Question

There is a river flowing west and a man swimming while making an angle to the river . So if we have to calculate the the time taken by the man to reach the opposite shore we take the y component of the velocity of the man. Why?​

Answer 1

The y component of the velocity of the man is "usually" taken to find the time taken by the man to reach the opposite shore, not necessarily.

Choosing the particular velocity component of man depends on the displacement component of the man given in the question!

$\begin{aligned}\sf{1.}\ \ &\textsf{One can consider the y component of the velocity of man if (s)he is aware}\\&\textsf{of the width of the river, i.e., perpendicular distance between the shores.}\end{aligned}$

$\begin{aligned}\sf{2.}\ \ &\textsf{One can consider the y component of the velocity of man if (s)he is aware}\\&\textsf{of the drift of the man. Drift is the horizontal (i.e., along the shore, in the}\\&\textsf{direction of the river flow) displacement of the man when he reaches the}\\&\textsf{opposite shore.}\end{aligned}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(-20,0){2}{\vector(0,1){20}}\multiput(0,0)(0,20){2}{\vector(-1,0){15}}\put(0,0){\vector(-3,4){15}}\multiput(-40,20)(0,-20){2}{\line(1,0){80}}\put(-10,22){ \sf{Drift} }\put(-30,10){ \sf{Width} }\put(-13,10){ \sf{v_m} }\put(-5,1){ \theta }\put(-16,-4){ \sf{v_r=v_m\cos\theta} }\put(2,10){ \sf{v_m-v_r=v_m\sin\theta} }\put(50,10){\vector(-1,0){10}}\put(50,10){\vector(0,1){10}}\put(40,6){ \sf{X} }\put(51,19){ \sf{Y} }\end{picture}$

$\sf{\left[\,v_m=velocity\ of\ man;\ v_r=velocity\ of\ river\,\right]}$

The two kinds of questions given below requires individual case / situation as mentioned above.

$\begin{aligned}\sf{1.}\ \ &\textsf{A man can cross a river of width h in still water with a velocity v in}\\&\textsf{minimum time. He makes an angle \theta with the river flow, if he swims}\\&\textsf{in flowing water with the same speed. Find the time taken by the man}\\&\textsf{to reach the opposite shore.}\end{aligned}$

Here we have to consider the vertical (y) component of motion of the man.

• Vertical displacement of man $=h$

• Vertical velocity of man $=v\sin\theta$

Hence the time is,

$\longrightarrow t=\dfrac{h}{v\sin\theta}$

If the velocity of river $(v_r)$ is given in both kinds of questions above, instead of the velocity of man $(v),$ we see that,

$\longrightarrow v_r=v\cos\theta$

$\longrightarrow v=\dfrac{v_r}{\cos\theta}$

$\longrightarrow v\sin\theta=v_r\tan\theta$

Hence the time is,

$\longrightarrow t=\dfrac{h}{v_r\tan\theta}$

$\begin{aligned}\sf{2.}\ \ &\textsf{A man can cross a river in still water with a velocity v in minimum time.}\\&\textsf{He makes an angle \theta with the river flow, if he swims in flowing water with}\\&\textsf{the same speed. If the man reaches the shore at a horizontal displacement}\\&\textsf{of x from his initial position, find the time taken by the man to reach the}\\&\textsf{opposite shore.}\end{aligned}$

Here the width of the river is not given but drift of the man is given. So we have to consider the horizontal (x) component of motion of the man.

• Horizontal displacement of man $=x$

• Horizontal velocity of man $=v\cos\theta$

Hence the time is,

$\longrightarrow t=\dfrac{x}{v\cos\theta}$

If the velocity of river $(v_r)$ is given in both kinds of questions above, instead of the velocity of man $(v),$ we see that,

$\longrightarrow v_r=v\cos\theta$

Hence the time is,

$\longrightarrow t=\dfrac{x}{v_r}$

So if the time taken by the man to reach the opposite shore has to be calculated, we "must not" take the y component, instead we must take the component of the motion of the man which is related to his displacement, either river width or drift, which is given in the question. The choice is optional if both displacements are given.