A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If the reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Answer

B, C

Explanation

Velocity of man, `V_m = 3 km/hr`
BD= horizontal distance for resultant velocity R
X- component of resultant
`R= 5+3 cos theta`
`t= 0.5/(3 sin theta)`
Which is same of horizontal component of velocity.

H=BD
`=(5-3cos theta) (0.5/3 sin theta )`
`=(5+3 cos theta )/(6 sin theta)`
For H to be minimum `((dH)/(d theta))=0`
`gt d/(d theta) ((5+3 cos theta )/(6 sin theta ))=0`
`rarr 18 (sin^2 theta + cos^2 theta)-30 cos theta = 0`
`rarr 30 cos theta = 18`
`rarr cos theta = - 18/30 = - 3/5`
`sin theta = sqrt ( 1- cos^2 theta0=4/5`
`:. H= (5+3cos theta)/(6 sin theta)`
`= (5+3( 3/5))/)6xx4/5= (25-9)/24`
` = 16/24 = 2/3 km`.