The impatient boatman

I solved this question by the concept of relative velocity.

Let A be the final position of the boat. Any general position of the boat is shown in the figure.

Velocity of approach of the boat to the light house is $v-vcos \theta$

So $-\frac { da }{ dt } =(v-vcos\theta )$

On setting the limits we get

$-\int _{ 2d }^{ \sqrt { { x }^{ 2 }+{ d }^{ 2 } } }{ da } =\int _{ 0 }^{ T }{ (v-vcos\theta )dt }$ ......................(1)

Now in this time interval the would cover $x$ km in horizontal direction

So $\frac { ds }{ dt } =(v-vcos\theta )$

On setting the limits we get

$\int _{ 0 }^{ x }{ ds } =\int _{ 0 }^{ T }{ (v-vcos\theta )dt }$ ........(2)

The term in the LHS of 1st equation and RHS of second equation is same. So eliminating that term we get

$x=\frac{3d}{4}$ . On putting values we get $x=0.75$

At any point in the trajectory of the boat,let the line joining it and the lighthouse BL make an angle 180- $\theta$ with the direction of river current.As the boatman tries to point the boat towards the lighthouse,velocity of boat w.r.t river ${ v }_{ B,R }$ at any time, will be along the line joining the boat and the light house.

if we find net component of velocity of the boat along this line at any time, ${ v }_{ BL }$ ,we get

${ v }_{ BL }$ = ${ v }_{ B,R }-{ v }_{ R }.cos(\theta )$ = ${ v }_{ R }-{ v }_{ R }.cos(\theta )$ $\because$ ${ v }_{ R }={ v }_{ B,R }$

this is same in magnitude as the net component of velocity of the boat along the river current, ${ { v }_{ x } }$

${ { v }_{ x } }={ v }_{ R }-{ v }_{ B,R }.cos(\theta )={ v }_{ R }-{ v }_{ R }.cos(\theta )$

$\therefore$ Distance traveled towards lighthouse=Distance traveled along the river.

$\therefore$ CD=(2 $\times$ d)-DL

Let CD='x'km Applying pythagoras theorem in $\Delta$ LCD, We get x=3 $\times$ d/4 Putting values, we get x=0.75