A man wishes to cross a river and come back to original position by swimming minimum distance.if the width of the river is 50m speed of river is 2m/s and velocity of man with respect to river is 4m/s , then time taken by man to come back to original position is
Answers
Answer:
Explanation:
(a) Shortest Path .
If the main has to move along the shortest path , then the path of this boat should be AB ( perpepndicular to river).
If the man rows vertically , the river will carry him to right , hence the man should incline his boat to the left. When his velocity is inclined , it has two components , Vx and Vy [compare with inclined velocity of cat].
The horizontal component Vx should cancel the speed the river .
Thus Vx=VBcosθ=VR
∴4cosθ=2,θ=60∘
Alternatively , we can use resultant vertical as in cat mouse problem (Method 3).
(b) Shortest time :
Note that when the man rows along shortest distance , he DOES NOT cross the river in the shortest time , because time taken to cross the river =widthe of riververtical velocity
As long as the boat is inclined at angle θ , vertical velocity is VB . To row in shortest time , the enitre velocity of boat should be used in vertical direction i.e., Man should row perpendicular to stream .
Note : When the man rows in shortest time , he does not reach point B but instead point B. Shortest distance is analogus to reaching a point on opposite bank (say house ) where shortest time is linked with only reaching opposite bank.
Answer:
let river velocity is
v
r
=5
i
^
A man is trying to cross a river least possible displacements by swimming at an angle of 143 to the stream
Let the velocity of the man is v
m
shown in the diagram v
m
sin53
0
=5
⇒v
m
=
4
25
The drift he suffers when he is crossing the same river in the least possible time is D when he cross the river perpendicular to the direction of the river flow , time taken by the man is t
⇒t=
v
S
=
25
1000
×4=160sec.
drift is D=v
r
t=5×160=800m
Explanation: