28. Physics | Kinematics | A Swimmer Motorboat Following The Light House | by Ashish Arora (GA)

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.

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