in this illustration, we are going to discuss

about, a situation in which, a swimmer motorboat, following the light house. now, here we are

given that there are 2 ports ay and b on 2 banks of a river with line ay b. which makes

an angle 60 degree with the river current that is root 17 meters per second. and a distance

between ay and b is 12 hundred meter and a light house, throws a light beam from port

ay to b. and a motor boat starts from 1 port to another such that it always follows the

light beam. if boat reaches b. and come back to ay in 5 minutes we are required to find

the velocity of boat with respect to water. lets first draw the situation and analyze

it carefully. say if this is the river. from point ay. light beam is thrown, to another

port b which is, at an angle 60 degree with the river current. we also given with, the

river current speed that is root 17 say in this direction river current u is their. so

to move along this path the boat has to travel, with the some speed such that if this speed

is v. than, u is added to it and finally the boat will move along this line. and say if,

boat velocity with respect to water is making an angle theta. then we can write, the velocity

of boat. along path. ay b is. here we can calculate velocity of boat along path ay b

when it is going from ay to b is equal to, this u. so if this angle is theta this is

60 degree here we can see. this velocity we can write as. v coz theta. plus u coz 60 degree.

similarly when it returns. then boat has to travel, with the, same speed v at an angle

theta but when the speed u is added to it, the resulting speed will be less compare to

earlier because, the river current component, along, the line b ay. will be subtracted from

v coz theta. so here again we can write. in return path. b ay. boat velocity along. line

b ay is, here we can see velocity of boat along line b ay in return path we can write

as v coz theta minus, u coz 60 degree. and we can also see as, boat is not moving in

direction normal to this line ay b the component of boat velocity, with respect to water and

river current. in normal directions must be equal so these will cancel out. so we can

write in direction. perpendicular to line. ay b. we use. v sine theta is equal to, u

sine 60 degree. the value of u is root 17 sine 60 is root 3 by 2 this can be written

as root 51 by 2 say this our equation 1. and we are also given the total time in going

and coming back. along line ay b is 5 minute so we can write. this distance we are given,

that it is 12 hundred meter. so we can write the time in going from ay to b is 12 hundred

divided by. this v coz theta. plus, u is root 17 coz 60 is 1 by 2 so this is root 17 by

2. plus the time in coming back is again 12 hundred divided by, v coz theta minus root

17 by 2. and this equal to 5 minute in second this will be 300. so if we simplify this equation.

this will be quadratic and v coz theta on simplifying we are getting. v square coz square

theta minus. 8 v coz theta minus, 17 by 4 is equal to, zero. if we solve, it for the

value of v coz theta here you can see v coz theta we are getting is minus of minus 8 is

8. plus minus root of, b square is 64. minus plus, 17, divided by 2. so, the 64 plus 17,

is 81, the root of 81 is 9. so here we can eliminate the negative sign because v coz

theta can not be negative. so it will be 17 by 2. if we take this as our equation 2. then,

if we continue over here we can see. 1 by 2 if we divided these 2 equations, this will

result the value of tan theta. which is, root 51 by, root 17 that is root of, 3 by 17. this

implies the angle theta at which. the boat should had, will be given as tan inverse of.

root of 3 by 17. and if this angle we use then, from equation 1. we can substitute and

easily calculate the value of velocity, which is. here root 51 by 2 upon sine theta on simplifying

will get it root 85 meters per second. that is a final result of our problem.