Question

pls help in the question
( don't show ur over talent) ​

Answer 1

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(0,20){2}{\line(1,0){50}}\put(25,0){\vector(0,1){12}}\put(16,12){\vector(1,0){9}}\put(25,0){\vector(-3,4){9}}\put(16,14){\scriptsize\text{ \sf{3\ km\,h^{-1}} }}\put(8,5){\scriptsize\text{ \sf{5\ km\,h^{-1}} }}\put(26.5,6){\scriptsize\sf{v}}\put(45,0){\vector(0,1){20}}\put(47,9){\scriptsize\sf{0.2\ km}}\end{picture}$

In the fig., $\sf{v}$ is the resultant velocity of the man due to the flowing of river, which points towards the shortest path between river banks.

Shortest distance between the banks is the perpendicular distance between them.

The magnitude of $\sf{v}$ is,

$\longrightarrow\sf{v=\sqrt{5^2-3^2}\ km\,h^{-1}}$

$\longrightarrow\sf{v=\sqrt{25-9}\ km\,h^{-1}}$

$\longrightarrow\sf{v=\sqrt{16}\ km\,h^{-1}}$

$\longrightarrow\sf{v=4\ km\,h^{-1}}$

The total distance travelled by the man for his round trip is twice the river width, i.e.,

$\longrightarrow\sf{d=2\times0.2\ km}$

$\longrightarrow\sf{d=0.4\ km}$

Hence the time taken for his round trip is,

$\longrightarrow\sf{t=\dfrac{d}{v}}$

$\longrightarrow\sf{t=\dfrac{0.4}{4}\ h}$

$\longrightarrow\sf{\underline{\underline{t=\dfrac{1}{10}\ h}}}$

Hence (1) is the answer.