crossing time , τ = 9.6 / sin37 ≈ 16 s
We know that , sin (90-α) = sin (90+α)
We get from eqn.(4) , 2θ = 90-α = 74o or α = 16o
Hence other angle (90+α ) that satisfies eqn.(4) , 2θ = 90+α = 106o or θ = 53o
For this angle θ = 53o , river crossing time , τ = 9.6 / sin53 ≈ 12 s
--------------------------------------------
Horizontal displacement x of swimmer in time t ,
x = ( - u cosθ ) t + (t2/4) ) ..................(5)
where u is swimmer's speed in still water and θ is the direction of angle made by swimmer with horizontal bank.
during this time t , vertical displacement y = ( u sinθ ) t or t = y / ( u sinθ )..............(6)
using eqn.(6) , we rewrite eqn.(5) as , x = { (-u cosθ) [ y / ( u sinθ ) ] + [ (1/4) y2 ] / ( u2 sin2 θ ) }
Above equation can be simplified as, 4 u2 sin2θ x = - ( 2 u2 sin2θ ) y + y2
If we substitute u = 5 m/s and θ = 37o in above eqn., we get trajectory as , y2 - 48 y = 36 x
If we substitute u = 5 m/s and θ = 53o in above eqn., we get trajectory as , y2 - 48 y = 64 x
Answered by Thiyagarajan K | 31 Oct, 2020, 04:31: PM
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