A man in a river boat must get from point `A` to point `B` on the opposite bank of the river (see figure).The distance `BC=a`.The width of the river `AC=b`.At what minimum speed `u` relative to the still water should the boat travel to reach the point `B`?The velocity of flow of the river is `v_(0)`

Options

A. `(v_(0)b)/(sqrt(a^(2)+b^(2)))`
B. `(v_(0)a)/(sqrt(a^(2)+b^(2)))`
C. `(v_(0)b)/(sqrt(2)a)`
D. `(v_(0)a)/(sqrt(2)b)`

Answer

a

Explanation

Suppose u is the speed of the boat relative to water, then velocity of the flow (w.r.t. bank) `v_(0)`.
`v_(x)=(u cos theta+v_(0))`, and perpendicu lar to flow will be `u_(y)=u sin theta`
Time to cross the river, `t=(b)/(u sin theta)` In the time, the distance travelled by the boat in the direction of flow.
`a=v_(x)t=(u cos theta+v_(0))(b)/(u sin theta)`
or `au sin theta=ub cos theta+v_(0)b`

`therefore u=(v_(0)b)/((alpha sin theta-b cos theta))`.............(i) u to be minimum.
`theta=0` or `(d)/(dtheta)[(v_(0)b)/(alpha sin theta-b cos theta)]=0`
or `tan theta=-(a)/(b) therefore cos theta=(b)/(sqrt(a^(2)+b^(2))`
On substituting these values in equation (i), we get
`u_("min")=(v_(0)b)/(sqrt(a^(2)+b^(2))`