The width of a river is D. A man can cross the river in time `t_1` in the absence of any river current. But in presence of a certain river current, the man takes a time `t_2` to cross the river directly. Show that the velocity of the current is
`v=D sqrt((1)/(t_1^2)-(1)/(t_2^2))`

Explanation

Let u=velocity of the man on the river.
`therefore t_1=D/u or, u=D/(t_1) ……(1)`
In presence of river current of velocity v, the effective velocity of the man will be the resultant w of u and v.
`vecw=vecu+vecv`
To cross the river directly, the man has to swin in such a direction that the resultant w is across the width of the river [Fig.2.53] . Clearly,

and the time taken to cross the river is `t_2 =D/w=D/(sqrt(u^2-v^2)).`
`therefore t_(2)^(2)=(D^2)/(u^2-v^2) or, u^2=v^2+(D^2)/(t_2^2) ..... (2)`
From (1) and (2) , `(D^2)/(t_1^2)=v^2+(D^2)/(t_2^2) or, v^2=D^2(1/(t_1^2)-1/(t_2^2))`
`therefore v=Dsqrt(1/(t_1^2)-1/(t_2^2)).`